Math 104 Lecture Notes

1. 01/18/11: Building Up to the Reals, Part One

1.1. Symbols.

We will start with a list of mathematical symbols that we will use throughout thecourse.

∀: for all∃: there exists⇒ or ⇐: implies⇐⇒ : if and only if∈: belongs to{x | condition}: set notation∩: intersection∪: unionq: disjoint union, or if A = B q C, then A = B ∪ C and B ∩ C = ∅S \A: set difference, or {x ∈ S |x /∈ A}

An example of a formula written (almost) completely in symbols:

(∀ε > 0)(∃δ > 0) s.t. (|x− y| < δ ⇒ |f(x)− f(y)| < ε)

1.2. The Natural Numbers, or N.

The main idea for the first part of the course will be R, or the real numbers. Thereare three main ways to prove their existence: decimal expansions, Dedekind cuts,or sequences of rationals. We will eventually use a hybrid of the second and thirdmethods, but that is for later. For now...N = {1, 2, 3, 4, 5, 6, . . . }∗. This is a commutative semiring with an identity (built

off 1 with operators + and ·). There is a notion of order, or x > y, contained inthe algebraic structure of N.

Axiom (Induction). If S ⊆ N and:

(1) 1 ∈ S(2) x ∈ S ⇒ (x+ 1) ∈ S

hold, then S = N.

In other words, if the two conditions above are true, then nothing can fail to be inS. Note that we will say 0 /∈ N.

1.3. The Integers, or Z.

Z = N ∪ {0} ∪ (−N). This is a commutative ring with multiplicative identity,meaning that it is an abelian group under (+, 0) and is associative, commutative,and distributive (i.e. x(y + z) = xy + xz) under multiplication. We also define0x = 0 ∀x.

Note that a group G is a set with an operation ? such that g ?h ∈ G for g, h ∈ G,along with these properties:

(1) G has an identity e, for which the following holds: e ? g = g ? e = g ∀g.(2) (g ? h) ? k = g ? (h ? k)(3) (∀g ∈ G)(∃h ∈ G) such that g ? h = e = h ? g.

∗Read the book for notes on their structure.

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Math 104, Spring 2011 Vaughan Jones Lecture 1

An abelian group is a group with the property that g?h = h?g. From the propertiesof a group, it follows that the h in (3) is unique, namely h = g−1 or −g, and that(a ? b = a ? c)⇒ b = c.

There is an order on Z, or some notion such that x > y. Namely,

x ≥ y ⇐⇒ x− y ∈ N ∪ {0} .

We then obtain three properties about this ordering:

(1) x ≤ x(2) (x ≤ y and y ≤ x)⇒ x = y(3) (x ≤ y and y ≤ z)⇒ x ≤ z

These are the axioms of a partially ordered set. Another example is the orderingon the subsets of a set with ⊆ instead of ≤.

1.4. The Rationals, or Q, an Ordered Field.

Q = {mn | (m,n) = 1 and m,n ∈ Z and n 6= 0}. Note that (x, y) is equivalent tothe expression gcd(x, y).

We will define an order on Q: Let P = {mn |m ∈ N, n ∈ N}. Then,

x ≤ y ⇐⇒ y − x ∈ P or y = x .

Definition. A field F is a set with two operations +, · and two distinguishedelements 0, 1 such that, for 0 6= 1,

• (F,+0) is an abelian group• (F \ {0}, ·, 1) is an abelian group• x(y + z) = xy + xz

As a precaution, we can extend · to all of F by 0x = 0 ∀x.Q is not only a field, however, but also an ordered field.

Definition. An ordered field is a field F with a subset P ⊆ F with the property

F = P q {0} q −P

where −P = {−p | p ∈ P}, and axioms

(1) P + P ⊆ P , where P + P = {p1 + p2 | p1, p2 ∈ P}(2) P · P ⊆ P, where P · P = {p1 · p2 | p1, p2 ∈ P}

Proposition. 1 ∈ P.

Proof. We will proceed by contradiction. Note that if x ∈ P, then (−1)x ∈ −P.Then, if 1 ∈ −P, then (−1) ∈ P, which is a contradiction because (−1) · (−1) = 1,but 1 /∈ P. �

Proposition. x−1 ∈ P.

Proof. We will proceed by contradiction, and assume that x−1 ∈ −P. If that is thecase, then x · (−x−1) = −1. Then, we obtain a contradiction because x ∈ P and(−x−1) ∈ P, but −1 /∈ P. �

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Thus, given an ordered field (F,+, ·, 0, 1,P), we can define:

x ≤ y [x < y] ⇐⇒ y − x ∈ P ∪ {0} [y − x ∈ P] .

This ordering gives us the following consequences, taken as axioms in other sourcessuch as the book:

(1) a ≤ b ⇐⇒ −b ≤ −a(2) c ≥ 0⇒ ac ≤ bc(3) a2 ≥ 0(4) (0 ≤ a, 0 ≤ b)⇒ ab ≥ 0(5) a > 0⇒ a−1 > 0(6) 0 < a < b⇒ 0 < b−1 < a−1

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2. 01/20/11: Building Up to the Reals, Part Two

2.1. More on Q.

Last time, we discussed ordered fields (F,+, ·, 0, 1,P) where F = P q {0} q −P,and defined an order x ≤ y ⇐⇒ y − x ∈ {0} ∪ P. Note that if x, y ∈ F , theneither x ≤ y or y ≤ x. This property is known as the Total Order Property.

Now, it’s time for our first proof.

Theorem. The equation x2 = 2 has no solution in Q.

Proof. By contradiction; suppose x ∈ Q satisfies x2 = 2. Since (−x)2 = x2 = 2,we may suppose x is positive. Write x = m

n with m,n ∈ N and (m,n) = 1. Since

x2 = 2, (mn )2 = 2, which means that:

(*) m2 = 2n2 .

Therefore, m2 is even. However, because the square of an odd number is odd, itfollows that m is even. Say m = 2k with k ∈ N. Substituting m = 2k in (*), wesee:

2k2 = n2∗

So n2 is even, and by a previous argument, n is even. However, this means that 2divides both m and n, contradicting (m,n) = 1. �

Note: When writing proofs, make sure that you are thinking about whether whatyou’re writing makes sense and whether you’re writing too much.

2.2. Terminology.

Let (X,≤) be a poset, and let S 6= ∅ be a subset of X.

Definition. S is bounded above if there is an x ∈ X with x ≥ s for all s ∈ S.Such an x is called an upper bound for S.

Definition. Same as above, but with below, x ≤ s, lower bound.

Definition. x ∈ X is a least upper bound for S if

(1) x is an upper bound.(2) If y is an upper bound for S, then y ≥ x.

The least upper bound of S is called the supremum and is denoted supS or ∨S. If|S| <∞, we can write S = {x1, x2, . . . , xn}. In this case, supS = max{x1, x2, . . . , xn}.

Definition. Same as above, but with greatest lower, lower bound, y ≤ x. Thegreatest lower bound of S is known as the infimum and is denoted inf S or ∧S. If|S| <∞, we can write S = {x1, x2, . . . , xn}. In this case, inf S = min{x1, x2, . . . , xn}.

Note: If supS or inf S exists, then it is unique.

Examples: {q ∈ Q | 0 < q < 1} has a supremum, 1. {q ∈ Q | q2 < 2} has an upperbound, but no supremum.

Definition. An ordered field is called complete if every subset S ⊆ F which isbounded above has a least upper bound.

∗Note that we removed the substitution step.

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2.3. The Reals, or R, the Complete Ordered Field.

R is a complete ordered field. Note that all complete ordered fields are isomor-phisms of one another, so there is only one complete ordered field.

Proposition (ε-characterization of the supremum). Let S ⊆ R, r ∈ R. Thenr = supS if and only if

(1) r is an upper bound of S, and(2) (∀ε > 0)(∃s ∈ S) with s > r − ε.

Proof. We will go in both directions.

(⇒) (1) is trivial. For (2), let ε > 0 be given. By contradiction, suppose nosuch s exists, i.e. ∀s ∈ S, s ≤ r − ε ⇒ r − ε is an upper bound. However,r − ε < r, so r is not the least upper bound, and we have a contradiction.

(⇐) By (1) we only have to show that r is the least upper bound. Suppose p isan upper bound for S (namely, p ≥ S), and p < r. Apply (2) with ε = r−pto obtain an s ∈ S such that s > r − (r − p) = p, which contradicts that pis an upper bound.

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Theorem. R is Archimedean, i.e. ∀a ∈ R∃n ∈ N such that n > a. Equivalently,∀a ∈ R (a > 0) ∃n ∈ N such that 1

n < a.

Proof. By contradiction, suppose there is an a ≥ N. Then, N is bounded above.By the completeness of R, we can let s = supN. By the ε-characterization of thesupremum with ε = 1, we obtain that ∃n ∈ N with n > s−1. However, this impliesthat n+ 1 > s, which is a contradiction. �

Theorem. “Q is dense in R”, i.e. ∀a, b ∈ R where a ≥ 0 and b > a, ∃q ∈ Q suchthat a < q < b.

Proof. We know that b − a > 0, so 1b−a > 0. Therefore, by the Archimedean

property, we can choose an m such that m > 1b−a , which is equivalent to saying

mb−ma > 1. Consider {n ∈ N |n > ma}. This set is nonempty by the Archimedeanproperty. Let p be its least element (as a subset of N). Then, p < mb, because ifp ≥ mb, then p − 1 would be greater than ma. Therefore, ma < p < mb, whichimplies a < p

m < b. �

Note: In this proof, we have implicitly assumed that a ≥ 0. However, this does notcause any problems for us.

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3. 01/25/11: Completeness and Sequences

3.1. Completeness.

Theorem. The equation x2 = 2 has a solution in R.

Proof. Let S = {x ∈ R |x2 = 2, x > 0}. Then, S is nonempty, because 1 ∈ S. S isalso bounded above, because if x, y ≥ 0, then

x2 < y2 ⇒ x2 − y2 < 0⇒ (x− y)(x+ y) < 0⇒ x− y < 0

so x < y. Thus, if y = 2, then y2 = 4 > 2, so 2 is an upper bound for S. Therefore,by the completeness of R, supS exists.

Let x = supS. We want to show x2 = 2.To show x2 cannot be less than 2, we will assume that x2 < 2 and try to show

that we can add ε > 0 to x such that (x+ ε)2 < 2, so that x is not an upper boundfor S.

Scratchwork: We have x2 + 2εx + ε2 < 2 ⇒ 2εx + ε2 < 2 − x2 = δ. So, we willtry to make 2εx and ε2 both less than δ

2 . Choose ε < δ4x for the first term and

ε < min{1, δ2} for the second term.

So, back to the proof, choose ε to be less than min{1, δ4x ,δ2}, where δ = 2−x2 > 0.

Then, we obtain:

(x+ ε)2 = x2 + 2εx+ ε2 < x2 + δ = x2 + 2− x2 = 2 .

Now, we must show that x2 cannot be greater than 2. We will use the ε-characterization of the supremum, and try to find an ε > 0 such that (x− ε)2 > 2,so that x is not the supremum of S.

Scratchwork: We have x2−2εx+ε2 > 2⇒ x2−2 > −ε2+2εx. Let δ = x2−2 > 0.We want to make 2εx less than δ

2 . We can choose ε < δ2x .

So, back to the proof, choose ε to be less than δ2x . Then, we get:

(x− ε)2 = x2 − 2εx+ ε2 > x2 − 2εx > x2 − δ = 2 .

We know that x − ε < x, but this implies that x − ε > s for all s ∈ S, which is acontradiction because x is supposed to be the least upper bound of S.

Therefore, x2 must equal 2, and thus x = supS is a solution to the aboveequation in R. �

This leads to the fact that every positive element in R is a square, a fact whichagain uses the completeness of R.

3.2. Infinity.

∞ is a very nice symbol, with the properties that ∞ > r and −∞ < r for allr ∈ R.

supS =∞ means that S is not bounded above.Recall interval notation: [a, b) = {x | a ≤ x < b}, (a, b] = {x | a < x ≤ b}, and so

on. Likewise, we define:

[a,∞) = {x | a ≤ x}(−∞, y) = {x |x < y}

and so on.

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3.3. Sequences.

Definition. A sequence beginning at m is a function from {n ∈ Z |n ≥ m} to R,or some other set, and is denoted (sn).

Note: We need for a sequence to be well-defined, such as (sn) = n2 or (sn) = (−1)n,not just something like (0, 12 ,

14 , 0, 2, . . . ).

Definition. limx→∞(sn) = s, or sn → s, is equivalent to:

(∀ε > 0)(∃N ∈ N) s.t. (n > N)⇒ |sn − s| < ε .

Note: There are two irrelevancies in this definition: n ≥ N and |sn − s| ≤ ε lead tothe same definition. Also, think of |a− b| as the distance between a and b. If youadd the condition that ε ∈ Q, it is still the same definition by the density of Q.

Example: limx→∞1+n1+3n = 1

3 .Scratchwork: Notice that for any n, it is the case that:

sn − s =1 + n

1 + 3n− 1

3=

3 + 3n− 1− 3n

(1 + 3n)3=

2

3 + 9n

and this is the same regardless of the absolute value sign. Therefore, we want2

3+9n < ε.

2

3 + 9n< ε⇒ 2

ε< 3 + 9n⇒ 2

ε− 3 < 9n⇒ n >

1

9(2

ε− 3)

Proof. Let ε > 0 be given. Choose any N ∈ N such that N > 19 ( 2

ε − 3). Then, forn > N :

|sn − s| =∣∣∣∣ 1 + n

1 + 3n− 1

3

∣∣∣∣ =2

3 + 9n<

2

3 + 9N< ε

�

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4. 01/27/11: Convergence, Monotone Sequences

4.1. Some Basic Notes.

If A ⊆ B ⊆ R and B is bounded above, then supA ≤ supB.

Proposition. If (sn) is a sequence where lim sn = s and lim sn = t, then s = t.

Proof. Suppose s 6= t and put ε = |s−t|50 into the definition of the convergence of

a sequence. Then, choose N such that n ≥ N ⇒ |sn − s| < |s−t|50 . Also, since

lim sn = t, choose M such that n ≥ M ⇒ |sn − t| < |s−t|50 . Choose k such that

k ≥M and k ≥ N .Then, |s−t| ≤ |s−sk|+|sk−t| by the triangle inequality. However, then |s−sk| <

|s−t|50 and |sk − t| < |s−t|

50 , which implies |s− t| < 2(|s−t|50

), a contradiction. �

As an aside: The Triangle Inequality states that for a, b, c ∈ R, |a−b| ≤ |a−c|+|b−c|.This follows from the slightly easier assertion that |a + b| ≤ |a| + |b|, in which

you can define a = a− c and b = b− c.

4.2. Bounded Sequences.

Definition. (sn) is bounded above if {sn |n ∈ N} is bounded above, and similarlyfor bounded below. (sn) is bounded if {sn |n ∈ N} is bounded above and below.

Proposition. A convergent sequence is bounded.

Proof. Suppose sn → s, or limx→∞ sn = s. Then by definition there exists an Nsuch that s− 1 < sn < s+ 1 for all n ≥ N .

Let M = max{s1, s2, . . . , sN , (s + 1)}. Clearly sn ≤ M for all n. Then, letm = min{s1, s2, . . . , sN , (s− 1)}. Clearly sn ≥ m for all n. �

Theorem. Given sn → s and tn → t, we have:

(1) sn + tn → s+ t.(2) sntn → st.(3) If tn 6= 0 for all n and t 6= 0, then lim sn

tn= s

t .

Proof. We will only prove (2), as the others are similar.Scratchwork: We are working with |sntn − st|, so we will write it as

|(sntn − snt) + (snt− st)| ≤ |sntn − snt|+ |snt− st| = |sn||tn − t|+ |sn − s||t|

However, we know that all four expressions are bounded, so we are ready, and theformal proof follows.

By the previous result, choose B > 0 such that |sn| < B for all n. Now sincetn → t, choose N1 such that (n ≥ N1) ⇒ |tn − t| < ε

2B where ε > 0 is given.Also, since sn → s, choose N2 such that (n ≥ N2) ⇒ |t||sn − s| < ε

2 . Now, chooseN = max{N1, N2}. Then, if n ≥ N :

|sntn − st| ≤ |sntn − snt|+ |snt− st| = |sn||tn − t|+ |t||sn − s| .

Therefore, since |sn||tn − t| ≤ B · ε2B = ε

2 because n ≥ N1 and |t||sn − s| ≤ ε2

because n ≥ N2, their sum is less than or equal to ε2 + ε

2 = ε. �

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Note: This theorem is very convenient! For instance:

1 + n

1 + 3n=

1 + 1n

3 + 1n

→ 1

3.

So what about when a sequence doesn’t converge? Well, first of all, an un-bounded sequence obviously doesn’t converge. But what about sn = (−1)n?

If this sequence converges, then for all ε > 0, then for all sufficiently large n,|sn − a| < ε. However, if n is odd, then |1 + a| < ε and if n is even, |1 − a| < ε.This means:

2 = |1− (−1)| ≤ |a− 1|+ |a+ 1| < 2ε

Set ε = 12 , and we have a contradiction.

4.3. Monotonic Sequences.

Definition. A sequence (sn) is nondecreasing if sn+1 ≥ sn for all n. (sn) isnonincreasing if an ≥ an+1 for all n. (sn) is monotonic if it is either nonde-creasing or nonincreasing.

Theorem. In an ordered field F , F is complete if and only if every bounded mono-tone sequence converges.

Proof. In next lecture. �

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5. 02/01/11: Monotone, limsup, Cauchy

5.1. Monotone Sequences, continued.

Theorem. In an ordered field F , F is complete if and only if every bounded mono-tone sequence converges.


(⇒) Let (sn) be monotonic and bounded. Without loss of generality, suppose(sn) is nondecreasing. Then, {sn |n ∈ N} is bounded above and nonempty.By completeness, there exists an s = sup{sN |n ∈ N}.

We will claim that sn → s. Let ε > 0 be given. We want to find anN such that if n ≥ N , then |sn − s| < ε. By the ε-characterization of thesupremum, there exists N ∈ N such that sN > s− ε. Then, for all n ≥ N ,we have:

sN ≤ sn ≤ s⇒ s− ε < sN ≤ snwhich implies |sn − s| < ε.

(⇐) We will again assume, without loss of generality, that (sn) is nondecreasing.Let S 6= ∅ be a subset of F that is bounded above by B. We must showthe existence of supS. We will construct inductively two sequences, (xn)and (yn), with the following properties:(1) xn ∈ S and yn is an upper bound for S.(2) xn+1 ≥ xn and yn+1 ≤ yn.(3) |xn+1 − yn+1| ≤ 1

2 |xn − yn|.To do this, let x0 ∈ S and y0 be any upper bound for S. Supposex1, x2, . . . , xn and y1, y2, . . . , yn are already constructed, and satisfy thethree properties. Then, either xn+yn

2 is an upper bound for S or it is not.

If it is, then set xn+1 = xn and yn+1 = xn+yn2 . If it is not, then there is an

s ∈ S such that s ≥ xn+yn2 . Set xn+1 = s and yn+1 = yn. Note then that

every xn stays in S and every yn is an upper bound for S, so (1) is satisfied.By construction, (2) is true, and so is (3). Therefore, the sequence we haveconstructed satisfies all of the given properties.

Note that (xn) is a nondecreasing sequence bounded above by y0 andthat (yn) is a nonincreasing sequence bounded below by x0. Therefore, byhypothesis, there are x, y ∈ F such that xn → x and yn → y. We will claimthat x = y. It is obvious from (3) that |xn− yn| ≤ 1

2n |x0− y0|. Suppose for

the moment we can show that 12n → 0. Then, by choosing sufficiently large

n, it will be the case that for any given ε > 0, |x − xn| < ε, |yn − y| < ε,and |xn − yn| < ε∗. Then, we can obtain that:

|x− y| = |x− xn + xn − yn + yn − y| ≤ |x− xn|+ |xn − yn|+ |yn − y| .Therefore, |x− y| < 3ε, so x = y.

However, we still need to show that 12n → 0. We know 1

2n is a nonin-creasing bounded sequence, so it does have a limit z ∈ F . Also, by theelementary properties of limits which are valid in any ordered field, thesequence (2 1

2n ) → 2z. However, the limit of (2 12n ) is the same as that of

12n . Therefore, 2z = z, which means that z = 0. Therefore, we now knowthat x = y.

∗This condition is what we need the convergence of 12n

for.

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Finally, we need to show that x = y = supS. y is an upper bound forS because yn ≥ s for all s ∈ S and for all n ∈ N, so y ≥ s for all s ∈ S aswell. Also, if there existed an upper bound t < y, then there would be anxn such that xn > t, because (xn) → x. Therefore, t cannot be an upperbound, and thus y is the least upper bound.

�

Note: The way we proved that x = y is equivalent to proving lim(xn − yn) = 0,which, for convergent sequences, implies that limxn = lim yn.

Definition.

limn→∞

sn = +∞ ⇐⇒ ∀M ∈ R, (∃N ∈ N) s.t. (n ≥ N)⇒ (sn ≥M).

limn→∞

sn = −∞ ⇐⇒ ∀M ∈ R, (∃N ∈ N) s.t. (n ≥ N)⇒ (sn ≤M).

Note: ∞ is not a number.

Proposition. Any unbounded monotone sequence tends to ∞ or −∞.

Proof. The proof is supplied in the textbook (10.4). �

5.2. lim sup, or lim, and lim inf, or lim.

Suppose (sn) is a sequence in R. Let Tn = {sk | k ≥ n}∗. It is obvious thatTn ⊇ Tn+1 ⊇ Tn+2 ⊇ . . . , and if A ⊆ B for any sets A and B, then it is the casethat supA ≤ supB and inf A ≥ inf B.

Therefore, the sequence (supTn) is nonincreasing and the sequence (inf Tn) isnondecreasing. By the completeness of R as well as the “suitable consideration” of±∞ in the unbounded cases, limn→∞ supTn and limn→∞ inf Tn always exist!

Examples: lim sup(−1)n = 1 and lim inf(−1)n = −1.Let sn be defined such that sn = 1− 1

n if n is even and sn = −n if n is odd. Then,lim sup(sn) = 1 and lim inf(sn) = −∞.Let sn be defined such that sn = 1 + 1

n if n is even and sn = −n if n is odd. Then,lim sup(sn) is still 1!

Remember: lim sup, lim inf always exist.

Theorem. Let (sn) be a sequence. Then:

(1) sn → s⇒ lim inf sn = lim sup sn = s(2) If lim inf sn = lim sup sn, then lim sn = lim inf sn = lim sup sn.

Proof. We will prove each part.

(1) We know that (∀ε > 0)(∃N) s.t. {TN} ⊆ (s− ε, s+ε). Therefore, supTN ⊆[s− ε, s+ ε], and supTn ⊆ [s− ε, s+ ε] for n > N . Because this is true forall ε, lim sup sn = s. The proof of lim inf is similar.

∗The T stands for “tail.”

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(2) Suppose lim sup sn = r = lim inf sn for some r ∈ R. Let ε > 0 be given.We want to show that sn → r. We know that:

∃N1 s.t. supTn < r + ε ∀n ≥ N1 and

∃N2 s.t. inf Tn > r − ε ∀n ≥ N2 .

Let N = max{N1, N2}. Then, if n ≥ N , sn ∈ Tn, so sn ≤ supTn < r + εand sn ≥ inf Tn > r − ε, which implies |sn − r| < ε.

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5.3. Cauchy Sequences.

Definition. Let (sn) be a sequence of real numbers. We say (sn) is Cauchy if:

(∀ε > 0)(∃N ∈ N) s.t. (m,n ≥ N)⇒ (|sn − sm| < ε) .

Note: This makes sense in spaces with an arbitrary notion of distance.

Proposition. A convergent sequence is Cauchy.

Proof. Assume sn → s for a sequence (sn). Let ε > 0 be given and choose N sothat (n ≥ N)⇒ |sn − s| < ε

2 . Then, if m,n ≥ N , we obtain:

|sm − sn| = |sm − s+ s− sn| ≤ |sm − s|+ |s− sm| <ε

2+ε

2= ε .

�

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6. 02/03/11: Cauchy Sequences, Subsequences

6.1. A Simple Proof.

Say we want to show that for 0 < r < 1, rn → 0.One way to show this is to assume rn → x. Then, we know that krn → kx. Let

k = r. Then, rrn → rx, but rn+1 has the same limit as rn. Therefore, rx = x andx = 0.

Another way is to show that ( 1r )n → ∞, as that implies rn → 0. Write 1

r as1 + b > 0. Then (1 + b)n ≥ 1 + nb, which tends to ∞.

6.2. Cauchy Sequences, again.

Last time, we learned about lim sup, lim inf, and Cauchy sequences. Recall that(sn) is Cauchy if and only if:

(∀ε > 0)(∃N ∈ N) s.t. (n,m ≥ N)⇒ |sn − sm| < ε .

Note: This definition makes sense in any ordered field.

Proposition. A Cauchy sequence is bounded.

Proof. Choose ε = 17 (arbitrarily, of course). Then, we know that

∃N ∈ N s.t. ∀n,m ≥ N, |sn − sm| < 17 .

This means that for n ≥ N , |sn − sN | < 17, or sN − 17 < sn < sN + 17. So for alln, it is the case that

sn ≤ max{s1, s2, . . . , sN−1, sN + 17}

and likewise,

sn ≥ min{s1, s2, . . . , sN−1, sN − 17} .

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Theorem. If (sn) is a sequence in R, then (sn) is Cauchy if and only if (sn)converges.

Proof. We will proceed in both directions.

(⇐) This was proved in the previous lecture.(⇒) Let (sn) be a Cauchy sequence in R. Note that because (sn) is bounded,

lim sup sn and lim inf sn exist and are real numbers. Therefore, it sufficesto show that lim inf sn = lim sup sn. Because lim inf ≤ lim sup in all cases,we only need to show that lim inf sn ≥ lim sup sn. To do this, we will showthat lim inf sn + ε ≥ lim sup sn for any ε > 0.

Let lim sup sn = x and lim inf sn = y. Let ε be given, and choose anN ∈ N such that |sn − sm| < ε for n,m ≥ N . Note that this impliessn − sm < ε. Then, because lim sup sn = x, there exists an n ≥ N suchthat sn > x− ε, and similarly there exists an m ≥ N such that sm < y+ ε.Therefore:

x− y = x− sn + sn − sm + sm − y < 3ε .

�13


Proposition. Let (sn) be a sequence in an (Archimedean) ordered field. Then, ifthere exist k, r (where 0 < r < 1) such that |sn − sn+1| < rnk for all n, then (sn)is Cauchy.

Proof. Let ε > 0 be given. Choose N such that n ≥ N ⇒ rn < 1−rk ε. This is

possible because rn → 0, which we may need Archimedean for. Without loss ofgenerality, we will assume that m > n. Then, for n,m ≥ N , we have:

|sn − sm| = |sn − sn+1 + sn+1 − sn+2 + · · ·+ sm−1 − sm| ≤|sn − sn+1|+ |sn+1 − sn+2|+ · · ·+ |sm−1 + sm| < krn(1 + r + · · ·+ rm−n) =

krn(

1− rm−n+1

1− r

)<

krn

1− r< ε

�

Theorem. If F is an ordered field, the following are equivalent:

(1) F is complete, i.e. every nonempty subset bounded above has a supremum.(2) Every nondecreasing bounded sequence converges.(3) Every Cauchy sequence converges (We may need F is Archimedean).

Proof. By our previous results, we only need to prove (3) ⇒ (1). To do this, wewill refer back to our proof of (2) ⇒ (1).

We will show that (xn), (yn) from the aforementioned proof are Cauchy. Thiswould show that by assuming Cauchy sequences converge, we can get to the com-pleteness of F .

We know that |xn − xn+1| < 12n |x0 − y0| because xn ≤ xn+1 ≤ yn, and similarly

for yn+1. Therefore, (xn) and (yn) are Cauchy, and limxn = x, lim yn = y for somex, y ∈ R. The rest of the proof follows in exactly the same way as the proof fornondecreasing bounded sequences. �

6.3. Subsequences.

Definition. For (sn), n ∈ N given, a subsequence of sn is a sequence of the form(tk | tk = snk) with nk < nk+1 < . . . for all k.

Example: Take the sequence (n). (2n) is a subsequence of (n), but (1) is not.However, if our sequence is {1, 2, 3, 1, 2, 3, . . . }, then (1) is a subsequence.

Proposition. If sn → s, then any subsequence of sn tends to s.

Proof. Let (snk) be a subsequence. Let ε > 0 and limn→∞ sn = s be given. ChooseN such that n ≥ N ⇒ |sn − s| < ε. Then, if k ≥ N , then |snk − s| < ε becausenk ≥ N . �

Theorem (Bolzano-Weierstrass). In R, any bounded sequence has a convergentsubsequence.

Proof Sketch. Take the interval in which the sequence lies, and divide it in half.Then, choose a half with infinitely many terms (there must be at least one) andpick a term in that interval. Then, divide that half in half and choose a half withinfinitely many terms again. Pick another term from that half, such that that termappears later in the sequence than the first. Repeat this sequence indefinitely. Thiswill lead to a limit for the subsequence that you will have picked. �

14


7. 02/08/11: Defining the Reals, Part One

7.1. Equivalence Relations.

Definition. Given a set S, an equivalence relation on S is a relation x ∼ y with 3properties:

(1) x ∼ x ∀x ∈ S (Reflexivity)(2) x ∼ y ⇒ y ∼ x (Symmetry)(3) x ∼ y and y ∼ z ⇒ x ∼ z (Transitivity)

Given x ∈ S, define [x] = {y | y ∼ x}.

Here are some facts about [x]:

(1) x ∼ y ⇐⇒ [x] = [y](2) [x] ∩ [y] = ∅ if x � y and [x] ∩ [y] = [x] if x ∼ y.

Therefore, S is the disjoint union of equivalence classes.However, if S =

∐i∈I Si, then define ∼ by:

x ∼ y ⇐⇒ ∃i ∈ I s.t. {x, y} ⊆ SiIt is easy to check that this relation is an equivalence relation, meaning that apartition on a set gives you an equivalence relation, not just the other way around.

7.2. Defining the Reals.

The Equivalence Relation. First, we will define a relation ∼ on sequences (sn) ofrationals, such that (sn) ∼ (tn) if and only if:

limn→∞

sn − tn = 0, i.e. (∀ε > 0, ε ∈ Q) ∃N s.t. n ≥ N ⇒ |sn − tn| < ε .

This is an equivalence relation. Note that in an Archimedean field, (∀ε > 0) can bereplaced by (∀ε > 0 | ε ∈ Q), by the denseness of the rationals.

We will show that if (sn) is Cauchy and (sn) ∼ (tn), then (tn) is Cauchy. Weknow that |sn − sm| < ε and want to show that this implies |tn − tm| < ε.

|tn − tm| = |tn − sn + sn − sm + sm − tm| ≤ |tn − sn|+ |sn − sm|+ |sm − tm| < 3ε

Therefore, (tn) is Cauchy.So, we will define F to be the set of equivalence classes of Cauchy sequences. We

will give it the structure of an ordered field and show it is complete.

Checking the Field Properties. First, we will define + such that [(sn)] + [(tn)] =[(sn + tn)]. We must check to make sure that this is well-defined. Suppose (sn) ∼(s′n) and (tn) ∼ (t′n). Then, we know that lim sn − s′n = 0 and lim tn − t′n = 0.Therefore:

lim(sn − s′n + tn − t′n) = lim((sn + tn)− (s′n + t′n)) = 0 + 0 = 0

and + is well-defined. Let [(0n)] be the additive identity and note −[(sn)] = [(−sn)],so inverses exist. + is also commutative, so this is an abelian group.

Next, we will define · such that [(sn)] · [(tn)] = [(sntn)]. Again, we want this tobe well-defined. First, we will check that (sntn) is also Cauchy:

|sntn − smtm| = |sntn − sntm + sntm − smtm| ≤ |sn||tn − tm|+ |tm||sn − sm| .15


Obviously, the rightmost side goes to zero, so (sntn) is Cauchy. Now, supposelim sn − s′n = 0 and lim tn − t′n = 0. We want to show that lim(sntn − s′nt′n) = 0.

|sntn − s′nt′n| = |sntn − s′ntn + s′ntn − s′nt′n| ≤ |tn||sn − s′n|+ |s′n||tn − t′n|Again, obviously, the rightmost side goes to zero, so multiplication is also well-defined. It is obviously commutative, associative, distributive over addition, andhas an identity, namely [(1n)]. However, the existence of inverses is not trivial. Forthis, we will make an important observation.

Proposition. Let (sn) be a Cauchy sequence of rationals. Then, either:

(1) sn → 0,(2) ∃N s.t. sn > ε ∀n ≥ N , or(3) ∃N s.t. sn < −ε ∀n ≥ N .

where ε ∈ Q and ε > 0.

Proof. Suppose sn does not converge to 0. Then, the negation of the definition ofconvergence gives us that:

∃ε > 0 s.t. ∀N ∈ N,∃r ≥ N s.t. |sr| ≥ ε .

By the Cauchiness of (sn), there exists an N s.t. |sn − sm| < ε2 ∀n,m ≥ N , where

ε is the same as the ε in the above expression. Choose the r guaranteed for this εsuch that r ≥ N and |sr| ≥ ε.

Suppose sr ≥ ε. Then, for n ≥ N , we have |sr − sn| < ε2 , so sn >

ε2 , which is

case (2).Now, suppose sr ≤ − ε

2 . Then, for n ≥ N , we have |sr − sn| < ε2 , so sn < − ε

2 ,which is case (3). �

Note that it is obvious to see that (1), (2), and (3) depend only on the class of (sn).If [(sn)] 6= [(0n)], then sn does not converge to 0, and we are in case (2) or (3).

Then, we can change finitely many terms, by the above proposition, in order tohave sn 6= 0 for all n. Then, [(sn)] · [( 1

sn)] = [(1n)], but we must also make sure

( 1sn

) is Cauchy. We obtain:∣∣∣∣ 1

sn− 1

sm

∣∣∣∣ =

∣∣∣∣sn − smsnsm

∣∣∣∣ < ε

B2

where |sn| < B, because Cauchy sequences are bounded.Therefore, finally, we have established that this is an abelian group under mul-

tiplication, and thus that F is a field which contains the rationals as constantsequences, i.e. q ∈ Q is defined as (qn).

Ordering the Field. We will define P as such:

P = {[(sn)] | ∃N, ε > 0 with sn > ε ∀n ≥ N for some, hence any (sn) ∈ [(sn)]} .

Essentially, P is the set of equivalence classes of Cauchy sequences that fall intocategory (2) of the above proposition. This trivially satisfies all of the conditionsnecessary for an order.

Observe immediately that F is Archimedean. For some [(sn)], sn < k for somelarge integer k, because Cauchy sequences are bounded. Thus, 2k − sn > 0 for alln, and (2kn) therefore bounds (sn) for all n. A similar argument can be used toestablish the nonexistence of infinitesimals.

16


8. 02/10/11: Defining the Reals, Part Two

8.1. Recap.

Recall that the intuition of this construction is that we are approximating realsby sequences of truncations of rational approximations of each real number; this iswhere the notion of equivalence classes comes from.

So far, we have that the additive group with identity [(0n)] and multiplicativegroup with identity [(1n)] exist and are abelian. Therefore, the previously defined Fis a field. Also, we have an order with P = {[(sn)] | case two of the proposition fromlast time}, so F is an ordered field. Furthermore, F is Archimedean because givena Cauchy, and thus bounded, [(sn)], there exists a q ≥ sn + ε for all n where q ∈ N,so in the field, “q” > [(sn)]. Thus, we will use the fact that in an Archimedeanfield, if all Cauchy sequences converge then it is complete in order to prove thecompleteness of F .

8.2. Defining a Concept of “Less Than”.

Note that |[(sn)]| = [(|sn|)]. We could define a notion of “less than” as such:

|[(sn)]| < ε⇒ ε− [(|sn|)] > 0⇒ ∃η > 0 s.t. for large n, ε− |sn| > η, or |sn| < ε− ηIn this definition, ε, η, sn are all rationals. This is a perfectly serviceable definition,but in order to get away from η, we will use an alternate definition.

Definition. “ [(sn)] is less than ε” is defined through these two implications:

(1) [(sn)] < ε⇒ ∃N ∈ N s.t. sn < ε ∀n ≥ N , and(2) ∃N ∈ N s.t. sn < ε ∀n ≥ N ⇒ [(sn)] ≤ ε.

Now, we are finally ready.

8.3. Convergence of Cauchy Sequences in F .

Theorem. Every Cauchy sequence in F converges.

Proof. What we have to do is given ([(sn)]m) Cauchy, come up with a [(tn)] wherelimm→∞[(sn)]m = [(tn)]. Our notation will be:

[(sn)]1 = s1,1, s2,1, s3,1, s4,1, . . .

[(sn)]2 = s1,2, s2,2, s3,2, s4,2, . . .

[(sn)]3 = s1,3, s2,3, s3,3, s4,3, . . .

...

[(sn)]k = s1,k, s2,k, s3,k, s4,k, . . . , sk,k, . . .

We are going to shoot for the diagonal, (sn,n). However, we are dealing withequivalence classes of Cauchy sequences, so we could have made a bad choice ofrepresentatives and (sn,n) may go to infinity. Therefore, we will change [(sn)]k sothat it only varies by 1

k in each term. We will do this by changing all elementsbefore N (Note, a finite number) so they all live within ε of the next. or, moreformally...

Proposition. Let (sn) be a Cauchy sequence. For all ε > 0, there exists a (tn)such that (sn) ∼ (tn) and |tn − tm| < ε for all n,m.

17


Proof. ∃N s.t. |sn − sm| < ε ∀n,m ≥ N . Then, fix:

t1 = sN , t2 = sN , . . . , tn−1 = sN , and tn = sn ∀n ≥ N .

It is obvious that [(tn)] = [(sn)]. �

So, because [(sn)]m is Cauchy, we can suppose:

(**) |sn,m − sk,m| <1

m∀k,m

Now, set the aforementioned tn = sn,n. Now, all that remains for us to do is toshow that [(sn)]m → [(tn)] as m→∞.

First, we will deal with the fact that [(tn)] must be Cauchy. We want to compare|tn− tm|, or |sn,n− sm,m|. Let ε > 0 be given, where in this case ε ∈ Q. Choose anN > 1

ε s.t. |[(sn)]m− [(sn)]k| < ε ∀m, k ≥ N . Also, choose a p s.t. |sp,m−sp,k| < ε.Now, suppose m, k ≥ N .

|tm − tk| = |sm,m − sk,k| ≤ |sm,m − sp,m|+ |sp,m − sp,k|+ |sp,k − sk,k| < 3ε

where the first term is bounded by (**), the second term is bounded by the choiceof N and the Cauchiness of ([(sn)]m), and the third term is also bounded by (**).Therefore, we are done, and [(tn)] ∈ F .

Now, we only need to show that limm→∞[(sn)]m = [(tn)], or in other words that∀ε > 0 ∃N s.t. m ≥ N ⇒ |[(sn)]m− [(tn)]| ≤ ε. Note that it is less than or equal toε because of the second implication in the definition of “less than” above. Let ε > 0be given. By the Cauchiness of (tn), we have ∃N > 2

ε s.t. ∀m,n ≥ N, |tn−tm| < ε2 .

Then, we obtain:

|tn − sn,m| = |sn,n − sn,m| = |sn,n − sm,m + sm,m − sn,m| ≤

|tn − tm|+ |sm,m − sn,m| ≤ε

2+ε

2= ε

Where the first term is bounded by the Cauchiness of (tn) and the second is boundedby (**). Therefore, |(tn)− [(sn)]m| ≤ ε.

Thus, since every Cauchy sequence in F converges and F is Archimedean, F iscomplete! �

8.4. Metric Spaces.

Definition. A metric space is a set X together with a function d : X ×X → R,known as a metric, where d(x, y) satisfies:

(1) d(x, y) ≥ 0 and d(x, y) = 0 ⇐⇒ x = y,(2) d(x, y) = d(y, x), and(3) d(x, y) ≤ d(x, z) + d(z, y) ∀x, y, z ∈ X.

Basically, all we are given when talking about metric spaces is a notion of distance.

Example: In R, a possible metric is d(x, y) = |x− y|.

Example: Take any X, and define d such that:

d(x, y) =

{0, if x = y;

1, if x 6= y.

This is known as the discrete metric.

18


9. 02/15/11: Metric Spaces, Part One

9.1. Metric Spaces on Rn.

Definition. A metric space is a set X and a function d : X ×X → R such that:

(1) d(x, y) ≥ 0 and d(x, y) = 0 ⇐⇒ x = y,(2) d(x, y) = d(y, x), and(3) d(x, y) ≤ d(x, z) + d(z, y)

Remember that some examples are |x − y| on R and the discrete metric from lastlecture.

Now consider Rn = {(x1, x2, . . . , xn) |xi ∈ R}. We will give this set severaldifferent metrics.

9.1.1. The Euclidian Metric. This metric is defined by:

d((x1, . . . , xn), (y1, . . . , yn)) =√

(x1 − y1)2 + (x2 − y2)2 + · · ·+ (xn − yn)2

For this metric, (1) and (2) are obvious. We will come back to (3).

We will now explore the concept of norms, with the intention of obtaining somenew metrics for Rn during this exploration.

Definition. A Norm on a (real) vector space V is a function || · || : V → R suchthat:

(i) ||v|| ≥ 0 and ||v|| = 0 ⇐⇒ v = 0,(ii) ||λv|| = |λ| ||v|| for λ ∈ R, and(iii) ||v + w|| ≤ ||v||+ ||w||.

9.1.2. The “Maximum” Norm. We will define this norm such that for V = Rn:

||(x1, . . . , xn)||∞ = max{|x1|, |x2|, . . . , |xn|}

(i) and (ii) are obvious. For (iii), use the triangle inequality on R.

9.1.3. The “Sum” Norm. We will define this norm such that for V = Rn:

||(x1, . . . , xn)||1 =

n∑i=1

|xi|

(i), (ii), and (iii) are all trivial for this norm.

Norms can also come from inner products (sometimes known as scalar products),which we will further investigate.

Definition. An inner product is a function 〈·, ·〉 : V × V → R such that:

(a) 〈v, v〉 ≥ 0 and 〈v, v〉 = 0 ⇐⇒ v = 0,(b) 〈v, w〉 = 〈w, v〉, and(c) 〈v + w, x〉 = 〈v, x〉+ 〈w, x〉 and 〈x, v + w〉 = 〈x, v〉+ 〈x,w〉.

19


9.1.4. The “Squaring” Norm. We will define this norm in terms of an inner product,such that for V = Rn:

||v||2 =

√√√√ n∑i=1

x2i

This norm has been defined from an inner product, where∑ni=1 x

2i = 〈v, v〉 for

v = (x1, x2, . . . , xn). While properties (i) and (ii) are immediate from the propertiesof an inner product, we must still verify (iii) for this norm. In other words, we mustshow that:

(*) ||v + w||2 ≤ ||v||2 + ||w||2 .

However, because both terms are positive, showing that the inequality holds whenboth sides are squared is equivalent to showing that it holds as presented in (*).Therefore, we have, by (c) (also known as bilinearity),:

||v + w||22 ≤ ||v||22 + 2 ||v||2 ||w||2 + ||w||22 ⇒

〈v + w, v + w〉 ≤ ||v||22 + 2 ||v||2 ||w||2 + ||w||22 ⇒

〈v, v〉+ 2 〈v, w〉+ 〈w,w〉 ≤ ||v||22 + 2 ||v||2 ||w||2 + ||w||22 ⇒| 〈v, w〉 | ≤ ||v||2 ||w||2

Note that the absolute value signs in the last inequality are valid because if it istrue without the absolute value signs, then it is true with them. This last inequalityis known as the Cauchy-Schwarz inequality, which is equivalent to:∣∣∣∑xiyi

∣∣∣ ≤√∑x2i

√∑y2i

So now, to show that || · ||2 is a norm, we will prove Cauchy-Schwarz.

Proof. Take v, w ∈ V . Consider 〈v + λw, v + λw〉 ≥ 0, λ ∈ R.

〈v + λw, v + λw〉 ≥ 0⇒〈v, v〉+ 2λ 〈v, w〉+ λ2 〈w,w〉 ⇒

||v||22 + 2λ 〈v, w〉+ λ2 ||w||22 ≥ 0⇒(λ ||w||2 +

〈v, w〉||w||2

)2

+ ||v||22 −〈v, w〉2

||w||22≥ 0

Because this is true for all λ, we can choose λ so that the first term is 0. Then, weobtain:

||v||22 −〈v, w〉2

||w||22≥ 0⇒ ||v||22 ||w||

22 ≥ 〈v, w〉

2 ⇒ | 〈v, w〉 | ≤ ||v||2 ||w||2 .

�

Therefore, since we have established the Cauchy-Schwarz inequality, || · ||2 is a norm.

The point of all of this norm-taking was that given a norm, we can make it ametric on Rn by d(v, w) = ||v − w||. For instance, || · ||2 is exactly the same as theaforementioned Euclidian metric on Rn!

20


9.2. Metric Space Concepts. Let (X, d) be a metric space.

Definition. The open ball centered at a of radius r for r > 0, a ∈ X is:

B(a, r) = {x ∈ X | d(a, x) < r}Example: In R, the open ball is denoted (a − r, a + r). In the discrete metric, theballs are either the single point or the whole space. In R2, || · ||2, B(a, r) is theopen circle of radius r around a. With || · ||∞, B(0, 1) is the square length 2 aroundO. With || · ||1, B(0, 1) is the interior of the diamond created by taking the linex+ y = 1 and reflecting it across all of the axes. B(0, 1) is known as the unit ball.

Definition. Given a sequence (xn) and an x ∈ X, (xn) converges to x if:

(∀ε > 0)(∃N ∈ N) s.t. d(xn, x) < ε ∀n ≥ NDefinition. Given a sequence (xn) and an x ∈ X, (xn) is Cauchy if:

(∀ε > 0)(∃N ∈ N) s.t. n,m > N ⇒ d(xn, xm) < ε

Definition. (X, d) is complete if and only if every Cauchy sequence converges.

Note: Convergence in Rn (with any of the norms defined above) is componentwiseconvergence. So, is Rn complete in its three metrics? Yes! The discrete metric isalso complete.

Definition. A subset U ⊆ X is open if:

∀x ∈ U ∃r > 0 s.t. B(x, r) ⊆ UProposition. B(a, r) is open.

Proof. Let x ∈ B(a, r) be given. Then, r > d(a, x), so r − d(a, x) > 0.We will claim that B(x, r − d(a, x)) ⊆ B(a, r). Suppose y ∈ B(x, r − d(a, x)).

Then, we know that d(x, y) < r − d(a, x) ⇒ d(a, x) + d(x, y) < r. So, by thetriangle inequality, d(a, y) < r. Therefore, for any y ∈ B(x, r − d(a, x)), y is inB(a, r). Therefore, B(x, r − d(a, x)) ⊆ B(a, r), and B(a, r) is open. �

As another example, every subset of the discrete metric is open.

Definition. A subset E ⊆ X is closed if and only if X \ E is open.

Note that closed does not mean not open!Example: In the discrete metric, every subset is both open and closed.

Proposition. E ⊆ X is closed if and only if (xn ∈ E)(xn → x)⇒ x ∈ E.


(⇒) Assume X \ E is open. Also, suppose (xn) ∈ E and xn → x, andtoward contradiction, that x /∈ E, or x ∈ X \ E. This implies that∃r > 0 s.t. B(x, r) ⊆ X \ E. But since xn → x, d(xn, x) < r for somen. Then, xn ∈ B(x, r) ⊆ X \ E, so xn ∈ (X \ E) ∩ (E), which is a contra-diction.

(⇐) Toward contradiction, suppose X \ E is not open. Then, ∃x ∈ X \ E suchthat B(x, 1

n ) ∩ E 6= ∅ ∀n. This means that for every n ∈ N, there exists

some yn ∈ E such that d(yn, x) < 1n . Thus, consider the sequence of yns,

which we will call (yn). (yn) then obviously converges to x, but then byassumption and hypothesis, x ∈ (X \ E) ∩ (E), which is a contradiction.

�

21


10. 02/17/11: Metric Spaces, Part Two

10.1. Review.

Given (X, d), U ⊆ X is open if ∀x ∈ U ∃r > 0 s.t. B(x, r) ⊆ U∗. If Ui where i ∈ Iis open for all i, then

⋃i∈I Ui is open.

E ⊆ X is closed if X \E is open†. If Ei is closed for i ∈ I, then⋂i∈I Ei is closed.

10.2. More Metric Space Concepts.

Definition. For x ∈ X, a neighborhood of x is a set S ⊆ X such that there existsan open set U ⊆ X with x ∈ U ⊆ S.

Definition. The closure of E is:

E =⋂

F closed,E⊆F

F

Observe that E is closed. This is the smallest closed set that contains E. Notethat B(a, r) is not necessarily C(a, r) = {x ∈ X | d(a, x) ≤ r}; for instance, this isnot true in the discrete metric space.

Note that subsets can be both open and closed! For example, define a metricspace where X = [−1, 1]∪ (5, 17). Then, each interval is an open and closed subset

of the space. Another example is if X = Q, then (−√

2,√

2) is open and closed.

Definition. The interior of S ⊆ X is S = {s ∈ S | ∃r > 0 s.t. B(s, r) ⊆ S}, or{s ∈ S |S is a neighborhood of S}.

Definition. The boundary of S is ∂S = S \ S.

10.3. The p-adic Metric.

We will define a metric on Z based on any p for p prime. We will call this a p-adicmetric.

Define a metric by a norm, |j|p. Write j = pr · m such that (m, p) = 1. Let|0|p = 0 and |j|p = 2−r where 2 can be any prime.

Then it is the case that

|j|p ≥ 0 and j = 0 ⇐⇒ |j|p = 0

as well as that

|i+ j|p ≤ max{|i|p, |j|p} ≤ |i|p + |j|p .

Furthermore, dp(i, j) = |i− j|p, and this metric makes Z into a metric space. Notethat, for instance, for p = 3, 3n → 0.

∗Note that the empty set is open.†Note that the empty set is closed.

22


Example:(1− 3)(1 + 3 + 32 + · · ·+ 3n) = 1− 3n+1 → 1

Therefore, 2 + 2 · 3 + 2 · 32 + · · ·+ 2 · 3n + · · · = −1.

We can also define | · |p on Q, where q = pr · mn for (m, p) = 1, (n, p) = 1. Then,|q|p = 2−r and dp(q1, q2) = |q1 − q2|p.

Note that we can write numbers in “decimal” form by writing them as such:

x =

∞∑i=1

ai · pi +

∞∑j=1

di · p−i := . . . a3a2a1.d1d2d3 . . .

Then, we get a “decimal” where as you add terms to the left it converges and asyou add to the right it diverges!

We can define a field with this metric by defining it like we did R, but with thisnew metric instead of the familiar one for R. This field is called the p-adic numbers.Note that this technique of completing a metric space (as we did when constructingR from Q) can be applied to any metric.

10.4. Even More Metric Space Concepts.

Definition. Let (X, d) a metric space and S ⊆ X. Then, x ∈ X is called anaccumulation point (or limit point or cluster point) of S if every neighborhoodof x contains a point in S other than x.

Note that E ⊆ X is closed if and only if it contains all of its accumulation points.

Definition. Let (X, d) be given, and consider T := {U |U open in X}. This col-lection of subsets is called a topology on X defined by d. A topology satisfies threeaxioms:

(1) ∅ ∈ T , X ∈ T .(2) If Ui ∈ T , then i ∈ I ⇒

⋃i∈I Ui ∈ T .

(3) For U1, U2, . . . , Un ∈ T , it is the case that⋂ni=1 Ui ∈ T .

Any property that can be expressed only using T , or open sets, is called a topo-logical property. For instance, we can say...

Proposition. Convergence is a topological property, or in other words:

limn→∞

xn = x ⇐⇒ ∀U open in X and x ∈ U, ∃N s.t. n ≥ N ⇒ xn ∈ U .


(⇒) Let U be open with x ∈ U given. Since U is open, ∃r > 0 s.t. B(x, r) ⊆ U .Therefore, ∃N s.t. n ≥ N ⇒ d(xn, x) < r ⇒ xn ∈ B(x, r) ⊆ U .

(⇐) Let ε > 0 be given. We know that B(x, ε) is open. Thus, ∃N s.t. n ≥ N ⇒xn ∈ B(x, ε)⇒ d(xn, x) < ε.

�

Therefore, changing the metric does not change the notion of convergence on a giventopology. Note that Cauchiness and completeness are not topological properties,but closures and interiors are.

Definition. Let U, V be open subsets of X. A topological space X is Hausdorffif given that x ∈ U , y ∈ V , and x 6= y, then U ∩ V = ∅.

23


11. 02/24/11: Continuity and Cardinality, Part One

11.1. More About Metric Spaces.

Proposition. x ∈ E ⇐⇒ (∀ε > 0) B(x, ε) ∩ E 6= ∅.


(⇒) We will proceed by contradiction. Suppose there exists an ε > 0 suchthat B(x, ε) ∩ E = ∅. Then, X \ B(x, ε) is a closed set containing E. Bythe definition of a closure, this implies x ∈ E ⊆ X \ B(x, ε). This is acontradiction because x ∈ B(x, ε).

(⇐) Let F be closed and E ⊆ F . Suppose x /∈ F . Then there exists an ε > 0such that B(x, ε) ⊆ X \ F . Because E ⊆ F , X \ F ⊆ X \ E. This is acontradiction, so for x ∈ F and ∀F where F is closed, E ⊆ F ⇒ x ∈ E.

�

Definition. Given (X, d) and A,B ⊆ X, we say that A is dense in B if B ⊆ A.

Note that AX ∩B = AB ∩B (where AX is the closure of A under X and AB is theclosure of A under B), by the ε-criteria for x ∈ E.

11.2. Continuity.

Definition (ε-δ). Given metric spaces (X, d) and (Y,D), a function f : X → Y issaid to be continuous at x if

(∀ε > 0)(∃δ > 0) s.t. d(x, y) < δ ⇒ D(f(x), f(y)) < ε .

Note that this is continuity at a single point !Example: f : R→ R such that

f(x) =

{0, if x ∈ Q;

x, otherwise.

This is a function which is continuous at 0 and discontinuous anywhere else.

Proposition. f is continuous at x ∈ X ⇐⇒ (xn → x)⇒ (f(xn)→ f(x)).Note: This is known as the limit definition of continuity at a point.


(⇒) Suppose xn → x and let ε > 0 be given. Choose a δ such that

d(x, t) < δ ⇒ d(f(x), f(t)) < ε .

Using δ in the definition of xn → x, get an N s.t. ∀n ≥ N , d(xn, x) < δ.This implies that d(f(xn), f(x)) < ε.

(⇐) Suppose f is not continuous at x ∈ X. This means that:

∃ε > 0 s.t. ∀δ > 0,∃t ∈ B(x, δ) s.t. d(f(t), f(x)) ≥ ε .

Choose a sequence xn ∈ B(x, 1n ) with d(f(xn), f(x)) ≥ ε. Then, we have a

contradiction, as xn → x but f(xn) 9 f(x).

�24


Note that this has become the commonly accepted definition of continuity. However,this is because of the shift in focus from topological spaces to metric spaces (areversal of a previous trend). In a general topological space, the limit definition ofcontinuity does not hold!

Definition. Given metric spaces X,Y and f : X → Y , f is continuous if f iscontinuous at every point of X.∗

Proposition. f : X → Y is continuous ⇐⇒ f−1(U) is open for all open U ⊆ Y .†


(⇒) Let U be open in Y . Take x ∈ f−1(U). This implies that f(x) ∈ U . By thedefinition of continuity, ∃B(x, δ) s.t. f(B(x, δ)) ⊆ U , so B(x, δ) ⊆ f−1(U).

(⇐) Take x ∈ X and let ε > 0 be given. We must show that

∃δ > 0 s.t. f(B(x, δ)) ⊆ B(f(x), ε) .

However, B(f(x), ε) is open, so f−1(B(f(x), ε)) is open by hypothesis. Also,x ∈ f−1(B(f(x), ε)), so ∃δ > 0 such that B(x, δ) ⊆ f−1(B(f(x), ε)), whichimplies that f(B(x, δ)) ⊆ B(f(x), ε).

�

Definition. (X, d) and (Y,D) are isometric if there exists a bijection f : X → Ywith D(f(x), f(y)) = d(x, y). This implies that f−1 is also an isometry.

Definition. f : X → Y is called a homeomorphism if f is a continuous bijectionwhose inverse is continuous.

Note: The last condition is necessary! For instance, the identity function from(R, discrete) to (R, usual) is continuous but f−1 is not.

11.3. Countability and Cardinality.

We will take a little detour from metric spaces in order to talk about Cardinality,so as to prepare for the guest lecture in two lectures from Professor Hugh Woodin.

First, we will explore the Cantor Set, defined as such:

S0 = [0, 1]

S1 =

[0,

1

3

]∪[

2

3, 1

]S2 =

[0,

1

9

]∪[

2

9,

1

3

]∪[

2

3,

7

9

]∪[

8

9, 1

]S3 =

[0,

1

27

]∪[

2

27,

1

9

]∪[

2

9,

7

27

]∪[

8

27,

1

3

]∪[

2

3,

19

27

]∪[

20

27,

7

9

]∪[

8

9,

25

27

]∪[

26

27, 1

]The Cantor set is defined as C :=

⋂n∈N Sn. Essentially, we start with [0, 1] and

remove the middle third each time, for every interval. The endpoints of the intervalsin the set are definitely countable, but the set itself as actually uncountable!

∗The related concept of uniform continuity is defined in a footnote in Lecture 16.†Reminder: Given sets A,B and f : A→ B, if:

S ⊆ A, define f(S) = {f(x) |x ∈ S}, and

T ⊆ B, define f−1(T ) = {x ∈ S | f(x) ∈ T}.In this case, f−1 is not the inverse of f ! f−1 always exists when applied to a set.

25


12. 02/24/11: Cardinality, Part Two

12.1. The Cantor Set, Continued.

Recall the Cantor set from last time. Here is an inductive way to define it.

Take an interval I = [a, b]. We will define C(I) = {[a, a + b−a3 ], [a + 2(b−a)

3 , b]}.So, given J0 = {[0, 1]}, we have Jn+1 =

⋃I∈Jn C(I). Then,

C =⋂n∈N

(⋃I∈Jn

I) .

Proposition. The interior of the Cantor set is empty, or C = ∅.

Proof Sketch. Choose n, ε with 3−n < ε10 . Then, for some n, x is in an interval of

width 3−n, so there is no ball around x. �

12.2. Random Facts.

There exists a continuous nondecreasing function f : [0, 1] → [0, 1] with f(0) = 0and f(1) = 1 such that f is constant on [0, 1] \ C. The graph of f is known as theDevil’s Staircase.

Russell’s Paradox. Given a set S, S ∈ S or S /∈ S. Define A = {S |S /∈ S}. DoesA belong to A? There is no way for A to belong to or not belong to itself, as bothresult in contradictions. Thus, Russell’s Paradox will act as a cautionary tale forthe discussions to come; we must be careful when we’re talking about sets of allsets.

12.3. Cardinality.

We will denote the cardinality of a set A as |A| (or #(A)).

Definition. Given sets A and B, |A| = |B| if there exists a bijection f : A→ B.

Definition. A is finite if there exists an n ∈ N such that |A| = |{1, 2, . . . , n}|. Inother words, for finite sets, |A| = |B| ⇐⇒ they have the same number of elements.A is infinite if it is not finite.

Definition. A is countable if A is finite or |A| = |N|, or equivalently, you canenumerate the elements of A as a sequence.

Proposition. S ⊆ N is countable.

Proof. Define (cn) as follows. Define c1 to be the smallest element of S, and defineS1 = S \ {c1}. Then, define cn+1 to be the smallest element of Sn, and defineSn+1 = Sn \ {cn}. Then, it is easy to check that {ck | k ∈ N} = S. �

Definition. |A| ≤ |B| (also written A - B) if there exists an injection f : A→ B.

26


Example: If S ⊆ T , then S - T . Therefore, this order on sets says that up to achange of name, A ⊆ B.

Proposition. A - B ⇐⇒ ∃ a surjection g : B → A.


(⇒) Given an injection f : A → B, define f such that for a ∈ A, g(f(a)) = aand for everything else, g(x) = a for some a.

(⇐) We want to find an injection f : A→ B. For each a ∈ A, choose an x ∈ Bwith g(x) = a, and define f(a) = x.

�

Proposition. Z, Z× Z, and Q are countable.

Proof. We will prove them one at a time.

(1) Z: We will arrange the elements of Z in a sequence, namely (cn) such thatc0 = 0, c2n−1 = n, c2n = −n for all n ∈ N.

(2) Z × Z: By the previous argument, |Z| = |N|, so it suffices to show thatN× N is countable. We will arrange the elements of N× N as such:

(1, 1) (1, 2) (1, 3) (1, 4) . . .

(2, 1) (2, 2) (2, 3) (2, 3) . . .

(3, 1) (3, 2) (3, 3) (3, 4) . . ....

......

.... . .

Now, proceed down each diagonal, starting at (1, 1), then moving to thenext diagonal containing (1, 2) and (2, 1), then the next, containing (1, 3),(2, 2), (3, 1), and so on. Alternately, define a function f : N× N → R suchthat f((m,n)) = 2m3n. By the unique prime factorization of integers, thisis certainly an injection.

(3) Q: Take the function f : Z × N → Q such that f((m,n)) = mn . This is

clearly a surjection.

�

12.4. Counting R.

We will define 2N = {sequences (an) such that an ∈ {0, 1}}, such as 01101001 . . .Note that this is the same as the set of all subsets of N, and in general, 2A is thesame as subsets of A, also known as the power set of A.

Theorem (Cantor). 2N is uncountable.

Proof. Suppose it is countable. Then, we can enumerate them as such:

(an)1 = a1,1 a2,1 a3,1 . . .

(an)2 = a1,2 a2,2 a3,2 . . .

(an)3 = a1,3 a2,3 a3,3 . . ....

......

.... . .

27


Construct bn = an,m + 1 (mod 2). Obviously, bn is in 2N, but it cannot be the nthterm in the sequence of sequences because it differs from the nth term in the nthplace. Therefore, it is not in the list, and we have a contradiction. �

Corollary. R is uncountable.

Proof. It suffices to define an injection f : 2N → R. Let f((an)) =∑∞n=1 an10−n∗.

We will claim that f is an injection. The proof is in the lecture notes on Cardinality,page 2. �

There is another proof in the notes, page 3, which shows that R has cardinality noless than |2N|.

Theorem (Schroeder-Berinstein). If |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.

Proof. Next lecture. �

By the remark after the Corollary and this Theorem, we can see that |R| = |2N|.

The Continuum Hypothesis asks whether there is a subset of R that is uncount-able but that does not have the same cardinality as 2N. Or in other words, if thereexists a set A such that |N| ≤ |A| ≤ |2N|. The answer is still not known.

∗Note that series are not defined yet. However, partial sums are Cauchy sequences, so they are

reals.

28


13. 03/01/11: Set Theory: Guest Lecture by Hugh Woodin

13.1. Schroeder-Bernstein.

Two prominent set theorists were Cantor and Godel, both of whom went insane.Anyway...

Definition. Two sets X,Y have the same cardinality if there exists a bijectionΠ: X → Y . We denote this as |X| = |Y |.

Caution: This definition is not as intuitive as it seems. If X is infinite and a ∈ X,then it can happen that X and X \ {a} have the same cardinality.

Definition. Suppose X,Y are sets. Then, the cardinality of X is less than or equalto the cardinality of Y if there exists an injection Π: X → Y . We denote this as|X| ≤ |Y |.

Now, we will turn our attention to the Schroeder-Bernstein Theorem.

Theorem (Schroeder-Bernstein). Let X,Y be sets and suppose |X| ≤ |Y | and|Y | ≤ |X|. Then, |X| = |Y |. In other words, suppose X,Y are sets and there existsinjections f : X → Y and g : Y → X. Then, there exists a bijection Π: X → Y .

Before proving the theorem, we will first develop some concepts and propertiesthat will help us along the way.

Consider a set Z and an injection h : Z → Z. Start with some point a ∈ Z, andkeep applying h to it. Then, one of three things will happen:

(1) We will get into a loop, so hn(a) = a for some n ∈ N.(2) We will reach a point where h−n(a) is not in the range of h, so we cannot

”back up” any farther, and effectively, the chain starts at a certain point.(3) We will never cycle, but we will be able to ”back up” endlessly, so for all

n ∈ N, h−n(a) is in the range of h.

Notice that case 3 is just a degenerate case of case 1, as it is simply just an endlessloop backwards. Now, we will define a concept using this intuition.

Definition. Suppose h : Z → Z is injective and suppose a ∈ Z. Then the orbit ofa given by h is the subset of Z generated by a, h, h−1 where h−1 : {range of h} → Z.More formally, define X0 = {a}. Then, define inductively:

xn+1 = {h(b) | b ∈ Xn} ∪ {c ∈ Z |h(c) ∈ Xn} ∪Xn

Then, the orbit of a given by h, denoted Za, is defined to be⋃∞n=0Xn where n ∈ N.

Example: X1 = {a, h(a)} ∪ {c} where h(c) = a if such a c exists. Furthermore,X2 = X1 ∪ {h(h(a))} ∪ {d} where h(d) = c if c existed and if d exists.

Example: Let Z = {0, 1, 2, . . . }, and let h(n) = n2 for n ∈ Z. Then, Z0 = {0},Z1 = {1}, and Z2 = {2, 4, 16, 256, . . . }, or in other words the even powers of 2. Z3

can also be seen as the even powers of 3.

Note that we can also denote the orbit of a given by h as Zha in order to explicitlystate the injection that we are dealing with.

Lemma 1. Suppose h : Z → Z is injective and suppose a, b ∈ Z. Let Zha and Zhbbe the orbits of a and b relative to h, as defined above. Also, suppose Zha ∩Zhb 6= ∅.Then, Zha = Zhb .

29


Proof. Exercise. �

Another intuitive definition follows from our examination of orbits above.

Definition. Suppose h : Z → Z is injective and suppose a ∈ Z. Let Zha be the orbitof a given by h. Zha has an eve-point if there exists a b in Zha such that b is notin the range of h.

Lemma 2. If an orbit has an eve-point, then the eve-point of that orbit is unique.


Lemma 3. Let h : Z → Z be injective and a ∈ Z. Suppose Zha has no eve-point.Then, h|Zha : Zha → Zha is a surjection, and hence a bijection.


Now, we are ready for the actual proof.

Proof (Schroeder-Bernstein). We are given injections f : X → Y and g : Y → X.We need to construct a bijection Π: X → Y . Consider the functions g ◦ f : X → Xand f ◦ g : Y → Y . These are both injections because the composition of twoinjections is also always injective. So now, we can consider the orbits of points in Xgiven by g◦f and the orbits of points in Y given by f ◦g. We will define Π: X → Yby defining Π|Xg◦fa

for each orbit of X given by g ◦ f . There are two cases for eachorbit:

(1) Xg◦fa has no eve-point. Then, for each b ∈ Xg◦f

a , Π(b) = f(b).

(2) Xg◦fa has an eve-point. Let a0 be the eve-point. Then, it follows that Y f◦gf(a)

must also have an eve-point. Let b0 be the eve-point of Y f◦gf(a), and define

Π(a0) = b0, Π(a1) = b1, and so on.

We can see that every point of every orbit is hit exactly once by Π, and thus Π isa bijection from X to Y . �

This theorem proves the rather trivial-looking statement that if |X| ≤ |Y | and|Y | ≤ |X|, then |X| = |Y |. However, while it may look trivial, it really is not.

Proposition. |R| = |P (N)|, where P (N) = {A |A ⊆ N}.

Proof. It suffices to show that |R| ≤ |P (N)| and |P (N)| ≤ |R|.(1) Clearly, there is a bijection Π: N→ Q, so we just need an injection f : R→

P (Q). Define f(r) := {q | q ∈ Q, q < r} for all r ∈ R. This is clearlyinjective, so |R| ≤ |P (Q)| = |P (N)|.

(2) We need an injection g : P (N)→ R. Define g(a), where a ∈ P (N), as such:

g(a) =

{∑i∈a( 1

10 )i if a 6= ∅0 if a = ∅

This is clearly injective, so |P (N)| ≤ |R|.�

Theorem (Generalized Cantor). Suppose X is a set. Then, |X| 6= |P (X)|, whereP (X) = {A |A ⊆ X}∗.

∗Note that this implies |X| < |P (X)| and |R| = |P (N)| 6= |N|.30


Proof. Suppose f : X → P (X). We need to produce an A ⊆ X such that A is notin the range of f . Define A = {a ∈ X | a /∈ f(a)}. Then, A cannot be in the rangeof f . To see this, suppose A = f(b). Then, we have:

b ∈ A ⇐⇒ b /∈ f(b) ⇐⇒ b /∈ Awhich is clearly a contradiction. �

This theorem implies that there are an infinite number of infinities, each one biggerthan some others.

Therefore, we will say ℵ0 = |N|. Where does |P (N)| = |R| fit in? Is it ℵ1?The Continuum Hypothesis states that |R| = ℵ1. We have gotten to the point

where we know that given our standard axioms, we can build universes where thisis true and universes where this is false. For another illustration of a conjecture likethis, consider the Parallel Postulate. It is true in R2, but false in a universe thatis a circle without the edge, despite that the usual axioms of geometry hold in thisnew space.

In recent times, we’ve gotten closer to an answer. Behind this question and a lotof other set theory lie many philosophical issues about the meaning of things thatdon’t exist...

31


14. 03/03/11: Compactness, Part One

14.1. Basic Concepts.

Let A ⊆ X, and let (X, d) be a metric space.

Definition. An open cover of A is a family {Uα |α ∈ I} of open subsets of Xsuch that A ⊆

⋃α∈I Uα.

Example: [0, 1] ⊆ R. Then, {(−1, 2)} is an open cover. {(x − ε, x + ε) |x ∈ [0, 1]}is also an open cover. Consider ε = 1

3 . Then, the cover is very redundant, as it isan uncountably large collection of intervals, and we could come up with somethingmore economical.

Definition. If {Uα |α ∈ I} is an open cover of A, a subcover is a subfamily{Uα |α ∈ J}, where J ⊆ I, that is also an open cover of A.

Definition. A ⊆ X is compact if for every open cover, there is a finite subcover.

It is the case that A ⊆ X is compact if and only if A ⊆ A with the inheritedmetric is compact. Therefore, the notion of compactness is an inherited notion, likeclosedness. Note that this is a very difficult definition to work with, if we want toshow a set is compact. However, this does make it very easy to show that a set isnot compact.

Example: Finite metric spaces are always compact for any A ⊆ X.

Example: Take R. {(x− ε, x+ ε) |x ∈ R} is an open cover of R. Furthermore, youcannot cover R with finitely many intervals, so R is not compact. Another exampleis, for Z, the open cover {(n− 3

2 , n+ 32 )}.

14.2. Basic Properties.

Proposition. In any metric space, a compact subset is closed and bounded∗.

Proof. We will prove the two points separately.

(1) First, we will prove boundedness. Take x ∈ A. Consider {B(x, n) |n ∈ N}.This family of sets is trivially an open cover, even of all of X. Take a finitesubcover {B(x, n1, . . . , B(x, nmax)}. Then,

⋃maxi=1 B(x, ni) = B(x, nmax).

A ⊆⋃maxi=1 B(x, ni) because it’s an open subcover, so A ⊆ B(x, nmax) and

A is bounded.(2) Now, we will prove closedness. Let x ∈ X \A, and define:

Un = X \ {y | d(x, y) ≤ 1n where n ∈ N} .

We will claim that the Uns cover A, or that every a ∈ A is in the complementof some {y | d(x, y) ≤ 1

n}. In other words, for some n, d(a, x) > 1n , or

n > 1d(a,x) . This is clearly true by the Archimedean property of R, so our

claim is true. Now, extract a finite subcover {U1, U2, . . . , UM}. Notice that⋃Mi=1 Ui = UM , so A ⊆ UM . Then, we know that:

{y | d(x, y) ≤ 1M } ⊆ X \A ⊆ UM

so it follows that B(x, 1M ) ⊆ X \ A. Thus, because x was arbitrary, there

is an open ball around every point in X \A, and A is closed.

∗A subset S ⊆ X is bounded if there exists an a ∈ X and an R > 0 such that S ⊆ B(a,R).

32


�

Proposition. If (X, d) is a compact metric space, then any closed subset of X iscompact.

Proof. Let A be closed, and let⋃α∈I Uα be an open cover of A. It is clear that

{Uα |α ∈ I} ∪ {X \A} is an open cover of X. By the compactness of X, we know

that X must then also be covered by⋃Nn∈N Uαn∪X \A, for some N ∈ N. Then, it is

clear that⋃Nn∈N Uαn is a finite subcover of A, because X \A cannot cover anything

in A. �

14.3. Sequential Compactness.

Definition. Given (X, d) and A ⊆ X, A is sequentially compact if every se-quence in A has a convergent subsequence in A.

Example: By the Bolzano-Weierstrass Theorem, every closed interval [a, b] is se-quentially compact. Furthermore, by extension, [a1, b1] × [a2, b2] × · · · × [an, bn] issequentially compact in Rn. To see this in R2, first get a convergent subsequencein the first coordinate, which we can do because if we consider just the sequence(an), it is a sequence in R. Then, fix that subsequence and get a convergent sub-subsequence in the second coordinate. Any subsequence of a convergent sequenceconverges, so both coordinates converge and we have a convergent subsequence inR2. This process can easily be generalized to any n.

Recall that when we defined compactness, we said that the definition was veryhard to work with. This definition is much easier, so we would like for them to beequivalent. Lucky for us, they are!

Theorem. Let (X, d) be a metric space. Then, X is compact if and only if X issequentially compact.

Proof. In the next lecture. �

33


15. 03/08/11: Compactness, Part Two

15.1. (Sequential) Compactness.Remember: E ⊆ X is compact if every open cover of E has a finite subcover, andE ⊆ X is sequentially compact if every sequence in E has a convergent subsequencein E.

Theorem. Given a metric space (X, d), X is compact if and only if X is sequen-tially compact.∗


(⇒) Let (xn) be a sequence in X. If {xn |n ∈ N} is finite, then some xn mustbe hit infinitely many times, so we would be done. Therefore, assume{xn} is infinite. Let x ∈ X. Either x is the limit point of a subsequenceof (xn), or there exists an r > 0 such that |B(x, r) ∩ {xn |n ∈ N}| ≤ 1.To see this, assume that the latter is not true for some x ∈ X. Then,for all r > 0, |B(x, r) ∩ {xn |n ∈ N}| > 1. In this case, we could takeever-decreasing values of r, and there would be at least one xn such thatd(x, xn) < r for every r > 0. Therefore, if you picked all of those terms,we would have a convergent subsequence of (xn). So, in other words, thereis a convergent subsequence of (xn) or

⋃x∈X B(x, r) forms an open cover

for X. By compactness, we can obtain a finite subcover of this open cover,so only finitely many B(x, r) cover X, and hence {xn}. Then, because forevery x, |B(x, r) ∩ {xn |n ∈ N}| ≤ 1, the finiteness of this subcover wouldimply that {xn} was also finite, which is a contradiction.

(⇐) Let {Uα} be an open cover of X.First, we will prove this statement:

∃R > 0 s.t. ∀x ∈ X,∃Uα s.t. B(x,R) ⊆ Uα .

Proceeding by contradiction, we will assume that no such R exists. Becauseno such R exists, in particular, for each n ∈ N, there exists an xn ∈ X suchthat B(xn,

1n ) * Uα for any Uα. Now, because X is sequentially compact,

we can get a convergent subsequence of this sequence which converges tosome x ∈ X, which we will denote (xnk) → x. We know that this x ∈ Uαfor some Uα, and because Uα is open, there exists an r > 0 such thatB(x, r) ⊆ Uα. Then, by the convergence of (xnk), there is some N ∈ Nsuch that for all n ≥ N , the following are true:(1) 1

n <r2 , and

(2) xn ∈ B(x, r).However, by (2), we have that B(xn,

r2 ) ⊆ B(x, r) ⊆ Uα, so in turn, by (1),

we have that B(xn,1n ) ⊆ Uα, which contradicts that we can find xn ∈ X

such that B(xn,1n ) * Uα, which means that our initial assumption that

such R does not exist was false. Therefore, our initial statement is true.†

Now, we will show that finitely many balls of radius R (as defined above)cover X. It is sufficient to show this because each ball is then containedin some Uα, so then we know that finitely many Uα cover X, and we will

∗Note that both of these are topological properties, but this theorem is not true in a general

topological space.†This is a result that is of independent interest, because it proves the existence of Lebesgue

numbers.

34


have extracted a finite subcover of a general open cover of X. We willproceed by contradiction, and assume that it is not the case that finitelymany balls of radius R cover X, and that it is impossible to generate afinite subcover. Then, we will construct a sequence (xn) as such. First,simply pick an x1 ∈ X. B(x1, R) does not cover X, so we can pick a pointx2 outside B(x1, R) such that d(x1, x2) > R

2 . Then, pick another point x3outside of B(x1, R)∪B(x2, R) such that d(xi, x3) > R

2 for i = 1, 2. Continue

this process, such that xn+1 is not in⋃ni=1B(xi, R) and d(xi, xn+1) > R

2for i = 1, 2, . . . , n, for all n ∈ N. Then, the latter statement is obviouslytrue for all subsequences of (xn). Therefore, this sequence cannot have anyconvergent subsequences, a contradiction of our initial assumption. Thus,it must be the case that finitely many balls of radius R cover X, so X mustbe compact.

�

Theorem (Heine-Borel). A subset of Rn is compact if and only if it is closed andbounded.

Proof. We will proceed in both directions.

(⇒) We already proved this statement for an arbitrary metric in the last lecture.(⇐) Let A ⊆ Rn be closed and bounded. By the boundedness of A, we know

that A ⊆ B(0, R + 1) for some R > 0. Then, it is clearly the case thatA ⊆ [−R,R]×[−R,R]×· · ·×[−R,R]. The right hand side of that expressionis sequentially compact (from last lecture) and hence compact (from thislecture), and we also proved last time that a closed subset of a compact setis also compact. Therefore, A is compact.

�

From this theorem, we obtain some very useful results, such as that the Cantor setis compact, the unit sphere is compact, and so on.

Now, we will build up to another major, but seemingly unrelated, result.

15.2. The Equivalence of Norms on Rn.

Proposition. Let X,Y be metric spaces such that X is compact. Then, given acontinuous function f : X → Y , f(X) is compact.

Proof. Let Uα be an open cover of f(X). Then, it is clear that f−1(Uα) covers X.Because f is continuous, we know that for any open set U ⊆ Y , f−1(U) is also anopen set. Thus, f−1(Uα) is an open cover of X, and by the compactness of X, forsome n ∈ N, we have that:

X ⊆n⋃i=1

f−1(Uαi)

This, in turn, implies that f(X) ⊆⋃ni=1 Uαi , so f(X) is compact. �

Proposition. Let X be a metric space. If X is compact, then any continuousfunction f : X → R attains its minimum and maximum values.

35


Proof. From the proposition above, we know that f(X) ⊆ R must be compact, soit is therefore closed and bounded. Thus, sup f(X) and inf f(X) exist, and by theclosedness of f(X), must lie inside f(X) itself. �

Corollary. Let X be a compact metric space. Then, given a continuous functionf : X → R such that f(x) > 0 for all x ∈ X, it is the case that there exists an ε > 0such that f(x) ≥ ε for all x ∈ X.

Proof. This follows immediately from the above proposition. �

Theorem. Any two norms on Rn are equivalent, or in other words, given twonorms | · | and || · || on Rn, there exist A,B > 0 such that for all x ∈ Rn,

A|x| ≤ ||x|| ≤ B|x| .

Proof. In next lecture. �

36


16. 03/10/11: Norms, Separability, Connectedness

16.1. Equivalence of Norms on Rn.

Theorem. Any two norms on Rn are equivalent, or in other words, given twonorms || · || and ||| · ||| on Rn, there exist A,B > 0 such that for all x ∈ Rn,

A||x|| ≤ |||x||| ≤ B||x|| .

Proof. We will make two trivial observations to start. First, note that if we replace||| · ||| with || · ||2, we will obtain the same result. Second, the statement is obviouslytrue for any A,B if x = 0, so we will assume that x 6= 0∗.

First, we will prove the first inequality. Let ei be a unit vector with a 1 in theith place. Then, x =

∑ni=1 xiei. Then, we can obtain that∣∣∣∣∣

∣∣∣∣∣n∑i=1

xiei

∣∣∣∣∣∣∣∣∣∣ ≤

n∑i=1

(|xi| · ||ei||) ≤

√√√√ n∑i=1

|xi|2 ·

√√√√ n∑i=1

||ei||2

by the triangle inequality, followed by the Cauchy-Schwarz inequality. Furthermore,√∑ni=1 |xi|2 = ||x||2, so this implies

1√∑ni=1 ||ei||2

||x|| ≤ ||x||2

which proves the first inequality. Observe that this result implies that || · || isuniformly continuous† for || · ||2, or, in other words, that the identity function from(Rn, || · ||) to (Rn, || · ||2) is uniformly continuous (and vice-versa).

Now, we will prove the second inequality. Let ε > 0 be given. Then, chooseδ = Aε, where A is the term we found above. Then, if ||x− y||2 < δ, then

||x− y|| ≤ 1

A||x− y||2 <

1

Aδ =

Aε

A= ε .

As noted above, this means that the identity function from (Rn, || · ||2) to (Rn, || · ||)is (uniformly) continuous. In particular, let Sn = {x ∈ Rn | ||x||2 = 1}. Then,clearly, the identity function from (Sn, || · ||2) to (Rn, || · ||) is also continuous. Weknow that for any x ∈ Sn, ||x||2 =

∑ni=1 x

2i = 1, so Sn is obviously closed and

bounded, and hence compact, by the Heine-Borel Theorem. Furthermore, || · || is anorm, so ||x|| > 0 for all x ∈ Sn. By the theorem from last lecture, this means thatthere exists an ε > 0 such that ||x|| > ε for all x ∈ Sn. So, if we take an arbitraryv 6= 0 in Rn, then we have:

ε = ε

∣∣∣∣∣∣∣∣ v

||v||2

∣∣∣∣∣∣∣∣2

<

∣∣∣∣∣∣∣∣ v

||v||2

∣∣∣∣∣∣∣∣⇒ ε||v||2 < ||v|| ⇒ ||v||2 <1

ε||v||

completing the proof. �

∗Note that 0, in this case, is the zero vector in Rn, not the real number 0.†Given metric spaces (X, d) and (Y,D), a function f : X → Y is uniformly continuous if for

all ε > 0, there exists a δ > 0 such that for all x, y ∈ X,

d(x, y) < δ ⇒ D(f(x), f(y)) < ε .

The difference between continuity and uniform continuity is that the δ in uniform continuity mustwork for all x, y.

37


Now, we will discuss two final metric space topics before diving into the subjectof Calculus.

16.2. Separability.

Definition. Let (X, d) be a metric space. X is separable∗ if there exists a count-able dense subset of X.

Example: R is separable because R ⊆ Q and |Q| = |N|.

Proposition. Every subset of a separable metric space is separable with the inher-ited metric.

Proof. Let (X, d) be a metric space, Q be the countable dense subset of X, andA ⊆ X. Consider {B(q, 1

n ) | q ∈ Q, n ∈ N}. This is a countable set, as Q × N is

countable. For each q, n, if B(q, 1n ) ∩ A 6= ∅, choose a point in it, which we will

denote xq,n. Then, let D = {xq,n | q ∈ Q, n ∈ N}. It is clear that D is countableand a subset of A. Therefore, we only need to show that it is dense, or in otherwords, given a ∈ A and ε > 0, that there exists an x ∈ D such that x ∈ B(a, ε).Choose an n ∈ N such that 1

n <ε2 . Because Q is dense in X, there exists a q ∈ Q

such that d(q, a) < 1n <

ε2 . This implies that B(q, 1

n ) ∩ A 6= ∅, so for some x ∈ D,

x ∈ B(q, 1n ). Therefore, we have:

d(x, a) ≤ d(x, q) + d(q, a) <1

n+

1

n< ε


The notion of separability might seem a bit random and unrelated, but in thewords of Professor Jones, “Paradise is a complete separable metric space.”

16.3. Connectedness.

Definition. Let (X, d) be a metric space. C ⊆ X is connected† if for open subsetsU, V of X:

C ⊆ U q V ⇒ C ⊆ U or C ⊆ V .

Note: We define the empty set to be connected.Example: All sets which are made up of just one point are obviously connected.

Proposition. The continuous image of a connected set is connected.

Proof. Take a continuous function f : X → Y , where X,Y are metric spaces. Then,we want to show that for a connected subset C of X and open subsets U, V of Y ,

f(C) ⊆ U q V ⇒ f(C) ⊆ U or f(C) ⊆ V .

Notice that C ⊆ f−1(U)q f−1(V ). Then, by the connectedness of C, C ⊆ f−1(U)or C ⊆ f−1(V ). Thus, f(C) ⊆ U or f(C) ⊆ V , so f(C) is connected. �

∗This is a topological notion.†This is also a topological notion.

38


Remember that for a continuous function, the inverse image of an open set is alsoopen.

Theorem. A subset of R is connected if and only if it is an interval.∗


(⇐) Let I ⊆ R be an interval. Let U and V be open sets in R, and supposeI ⊆ U q V . Now, we will proceed by contradiction and assume I ∩ U 6= ∅and I ∩V 6= ∅. Choose s ∈ I ∩U and t ∈ I ∩V . Without loss of generality,assume s < t. Consider the set {x | [s, x] ⊆ U}. This set is clearly boundedabove by t. Also, s is in the set, because s ∈ U . Therefore, the supremumexists, so we will say a = sup{x | [s, x] ⊆ U}. Because U is open, if awas in the set, it would be the case that (a − ε, a + ε) ⊆ U , and hence[s, a+ ε

2 ] ⊆ U . However, a was the supremum of U , so this could not be thecase, and a /∈ U . Similarly, a /∈ V , because if it was, then (a− ε, a+ ε) ⊆ V ,which would imply a − ε

2 ∈ V , a contradiction because anything less thana must be in U and we assumed that U ∩ V = ∅. Therefore, a /∈ U anda /∈ V . However, this creates a contradiction with our initial assumption,because [s, t] ⊆ I and s < a < t. Thus, any interval in R is connected.

(⇒) Let A ⊂ R be connected. We will claim that then, it has the propertythat if s, t ∈ A where s < t, then [s, t] ⊆ A. To see this, assume that thisis not true, and that for some s < x < t, x /∈ A. Then, it is clear thatA ⊆ (−∞, x) ∪ (x,∞), a disjoint union of two open sets. However, we alsoknow that s ∈ (−∞, x) and t ∈ (x,∞), which is a contradiction because Ais connected. Therefore, our claim is true, and A must be an interval.

�

Corollary. Let f : [a, b] → R be a continuous function. Then, f([a, b]) is also aclosed interval, [c, d].

Proof. This follows immediately from the two previous results. �

This is a very powerful corollary, as it shows that a continuous function on aninterval simply stretches, shrinks, and/or shifts the interval!

∗Interval, here, means any sort of interval, namely any one from this list: (a, b), (a, b], [a, b),

[a, b], (−∞, b), (−∞, b], (a,∞), [a,∞).

39


17. 03/15/11: More (Path) Connectedness, Series and Convergence

17.1. More on Connectedness.

Recall that E ⊆ (X, d) is connected if for open subsets U, V of X:

E ⊆ U q V ⇒ E ⊆ U or E ⊆ V .

We have already shown that in R, the only connected sets are intervals, as well asthat the continuous image of a connected subset is connected. Furthermore, notethat connectedness is an inherited property, like compactness.

Proposition. Let A,B be connected subsets of X. If A ∩ B 6= ∅, then A ∪ B isconnected.

Proof. Let U, V be open sets in X, and suppose A∪B ⊆ U qV . Then, A ⊆ U qVand B ⊆ U q V . Without loss of generality, assume A ⊆ U . Then, A∩B ⊆ U . Weknow that A∩B 6= ∅, so there exists a b ∈ B such that b ∈ U . By the connectednessof B it follows that B ⊆ U . Therefore, A ∪B ⊆ U , completing the proof. �

Now, given a metric space X, define a relation ∼ on X by:

x ∼ y ⇐⇒ {x, y} ⊆ Afor some connected subset A of X. The properties of reflexivity and symmetry aretrivial, so we only need to check transitivity. Assume x ∼ y and y ∼ z. Then,{x, y} ⊆ A and {y, z} ⊆ B, for connected subsets A,B of X. Note that y ∈ A∩B,so A∪B is connected by the previous proposition, and {x, z} ⊆ A∪B. Therefore,x ∼ z, and this is an equivalence relation.

Definition. The equivalence classes as defined above are called the connectedcomponents of X.

Note: The connected components are themselves connected and are maximal ele-ments for the property of being connected.Example: In Q, the connected components are just individual points.

17.2. Path-connectedness.

Definition. Let (X, d) be a metric space, and x, y ∈ X. A path∗ from x to y is acontinuous function c : [0, 1]→ X such that c(0) = x and c(1) = y.

Definition. (X, d) is path-connected if for all x, y ∈ X, there exists a path fromx to y. Equivalently, there exists a continuous function c : [a, b] → X such thatc(a) = x and c(b) = y.

Proposition. If a metric space is path-connected, then it is connected.

Proof. Let (X, d) be a metric space, and let U, V be open sets in X. SupposeX = U q V . Notice that if either U or V is the empty set, then the propositionis trivially true. Thus, assume that U and V are nonempty. Proceeding towardcontradiction, let x, y ∈ X and choose x ∈ U and y ∈ V . Now, let c : [0, 1]→ X be acontinuous function such that c(0) = x and c(1) = y. Then, c−1(U) is a nonemptyopen subset of [0, 1] as 0 ∈ c−1(U), and similarly, V , is also a nonempty open

∗This is a topological notion.

40


subset of [0, 1] because 1 ∈ c−1(V ). Also, we have that c−1(U) ∪ c−1(V ) = [0, 1]and c−1(U) ∩ c−1(V ) = ∅ because c is a well-defined function. However, because0 ∈ U and 1 ∈ V , this contradicts that [0, 1] is a connected set, so we have acontradiction and X must be connected. �

However, the converse is not true! For an example of this, see the Topologist’s sinecurve (or the discrete analog, with sawtooth forms instead of smooth curves).

Like with connected sets, we can define components on path-connected sets.

Definition. Let (X, d) be a metric space. Define a relation ∼ on X such that forany x, y ∈ X, x ∼ y if and only if there exists a path from x to y.

We will check that this is an equivalence relation.

(1) Reflexive: It is obvious that x ∼ x, because c(t) = x where 0 ≤ t ≤ 1 isclearly a path from x to x.

(2) Symmetric: Assume x ∼ y. Then, given a path c : [0, 1] → X such thatc(0) = x and c(1) = y, define another function d : [0, 1] → X such thatd(t) = c(1− t) where 0 ≤ t ≤ 1. This is clearly a path from y to x.

(3) Transitive: Assume x ∼ y and y ∼ z. Then, there exist two paths,c : [0, 1] → X such that c(0) = x and c(1) = y, and d : [0, 1] → X suchthat d(0) = y and d(1) = z. Define cd : [0, 2]→ X such that:

cd(t) =

{c(t) if 0 ≤ t ≤ 1,

d(t− 1) if 1 ≤ t ≤ 2

Now, we just need to show that this new function is continuous. This isobvious for all t where 0 ≤ t < 1 and 1 < t ≤ 2, because c and d werethemselves continuous. Thus, we only need to verify continuity at t = 1,which can be done using ε-δ. Therefore, x ∼ z.

Definition. The equivalence classes as defined above are called the path compo-nents of X.

That concludes our time with metric spaces. We will now move on to Calculus!Everything in this part of the class will be done in R.

17.3. Convergence of Series.

Definition. Let (an) be a sequence. Then,

∞∑n=0

an = L ⇐⇒ limn→∞

sn = L

where sn =n∑k=0

an.∗ We will say that∞∑n=0

an converges if∞∑n=0

an = L and L 6= ±∞.

Example:∞∑n=0

rn = 11−r for |r| < 1.

∗L can be ∞ or −∞.

41


Example:∞∑n=1

1n(n+1) = 1, because 1

n(n+1) = 1n −

1n+1 , so writing out sn, we get

1− 1n+1 , which tends to 1.

Here are a couple of remarks about series:

(1) If∑an converges, then an → 0 because an = sn − sn−1.

(2) If∑an and

∑bn converge, then

∑(an + bn) converges to

∑an +

∑bn by

the limit theorems.

Theorem (Comparison Test). For 0 ≤ an ≤ bn, if∑bn converges, then so does∑

an.

Proof. sn =∑nk=0 ak is nondecreasing if all of the terms an are positive. Thus,

sk ≤n∑k=0

bk ≤∞∑k=0

bk .

So, {sn |n ∈ N} is bounded above, and a bounded monotone sequence alwaysconverges. Therefore,

∑an converges. �

The textbook gives an alternate proof for this test, involving the fact that anyseries whose partial sums form a Cauchy sequence converges (known as the Cauchycriterion). This is the case because any Cauchy sequence in R converges. Thismethod is also used to prove the theorem regarding absolute convergence as wellas the Alternating Series Test below.

Example:∑

1n2 converges because 1

n2 <1

n(n−1) for large n.

Definition.∞∑n=0

an converges absolutely if∞∑n=1|an| converges.

Proposition. If a series converges absolutely, then it converges.

Proof. Notice that 2an = (|an| + an) − (|an| − an). Both of these two terms aregreater than or equal to zero, and by the triangle inequality, we have:

|an|+ an ≤ 2|an| and |an| − an ≤ 2|an| .Furthermore, because

∑an converges absolutely,

∑2|an| also converges. Then, by

the theorem, because the two equations above hold,∑(|an|+ an)− (|an| − an) =

∑2an = 2

∑an

also converges. �

Theorem (Alternating Series Test). The series∞∑n=0

(−1)n an, where an ≥ 0, con-

verges if an ≥ an+1 and limn→∞

an = 0.

Proof. Supplied in the Textbook, Theorem 15.3. �

Notice that if∑

(−1)n an = L where L 6= ±∞, then

∣∣∣∣ k∑n=1

an − L∣∣∣∣ ≤ |ak+1|. Fur-

thermore, it is the case that the sequence of even partial sums is decreasing andthat the sequence of odd partial sums is increasing.

42


18. 03/17/11: More Convergence, Power Series, Differentiability

18.1. More Convergence.

Theorem (Ratio Test). Let∑an be a series such that an > 0 for all n ∈ N. Then,

if lim sup an+1

an< 1, then

∑an converges.

Proof. Let r = lim sup an+1

an. Choose an R such that r < R < 1. Then, there exists

an N ∈ N such that sup{aN+1

aN, aN+2

aN+1, . . . } < R. This implies that aN+k+1

aN+k< R for

all k ≥ 0. Thus,

aN+1 < RaN , aN+2 < RaN+1 < R2aN , . . . , aN+k < RkaN .

So, consider∑∞j=N aj . This converges by comparison with

∑∞j=N R

jaN because

aj < RjaN . Therefore,∑an converges. �

Example:∑

n2n is a series that we can easily use the Ratio Test on.

Theorem (nth Root Test). Let∑an be a series such that an ≥ 0 for all n ∈ N.

If lim sup n√an < 1, then

∑an converges.

Proof. Let r = lim sup n√an. Choose an R such that r < R < 1. Using ε = R − r

in the definition of lim sup, we can see that there exists an N ∈ N such that

sup{(aN )1N , (aN+1)

1N+1 , . . . } < R. For n ≥ N , we have (an)

1n < R, so an < Rn.

Then, we can conclude that∑an converges by following the same procedure as

with the Ratio Test above. �

18.2. Power Series.

In general, a power series is a particular kind of series, of the form∞∑n=1

an(x−x0)n.

However, we usually just say x0 = 0 and write the power series as∞∑n=1

anxn.

Theorem. Let∑anx

n be a power series. Also, let

β = lim sup n√|an| and R =

1

β.

Then∗,

(1) If |x| < R,∑anx

n converges absolutely.(2) If |x| > R,

∑anx

n does not converge.

Proof. We will prove the two cases.

(1) We want to show that∑|anxn| converges. We can write because |anxn| =

|an||xn|, we must show that∑|an||xn| converges. We will use the nth root

test.

lim sup n√|an||xn| = lim sup( n

√|an| |x|) = |x| lim sup n

√|an| = |x|β .

So, the power series converges if |x|β < 1, or |x| < 1β = R†.

∗Note that β can be infinite.†R is known as the radius of convergence.

43


(2) Assume xβ > 1. Then, working backwards from the chain of equalities

above, we have that lim sup n√|an||xn| > 1. If n

√|an||xn| > 1, then it is

clearly the case that |an||x|n > 1. However, this would mean that the termsof the power series do not tend to zero, so the series as a whole does notconverge.

�

Here are some very useful power series to know.

· “ex” = exp(x) =

∞∑n=0

xn

n!

· sin(x) =

∞∑n=0

(−1)nx2n+1

(2n+ 1)!

· cos(x) =

∞∑n=0

(−1)nx2n

(2n)!

· 1

1− x=

∞∑n=0

xn for |x| < 1.

· log(1− x) = −∞∑n=1

xn

nfor |x| < 1.

18.3. Differentiability.

Let be a function f : (x, y)→ R where x, y ∈ R, and let a ∈ (x, y).

Definition. f is differentiable at a if limx→a

f(x)−f(a)x−a = L for some L ∈ R. Then,

we will write L = f ′(a).

However, we have not really defined what limx→a f(x), or in other words, the limitof a function at a, means, so we will do that now, for general metric spaces.

Definition. Let f : X → Y where (X, d) and (Y,D) are metric spaces, and leta ∈ X and y ∈ Y . Then:

limx→a

f(x) = y ⇐⇒ (∀ε > 0)(∃δ > 0) s.t. 0 < d(x, a) < δ ⇒ D(f(x), y) < ε .

An alternate definition of this concept is that if limx→a

f(x) = y, then:

F (x) =

{f(x), if x 6= a

y, if x = a

is continuous.

Theorem. If f is differentiable at a, then f is continuous at a.∗

∗Note that this does not apply for general metric spaces! We are returning to the definitions

of f and a given in the very beginning of this section.

44


Proof. We can prove this using the ε-δ definition of the limit of a function. However,we will try using the alternate F (x) method instead.

Because we know that limx→a

f(x)−f(a)x−a = L for some L ∈ R, we can define

F (x) =

{f(x)−f(a)

x−a , if x 6= a

L, if x = a.

where F (x) is continuous. We want to show that f(x) is continuous at a. Inparticular, we know that F (x) is continuous at x = a. Thus, for x 6= a, we have:

F (x) =f(x)− f(a)

x− a⇒ f(x) = f(a) + F (x)(x− a) .

Then, because F (x) is continuous at a, for any ε > 0, there exists a δ > 0 such that|a− x| < δ ⇒ |L− F (x)| < ε. This means:

|L− F (x)| =∣∣∣L− f(x)−f(a)

x−a

∣∣∣ = |L(x− a) + f(a)− f(x)| ≤(1)

|Lδ + f(a)− f(x)| < ε .(2)

In order to prove the continuity of f(x) at a, we must find some γ such that|a − x| < γ ⇒ |f(a) − f(x)| < ε. We will proceed casewise, for when L is positiveand negative.

Let L be positive and ε > 0 be given. Then, set γ = δ. Assuming |a−x| < γ = δ,we obtain:

|f(a)− f(x)| < |Lδ + f(a)− f(x)| < ε .

Therefore, f(x) is continuous in this case.Now, let L be negative and ε > 0 be given. Using a simple change of variables,

we will simply write L = −L, and keep L > 0. Notice that, in the series ofinequalities (1) above, | − L − F (x)| = |F (x) + L|. Therefore, we can write (2) as|f(x)− f(a) + Lδ| < ε. Then, letting γ = δ, and assuming |a− x| < γ, we obtain:

|f(a)− f(x)| = |f(x)− f(a)| < |f(x)− f(a) + Lδ| < ε .

Therefore, f(x) is also continuous in this case, completing our proof. �

Now, we will simply list some basic properties of derivatives. We will not provethese because the proofs are fairly elementary, and involve only simple limit compu-tations. However, the proofs are in the textbook and any other elementary calculustextbook. We will just assume that this theorem is true, and use its results liberally.

Theorem. Let f, g be functions on an open interval in R and let a be in thatinterval. Assume that f, g are differentiable at a. Then, the following are true:

(1) (cf)′(a) = cf ′(a), where c is a constant.(2) (f + g)′(a) = f ′(a) + g′(a).(3) (fg)′(a) = f(a)g′(a) + f ′(a)g(a).

(4) If g(a) 6= 0, then(fg

)′(a) =

f ′(a)g(a)− f(a)g′(a)

g(a)2.

For the quotient rule (part 4 above), notice that because g(a) 6= 0 and g is contin-uous at a, there exists an open interval containing a such that g(x) 6= 0 for all x inthat interval, making the theorem in fact valid as stated.

45


19. 03/29/11: Theorems About the Derivative

19.1. Review.

Suppose we have a function f : (a, b) → R where (a, b) is an open interval. If

limx→af(x)−f(a)

x−a = L, then L = f ′(a), and f is differentiable on a. Equivalently,

we can define a function F (x) such that F (x) = f(x)−f(a)x−a if x 6= a and F (x) = L if

x = a. Then, F is differentiable at a if and only if F is continuous at a.

Example: f(x) = x2χ where χ(x) is 1 if x ∈ Q and 0 if x /∈ Q. f is differentiableat x = 0. Because all sorts of functions are differentiable at a single point, thisconcept isn’t really useful. What we really want is differentiability in the wholeinterval.

19.2. The Main Results.

Theorem (Chain Rule). Let I and J be open intervals, and let f : I → R andg : J → R be functions such that if a ∈ I, then f(a) ∈ J . If f is differentiable at aand g is differentiable at f(a), then g ◦ f is differentiable at a, and

(g ◦ f)′(a) = g′(f(a))f ′(a) .

Note: This is just the familiar chain rule, or in other words, dydx = dy

du ·dudx .

Proof. Define a function h : I → R such that:

h(y) =

{g(y)−g(f(a))y−f(a) , if y 6= f(a),

g′(f(a)), if y = f(a).

Then, h(y) is clearly continuous because g is differentiable at f(a). Also, define afunction GF : I → R such that:

GF (x) =

{g(f(x))−g(f(a))

x−a , if x 6= a,

g′(f(a))f ′(a), if x = a.

Now, it suffices to show that GF is continuous at x = a.

Notice that (h◦f)(x) = g(f(x))−g(f(a))f(x)−f(a) . Then, we haveGF (x) = h◦f(x)· f(x)−f(a)x−a

when x 6= a and GF (x) = g′(f(a))f ′(a) when x = a. We know that (h ◦ f)(x) goes

to g′(f(a)) as x approaches a by the definition of h and f(x)−f(a)x−a approaches f ′(a)

as x approaches a by the differentiability of f , so we are done. �

Theorem. Let I be an open interval in R, and let f : I → R. Suppose x0 ∈ Iand that f(x0) is a maximum [or minimum] on I. If f is differentiable at x0, thenf ′(x0) = 0.

Proof. Suppose without loss of generality that f ′(x0) > 0. Then, we will claim thatit is the case that for all ε > 0, there exists an x such that x0 < x < x + ε andf(x) > f(x0). Assume toward contradiction that this is not the case. Then, we

have, for every ε, some x such that f(x) ≤ f(x0). This implies that f(x)−f(x0)x−x0

≤0, which furthermore implies that lim

x→x0

f(x)−f(x0)x−x0

≤ 0, which is a contradiction

because f ′(x0) > 0. Therefore, because there exists an x such that f(x) > f(x0),f(x0) cannot be a maximum. A similar argument can be used to see that f(x0)cannot be a minimum. �

46


Theorem (Rolle). Let f : [a, b]→ R be continuous and differentiable on (a, b), andsuppose f(a) = f(b). Then, there exists an x ∈ (a, b) such that f ′(x) = 0.

Proof. By the compactness of [a, b], f attains its maximum (and minimum) on[a, b]. If the maximum is attained at an endpoint, then it must attain its minimuminside (a, b) or the function must be constant (as the minimum would then be equalto the maximum). In the former case, the previous theorem proves our claim, andin the latter, the derivative is 0 everywhere. �

Theorem (Mean Value). Let f : [a, b] → R be continuous and differentiable on(a, b). Then, there exists an x ∈ (a, b) such that

f ′(x) =f(b)− f(a)

b− a

Proof. Let t(x) = f(x) −(f(a) + (x− a) f(b)−f(a)b−a

). Notice that t(a) = 0 and

t(b) = 0. Also, t′(x) = f ′(x)− f(b)−f(a)b−a . So, by Rolle’s Theorem, there exists an x

such that 0 = f ′(x)− f(b)−f(a)b−a , or f ′(x) = f(b)−f(a)

b−a . �

Corollary. Let f : (a, b) → R and suppose f ′(x) = 0 for all x ∈ (a, b). Then, f isa constant function.

Proof. If p > q in (a, b), then 0 = f(p)−f(q)p−q , which implies that f(p) = f(q) by the

Mean Value Theorem. �

Corollary. If f : (a, b) → R is differentiable and f ′(x) > 0 for all x ∈ (a, b), thenf is strictly increasing.

Proof. Notice that f ′(x) = f(p)−f(q)p−q for some p > q in (a, b), so f(p) > f(q). �

Note that this corollary can be generalized to all of the variations of >, namely≥, <, and ≤, for which the corollary would state increasing, strictly decreasing,and decreasing, respectively.

19.3. Taylor’s Theorem.

Let f : (a, b) → R be differentiable. Then, we can talk about f ′ as a function,namely f ′ : (a, b)→ R. So, we could potentially differentiate this function and get(f ′)′(x), or f ′′(x), and so on, where the nth derivative of f at x is denoted f (n)(x),and is defined to be (f (n−1))′(x).

Definition. Cn(a, b) := {f : (a, b)→ R | f ′, f ′′, . . . , f (n) are continuous on (a, b).}

Notice that C∞(a, b) =⋂∞n=1 C

n(a, b) are the smooth functions on (a, b).

As a side note, when dealing with Taylor series, we will always expand the seriesaround 0. Refer to the textbook (p. 174) for a brief explanation of shifting seriesin order to expand them around arbitrary points.

47


Definition. Given a function f ∈ C∞(a, b) such that a < 0 < b, the Taylor series∗

of f is defined as∞∑n=0

f (n)(0)xn

n!

This is clearly a power series, but we would like to know if it converges, what itconverges to, and if it actually has anything to do with the function. Luckily, wehave a Theorem that tells us just that.

Theorem (Taylor). Suppose f has n derivatives on (a, b), where a < 0 < b. Then,for x ∈ (a, b), there exists a y between 0 and x† such that

f(x) = f(0) + f ′(0)x+f ′′(0)x2

2!+ · · ·+ f (n−1)(0)xn−1

(n− 1)!+fn(y)xn

n!

Proof. Let M be the solution to the equation

f(x) =

n−1∑k=0

f (k)(0)xk

k!+Mxn

n!

for some fixed nonzero x. Furthermore, define the function

g(t) =

n−1∑k=0

f (k)(0) tk

k!+Mtn

n!− f(t)

Notice that g(x) = 0. Furthermore, g(0) = 0 and g(k−1)(0) = 0 for all k suchthat 2 ≤ k ≤ n. So, by Rolle’s Theorem, there exists an x1 between 0 and xsuch that g′(x1) = 0. Then, again by Rolle’s Theorem, because g′(0) = 0 andg′(x1) = 0, there exists an x2 between 0 and x1 such that g′′(x2) = 0. Continuing,we get that there exists an xn−1 between 0 and xn−2 such that g(n−1)(xn−1) = 0.Finally, there exists a y between 0 and xn−1 such that g(n)(y) = 0. However,g(n)(y) = M − f (n)(y), so M = f (n)(y) for some y between 0 and x, proving thetheorem. �

∗Technically, the Maclaurin series†This means that if 0 < x, then 0 < y < x, and if x < 0, then x < y < 0.

48


20. 03/31/11: Differentiation of Power Series

20.1. Consequences of Taylor’s Theorem.

Recall that Taylor’s Theorem states that for any n-times differentiable functionf(x), there exists a y between 0 and x such that:

f(x) = f(0) +f ′(0)x

1!+ · · ·+ f (n−1)(0)xn−1

(n− 1)!+f (n)(y)xn

n!

A simple corollary of this theorem follows.

Corollary. If there exists a bound on the set {f (n)(y) | y varies as n varies}, thenthe Taylor series of f converges to f .

To see this, notice that if there is a bound on the set, as n increases, xn

n! convergesto 0.

Taylor’s Theorem is very powerful, as it is what allows us to compute exactvalues of differentiable functions!

20.2. Term-by-Term Differentiation of Power Series.

Theorem. Let∑anx

n be a power series with radius of convergence R. Definef(x) =

∑anx

n on (−R,R). Then, the following are true:

(1)∑∞n=0 nanx

n−1 converges.(2) f is differentiable on (−R,R).(3) f ′(x) =

∑∞n=0 nanx

n−1 on (−R,R).

Before we get to the bulk of the Theorem, we will make a few easy observations.

Recall that R−1 = lim |an|1n for any power series. So, for the series in part (1)

above, we have R−11 = lim(n|an|)1n . However, we can show that limn→∞ n

1n = 1,

so by the limit theorems, R = R1, so if∑anx

n converges, then∑nanx

n−1 alsoconverges.

Also, the following identity is true for any a, b ∈ R such that a− b 6= 0:

(3)an − bn

a− b= an−1 + an−2b+ · · ·+ abn−2 + bn−1

Notice that the right hand side is an expression with n terms.The following is the infinite analogue to the Triangle Inequality:

(4)

∣∣∣∣∣∞∑n=0

cn

∣∣∣∣∣ ≤∞∑n=0

|cn|

Finally, recall the very simple identity

(5) 1 + 2 + · · ·+ n =n(n− 1)

2Now, we are ready to start.

Proof. We have already proved (1) in the first paragraph above, and proving (2)and (3) are basically equivalent.

In order to prove (2) and (3), it suffices to show that

(6) limy→x

∣∣∣∣∣∞∑n=0

(an(yn − xn)

y − x− nanxn−1

)∣∣∣∣∣ = 0

49


If R is finite, then choose an r such that 0 < r < R, |x| < r < R, and |y| < r < R.If R is infinite, then this step is not necessary, and we can simply pick any r.

Now, by (4), we have that (6) is equal to

limy→x

∣∣∣∣∣∞∑n=0

an(yn−1 + yn−2x+ yn−3x2 + · · ·+ yxn−2 + xn−1 − nxn−1)

∣∣∣∣∣Because (3) is n terms long, we can split up the last nxn−1 and distribute them toevery other term to obtain:

limy→x

∣∣∣∣ ∞∑n=0

an((yn−1 − xn−1) + x(yn−2 − xn−2)

+ x2(yn−3 − xn−3) + · · ·+ xn−2(y − x))

∣∣∣∣Then, if we factor out a y−x from each term and apply (3) to each one individually,we obtain

limy→x|y − x|

∣∣∣∣ ∞∑n=0

an((yn−2 + yn−3x+ · · ·+ xn−2)

+ x(yn−3 + yn−4x+ · · ·+ xn−3) + · · ·+ xn−2)

∣∣∣∣Notice that every term in the above expression is less than rn−2 because |x| < r < Rand |y| < r < R and there are n − 1 terms in the first parenthesized expression,n − 2 in the second, and so on until there is only 1 term in the last expression.Therefore, by (5), the whole expression above is less than or equal to:

limy→x|y − x|

∣∣∣∣∣∞∑n=0

|an|n(n− 1)

2rn−2

∣∣∣∣∣Now, remember that the radius of convergence of

∑∞n=0 bnx

n is equal to B−1, where

B = n√|bn|. It can be shown that n

√n(n−1)

2 converges to 1, so the coefficients of

this power series converge simply to |an|. Therefore, because r < R, the wholesecond term is a convergent power series, so the entire term can converges as y goesto x, completing the proof. �

20.3. Elementary Functions.

Now, we can define all of these nice functions we’ve previously seen, and provetheir familiar properties.

• exp(x) =

∞∑n=0

xn

n!

• cos(x) =

∞∑n=0

(−1)n x2n

(2n)!

• sin(x) =

∞∑n=0

(−1)n x2n+1

(2n+ 1)!

50


20.3.1. The exponential function is never negative. Suppose there exists an a suchthat exp(a) = 0. We can assume a < 0 because if a > 0, then we can reach acontradiction immediately through direct calculation. Consider the real number A,defined to be sup{a | exp(a) = 0}. Clearly, exp(A) = 0. Consider exp(A + x) forsome x ∈ R. It is easy to see by direct calculation that the derivative of exp(x) isexp(x). Therefore, by the Chain Rule, the derivative of exp(A+ x) = exp(A+ x).

Now, consider the Taylor series expansion of exp(A+ x), or

F (x) = exp(A+ x) = F (0) + F ′(0)x+ · · ·+ F (n)(y)xn

n!

for some y between 0 and x. Notice that exp(A+x) is increasing by direct calcula-tion, and every derivative is exactly exp(A+ x). Therefore, f (n)(y) = exp(A+ x),so we have a bound on y, and the Taylor series converges to F . However, it isthe case that F (0) = exp(A + 0) = exp(A) = 0, so plugging that in to the Tay-lor series, we get that exp(A + x) = 0 for all x. This is clearly false, becauseexp(A+ (−A)) = exp(0) = 1. Thus, we have a contradiction, and exp(x) 6= 0.

20.3.2. Adding real numbers in the exp function. Consider the following equation.

d

dx

(exp(x+ a)

exp(x)

)=

exp(x+ a) exp(x)− exp(x+ a) exp(x)

(exp(x))2= 0

By a corollary to the Mean Value Theorem, this implies that exp(x+a)exp(x) is constant.

Therefore, we have exp(x + a) = K exp(x) for all x ∈ R and some K ∈ R. Letx = 0. Then, exp(a) = K, so exp(x+ a) = exp(a) exp(x), a familiar identity.

20.3.3. The limit as exp approaches zero. exp(R) is connected because exp is acontinuous function, and the range does not contain 0, so the function is alwayspositive.

Now, choose ε > 0 such that exp(−ε) < 1. Then, it is the case that, for anyn ∈ N,

exp(−nε) = (exp(−ε))n

which is arbitrarily small as n goes to ∞. Therefore, limx→−∞ exp(x) = 0.

20.3.4. exp is a bijection from R to (0,∞). This follows immediately from theobservation that exp is strictly increasing.

20.3.5. The log function. We can define log(x) as the inverse function of exp(x),where log has domain (0,∞). Therefore, log exp(x) = x, exp log(x) = x, andlog(xy) = log(x) + log(y). Also, it is the case that the derivative of log is 1

x by theChain Rule on log exp(x).

20.3.6. Arbitrary exponential functions and e. We will define ar = exp(r log(a))for a, r ∈ R such that a > 0. Also, we will define the real number e = exp(1).Therefore, we can see that ex = exp(x).

20.3.7. The Pythagorean trigonometric identity. Consider the following equation.

d

dx(cos2(x) + sin2(x)) = −2 cos(x) sin(x) + 2 sin(x) cos(x) = 0

Thus, as before, cos2(x) + sin2(x) is a constant function. Plugging in 0, we get thatcos2(x) + sin2(x) = 1.

51


20.3.8. Pi. Because cos(1) > 0 and cos(2) < 0 by the error of the alternating seriesfunction, it is the case that there exists an x between 1 and 2 such that cos(x) = 0.We will define π

2 = inf{x |x > 0, cos(x) = 0}.

There are many more properties that we can prove at this stage, but they areleft as exercises.

20.4. Some Extra Tidbits. ∗

This section includes some counterexamples and odd functions related to differen-tiability, as well as another theorem.

(1) limx→0 sin 1x does not exist.

(2) The function

f(x) =

{x sin 1

x , if x 6= 0

0, if x = 0

is continuous at 0. However, it is not differentiable at 0, as

limδ→0

f(δ)− f(0)

δ= limδ→0

δ sin 1δ

δ= limδ→0

sin1

δ

which does not exist.(3) The function

g(x) =

{x2 sin 1

x2 , if x 6= 0

0, if x = 0

is everywhere differentiable, as if x = 0, then limδ→0δ2 sin 1

δ

δ = 0. However,

consider its derivative. If x 6= 0, then dgdx = 2x sin 1

x − cos 1x , and if x = 0,

then dgdx = 0, from the above calculation. Therefore, g′(x) is not continuous

at 0.

Theorem. Let f be differentiable on (a, b), where a < x1 < x2 < b, and let c ∈ R.If f ′(x1) < c < f ′(x2), then there exists a y such that x1 < y < x2 and f ′(y) = c.

Proof. Consider the function g(x) = f(x)− cx. Clearly, g is differentiable on (a, b),and its derivative is f ′(x)− c. [x1, x2] is compact, so the restriction of g to [x1, x2]attains its minimum and maximum. Therefore, for some y ∈ [x1, x2], g′(y) = 0, orf ′(y) − c = 0. If y was equal to either x1 or x2, then f ′(x1) = c or f ′(x2) = c,which contradict our assumptions. Thus, y must not equal either of those points,and we are done. �

∗This section was originally included in lecture 24 by Professor Jones. However, I have moved

it here so that the flow of the notes is not interrupted.

52


21. 04/07/2011: Uniform Convergence: Guest Lecture by ScottMorrison

21.1. Introduction.

The next two lectures will be about approximating continuous functions. We willtry to answer two questions:

(1) When fn, a sequence of continuous functions, converges to another functionf , how can we tell if f is continuous?

(2) When can we approximate a continuous function by so-called “nice func-tions”? For instance, can we find a sequence of polynomials (fn) such thatfn → f?

We will answer the first question in this lecture, and the second question in thenext.

21.2. The Supremum Norm.

We will start with a non-example.

Non-Example: Define fn : [0, 1]→ [0, 1] by

fn(x) =

{nx, if x ∈ [0, 1

n ]

1, if x ∈ [ 1n , 1]

We can see that fn(x) converges pointwise to the function

f(x) =

{0, if x = 0

1, if x ∈ (0, 1]

which is clearly not continuous. So what went wrong?The answer is that as x approaches 0, fn(x) takes longer and longer to converge

to f(x).To see this more clearly, take ε = 1

2 . What N ensures that |fn(x) − f(x)| < εfor all n ≥ N? Well...

|fn(x)− f(x)| < 1

2⇒ 1

n< 2x

So, we can take N = d 12xe, but that’s the best we can do. As x→ 0, N →∞.

To avoid this problem, we will define uniform convergence. However, first, let’sdefine the supremum norm.

Definition. Let f : X → R be a function. We will define a norm on the set of allfunctions from X to R, namely

||f ||∞ = supx∈X|f(x)|

This notion can be extended to a metric on the functions from X to R, namely

d(f, g) = ||f − g||∞ = supx∈X|f(x)− g(x)|

Of course, we can generalize this even further by simply replacing the R with anymetric space C, and the familiar metric on R with a metric on C.

For functions from R to R, d(f, g) < ε means that g never gets more than ε awayfrom f .

53


Definition. A sequence of functions fn converges uniformly if it converges withrespect to the metric defined above, i.e.

(∀ε > 0)(∃N ∈ N) s.t. (∀n ≥ N), d(fn, f) < ε

Notice that the statement

(∀ε > 0)(∃N ∈ N) s.t. (∀n ≥ N) and (∀x ∈ X), D(fn(x), f(x)) < ε

is equivalent to the definition above, where D is the metric on the codomain of fand fn. However, if we find such an ε for the latter statement and apply it to theformer, we must change the condition so that d(fn, f) ≤ ε, because the d in thefirst statement takes the supremum of values of fn(x) and f(x).

So what went wrong in our first example? We had an N for each x, but we couldnot find an N which satisfied the condition for all x.

Theorem. If (fn) is a sequence of continuous functions converging uniformly to afunction f , then f is continuous.

Proof. Let x0 ∈ X and ε > 0 be given. We want to find some δ > 0 such that for allx with D1(x, x0) < δ, D2(f(x), f(x0)) < ε, where D1 is the metric on the domainof fn and D2 is the metric on the range of fn.

Notice that

D2(f(x), f(x0)) ≤ D2(f(x), fn(x)) +D2(fn(x), fn(x0)) +D2(fn(x0), f(x0))

for all n. Because fn converges to f uniformly, we can find an N such thatD2(fN (y), f(y)) < ε

3 for all y. Also, because fN is continuous, we can choose aδ > 0 such that D2(fN (x), fN (x0)) < ε

3 as long as D1(x, x0) < δ. Using this δ, wehave:

D2(f(x), f(x0)) <ε

3+ε

3+ε

3= ε


Exercise: Can you find (fn) which are uniformly continuous and converge uniformlyto f , where f is not uniformly continuous? If not, can you improve this proof toshow that if every fn is continuous, then f must be uniformly continuous?

21.3. An Explicit Construction.

Let’s now use these concepts to construct a continuous surjective function fromthe Cantor set to [0, 1]. Note that by varying this construction, you can actuallymap the Cantor set to any compact metric space, like [0, 1]× [0, 1].

The basic idea will be to construct a sequence of functions fn : [0, 1] → [0, 1]which are “approximately surjective” when restricted to C, the Cantor set, andshow that they converge uniformly to some f which must be continuous. We willthen use this continuity to show that f is actually surjective when restricted to C.

Recall that C = M0 ∩M1 ∩M2 ∩ . . . where M0 = [0, 1], M1 = [0, 13 ]∪ [ 23 , 1], andso on, such that

Mk =1

3Mk−1 ∪

(2

3+

1

3Mk−1

)54


We will define fn such that fn|Mn : Mn → [0, 1] is surjective, or in other words,such that fn when restricted to Mn is surjective for each n. Let f0 : [0, 1] → [0, 1]be any continuous function with f0(0) = 0 and f0(1) = 1. Then, define

f1(x) =

12f0(3x), if x ∈ [0, 13 ]12 , if x ∈ [ 13 ,

23 ]

12 + 1

2f0(3x− 2), if x ∈ [ 23 , 1]

f1|M1 is surjective, as f1|[0, 13 ] = 12f0 is surjective onto [0, 12 ] and f1|[ 23 ,1] = 1

2 + 12f0

is surjective on [ 12 , 1]. Therefore, extending the construction, we get, for any k > 0:

fk(x) =

12fk−1(3x), if x ∈ [0, 13 ]12 , if x ∈ [ 13 ,

23 ]

12 + 1

2fk−1(3x− 2), if x ∈ [ 23 , 1]

Now, we must check that (fn) converges pointwise. To do this, we will need twoPropositions.

Proposition. |fn(x)− fn−1(x)| ≤ 2−n+1 for all x ∈ [0, 1].

Proof. This is clearly true for n = 1, as f1(x)− f0(x)| ≤ 1. Now, we will induct onn. Then, we have

|fn(x)− fn−1(x)| =

12 |fn−1(3x)− fn−2(3x)|, if x ∈ [0, 13 ]

0, if x ∈ [ 13 ,23 ]

12 |fn−1(3x− 2)− fn−2(3x− 2)|, if x ∈ [ 23 , 1]

Then, |fn(x)− fn−1(x)| ≤ 12 · 2

−n+2 = 2−n+1, so we are done. �

From this Proposition, we can see that (fn) is Cauchy for each x, so there exists apointwise limit, which we will call f .

Proposition. |fn(x)− f(x)| ≤ 2−n+2 for all x ∈ [0, 1].

Proof. First, we can see that, for n,m ∈ N and n < m:

|fn(x)− fm(x)| ≤ |fn(x)− fn+1(x)|+ |fn+1(x)− fn+2(x)|+ · · ·+ |fm−1(x) + fm(x)| ≤ 2−n+2

by the previous Proposition. Therefore, by the geometric series formula, we have

|fn(x)− f(x)| = limm→∞

|fn(x)− fm(x)| ≤ 2−n+2

which completes the proof. �

Therefore, because 2−n+2 can get arbitrarily small as n increases, (fn) convergesuniformly to f , and f is continuous by the previous Theorem. Because f is acontinuous function from [0, 1] to [0, 1] and clearly f(0) = 0 and f(1) = 1, by theMean Value Theorem, f is surjective on [0, 1].

So, all that is left is to show that the restriction of f to C is surjective. Takex ∈ [0, 1] and pick a y ∈ [0, 1] so that f(y) = x. If y ∈ C, then we’re done.Otherwise, y ∈ [0, 1] \Mk for some k. However, [0, 1] \Mk is simply a collection ofopen intervals. Therefore, we will make a claim.

Proposition. For all n ≥ k, fn is constant and independent of n on the closureof each interval of [0, 1] \Mk.

55


Proof. f1 is constant on [0, 1] \M1 = [ 13 ,23 ]. If fn−1 is constant on [a, b], then fn is

constant on [ 13a,13b] and [ 23 + 1

3a,23 + 1

3b], because Mk = 13Mk−1 ∪ ( 2

3 + 13Mk−1), so

[0, 1] \Mk =1

3([0, 1] \Mk−1) ∪

(1

3,

2

3

)∪(

2

3+

1

3([0, 1] \Mk−1)

)�

Therefore, by this proposition, we can pick y′ such that y′ is an endpoint of theinterval which y belongs to, so f(y′) = x. However, the endpoint of any suchinterval is in the Cantor set, so f is surjective on C, and we are done.

56


22. 04/12/2011: Weierstrass Approximation: Guest Lecture by ScottMorrison

22.1. Banach Spaces.

Recall the supremum norm on the functions from [0, 1] to R, namely

||f || = supx∈[0,1]

|f(x)|

and the associated metric defined by

d(f, g) = ||f − g|| = supx∈[0,1]

|f(x)− g(x)|

as well as the generalization of this norm into C(X), the set of continuous functionson X:

d(f, g) = supx∈X

d(f(x), g(x))

Definition. A vector space with a norm such that the vector space is complete iscalled a Banach space.

Theorem. C([0, 1]), the set of continuous functions from [0, 1] to R, is a Banachspace; in particular, the supremum norm makes this space complete.

Proof. Suppose (fn) is a Cauchy sequence of continuous functions. Certainly, fnhas a pointwise limit f because of the completeness of R. In other words, we knowthat for all ε > 0, there exists an N ∈ N such that n,m ≥ N implies

supx∈[0,1]

|fn(x)− fm(x)| < ε

Then, for any given x, |fn(x)− fm(x)| < ε, so (fn(x)), a sequence in R, is Cauchyand it converges to some number which we will call f(x). So, let’s check that (fn)actually converges to this f uniformly. Because this sequence is Cauchy, for anyε > 0, we can choose N such that ||fn − fm|| < ε

2 for all n,m ≥ N . This meansthat |fn(x) − fm(x)| < ε

2 for all x ∈ [0, 1]. Taking the limit as m goes to infinity,namely limm→∞(fm(x)), we get |fn(x)− f(x)| ≤ ε

2 < ε for all x ∈ [0, 1]. Therefore,||fn− f || < ε, so (fn) converges uniformly to f . By the theorem in the last lecture,this f is continuous, so f ∈ C([0, 1]), and we are done. �

22.2. Subalgebras.

Definition. A ⊆ C([0, 1]) is a subalgebra of C([0, 1]) if it is closed under addition,multiplication, and scalar multiplication.

Example: The polynomials are a subalgebra of C([0, 1]).

Example: Linear combinations of the trigonometric functions, namely

{cos(πnx) | 0 ≤ n <∞} ∪ {sin(πnx) | 0 ≤ n <∞}are also a subalgebra of C([0, 1]).

The main exploratory question for this lecture is “When is a subalgebra ofC([0, 1]) dense?” There is a Theorem that tells us exactly that, as well as aneasier corollary that can be proved independently. We will start, however, withsome definitions.

57


Definition. If X is a metric space, Y ⊆ X is dense in X if for all x ∈ X, thereexists a sequence (yn) ⊆ Y such that (yn) converges to x. Equivalently, Y ⊆ X isdense in X if the union of Y and the set of all of its limit points is all of X. GivenY ⊆ X, the closure of Y , denoted Y , is the union of Y and the set of all of itslimit points in X.

Definition. Let X be a metric space. A subalgebra A ⊆ C(X) separates pointsif for all distinct x, y ∈ X, there exists a function f ∈ A such that f(x) 6= f(y).

The following two theorems are our main results.

Theorem (Weierstrass Approximation). The polynomials are dense in C([0, 1]),i.e. every continuous function on [0, 1] is the uniform limit of a sequence of poly-nomials.

Theorem (Stone-Weierstrass). If X is a compact metric space and A ⊆ C(X)separates points and contains the function 1 where 1(x) = 1 for all x, then A isdense in C(X).

Notice that the Stone-Weierstrass Theorem implies the Weierstrass Approxima-tion Theorem, because [0, 1] is compact and the function f(x) = x separates points.

Proposition. If A ⊆ C(X) is a subalgebra which does not separate points, then Ais not dense in C(X).

Proof. It is certainly the case that C(X) separates points, as given distinct x, y ∈ X,we can take f(z) = d(x, z) where d is any metric on X. Then, f is continuous, andf(x) = 0 while f(y) > 0.

Assume toward contradiction that A ⊂ C(X) does not separate points. Then,in particular, for some x, y ∈ X, g(x) = g(y) for all g ∈ A. We showed above thatthere exists a function f ∈ C(X) such that d(f(x), f(y)) = L > 0. We will showthat this f is not a uniform limit of functions in A if (fn) is a sequence in A.

||fn − f || = supz∈X|fn(z)− f(z)| ≥ max{|fn(x)− f(x)|, |fn(y)− f(y)|(7)

≥ 1

2(|fn(x)− f(x)|+ |fn(y)− f(y)|)(8)

≥ 1

2|f(x) + f(y)|(9)

≥ 1

2|f(x)− f(y)| = L

2(10)

In the above, (9) follows from the equality of fn(x) and fn(y) as well as the triangleinequality, and (10) follows from the positive definiteness of f = d. Therefore, it isclear that (fn) cannot converge to f , and A is not dense in C(X). �

Therefore, separating points is a necessary and sufficient condition on A in orderfor it to be dense in C(X). Now, we will begin the task of proving the WeierstrassApproximation Theorem.

22.3. Proof of the Weierstrass Approximation Theorem.58


This proof will consist of four propositions.

Proposition. f(x) = 1−√

1− x can be uniformly approximated by polynomials.

Proof. Define a sequence of polynomials as such: let P0(x) = 0 and define Pn+1

recursively as Pn+1(x) = 12 (Pn(x)2 + x) for all n ∈ N. These are all clearly polyno-

mials. Also, observe that 0 ≤ Pn(x) ≤ 1 for x ∈ [0, 1]. To see this, we can simplyproceed by induction on n. Also, it is the case that

Pn+2(x)− Pn+1(x) =1

2(Pn+1(x) + Pn(x))(Pn+1(x)− Pn(x))

From this, we can see that Pn+1(x) ≥ Pn(x), which again can be proved by in-duction. From these observations, it is clear that (Pn(x)) is bounded above andmonotonic, and therefore converges to some number in R for each x, which we willdenote as g(x).

Consider that

limn→∞

(Pn+1(x)− 1

2(Pn(x)2 + x)) = 0

by construction. Therefore, g(x)− 12 (g(x)2 + x) = 0, and solving this equation for

g(x) allows us to see that g(x) = f(x) = 1−√

1− x for all x ∈ [0, 1].Therefore, we now have a sequence of polynomials converging pointwise to f(x),

so it remains to show that it, in fact, converges uniformly.As mentioned above, (Pn(x)) is a nondecreasing sequence in n. However, it can

also be seen that (Pn(x)) is a nondecreasing sequence in x. Therefore, becausethe function x is also nondecreasing, 1

2 (Pn(x)2 + x) = Pn+1(x) is nondecreasing.Finally, from this, we can see that Pn+1(x) − Pn(x) is also nondecreasing in x, afact that can be proved by induction (again). Thus, we have that

Pn+1(x)− Pn(x) ≤ Pn+1(1) + Pn(1)

and because this is a telescoping series, this implies that

Pm(x)− Pn(x) ≤ Pm(1)− Pn(1)

for m > n. Now, taking the limit as m goes to infinity, we get

f(x)− Pn(x) ≤ f(1)− Pn(1)

which is equivalent to the statement

||f − Pn|| ≤ f(1)− Pn(1)

because Pn+1(x) − Pn(x) is nondecreasing in x. Because (Pn(1)) converges tof(1), we can make the right hand side arbitrarily small. Therefore, Pn convergesuniformly in the supremum norm to f . �

Proposition. The function f(x) = |x| can be uniformly approximated by polyno-mials.

Proof. Notice that

ε = x2 + ε− |x|2 = (√x2 + ε− |x|)(

√x2 + ε+ |x|)

Then, this implies that√x2 + ε− |x| = ε√

x2 + ε+ |x|≤ ε√

ε= ε

12

59


Now, if 1 −√

1− x can be uniformly approximated by polynomials, then√

1− x,√x, and

√x2 + ε can also be uniformly approximated by polynomials by simple

manipulations. So, for each n ∈ N, choose polynomials (fn,m) such that (fn,m)

converges to√x2 + 1

n as m goes to ∞.∗

Define gn = fn,N where N is big enough so that ||fn,N −√x2 + 1

n || <1n . We

will claim that gn converges uniformly to |x|, as:

||gn − |x||| ≤ ||gn −√x2 +

1

n||+ ||

√x2 +

1

n− |x||| ≤ 1

n+

1√n

where the last inequality follows from the assumption above as well as the very firstseries of inequalities. Therefore, because 1

n + 1√n

obviously goes to 0 as n goes to

infinity, gn is a sequence of polynomials that converges uniformly to |x|. �

Proposition. Every piecewise linear function is a uniform limit of polynomials.

Proof Sketch. You can write any piecewise linear function in the form

f(x) = c+

k∑n=1

an|x− bn|

where c ∈ R, k ∈ N, an, bn ∈ R. Then, we can apply the previous Proposition toobtain the desired result. �

Proposition. Every continuous function is a uniform limit of piecewise linearfunctions.

Proof Sketch. Every continuous function on [0, 1] is uniformly continuous, because[0, 1] is compact. Thus, choose N ∈ N such that |f(x) − f(y)| < ε if |x − y| ≤ 1

N .Now, define a function g to be the piecewise linear function which agrees with fat the points k

N where 0 ≤ k ≤ N , and is linear in between. Then, as N goes toinfinity, g converges to f . �

This proposition proves the Theorem.

∗Note that what we’re proving here is that if (xn) is a sequence of limit points of X thatconverges, then the limit point of that sequence is also a limit point of X. It may actually bebetter to prove this fact independently.

60


23. 04/14/2011: The Stone-Weierstrass Theorem

23.1. Review.

Theorem (Stone-Weierstrass). Let (X, d) be a compact metric space and A be asubalgebra of C(X,R). If

(1) A contains the constant function 1, where 1(x) = 1 for all x ∈ X, and(2) A separates the points of X, namely for all x 6= y in X, there exists a

function f ∈ A such that f(x) 6= f(y),

then A = C(X,R), i.e. A is dense in C(X,R).

A review of the relevant concepts follows.

Remember that for any continuous functions f, g : X → R, we define:

• f + g = (f + g)(x) = f(x) + g(x)• fg = (fg)(x) = f(x)g(x)• λf = (λf)(x) = λ(f(x))

An algebra has various properties with the above operations. A subalgebra is asubset of this algebra that is closed under these operations.

Example: Let X = [0, 1]. Then, {f | f(0) = f(1), f continuous} is a subalgebra ofthe continuous functions from [0, 1] to R, denoted C([0, 1],R).

Example: Likewise,k∑

n=0

anxn

where x ∈ [0, 1] and an ∈ R is a subalgebra of C([0, 1],R).

Also, remember that on C(X,R), there is a norm called the supremum normdefined by

||f || = sup{|f(x)| |x ∈ X}which makes it a complete metric space. A complete normed metric space is calleda Banach space.

23.2. Proof of the Stone-Weierstrass Theorem.

Remember from last lecture that in C([0, 1])∗,√x is a uniform limit of polynomials.

From this, we can get the following result.

Proposition. If f ∈ A, where A is a subalgebra of C([0, 1]), then |f | ∈ A.

Proof. Notice that |f | =√f2, and that because A is a subalgebra of C([0, 1]) (and

hence closed under multiplication), f2 is in A. Therefore, it suffices to show thatgiven any function g ∈ A such that g(x) ≥ 0 for all x ∈ X,

√g is in A.

Rephrasing the last statement, we want to show that for all ε > 0, there existsan h ∈ A such that |h(x) −

√g(x)| < ε for all x ∈ X. Without loss of generality,

because A is closed under scalar multiplication, we can assume that ||g|| ≤ 1.Now, from the statement proved last time, namely that

√x is a uniform limit of

∗This is the same as C([0, 1],R).

61


polynomials, there exists a polynomial p on [0, 1] such that |p(r) −√r| < ε for all

r ∈ [0, 1]. Therefore, we can simply replace r with g(x) in order to obtain

|p(g(x))−√g(x)| < ε

for all x ∈ X. However, p(f(x)) = (p(f))(x), so we can simply set h to be p(f),completing the proof. �

Corollary. If f and g are functions in A, then Mf,g(x) and mf,g(x) are in A,where

Mf,g(x) := max{f(x), g(x)}mf,g(x) := min{f(x), g(x)}

Proof. This is immediate from the previous proposition, as

Mf,g(x) =1

2(f(x) + g(x) + |f(x)− g(x)|)

mf,g(x) =1

2(f(x) + g(x)− |f(x)− g(x)|)

and A is closed under addition. �

Notice that if A is a subalgebra of C(X,R), then A is also a subalgebra of C(X,R).Also, the above Corollary can be generalized to any number of finite functions byinduction.

Proposition. Suppose A is a subalgebra of C(X,R) that separates points and con-tains the constant functions. Then, for every f ∈ C(X,R), x ∈ X, and ε > 0, thereexists a function gx ∈ A such that gx(z) < f(z) + ε for all z ∈ X and gx(x) > f(x).

Proof. By hypothesis, because A separates points, for every y 6= x, there is somefunction hx,y such that hx,y(x) 6= hx,y(y). Now, because the constant function isin A and it is a subalgebra, we can choose constants a and b such that hx,y(z) =

ah(z) + b and the following equations hold:

hx,y(x) = f(x) +ε

2

hx,y(y) = f(y)− ε

2

Now, for every y 6= x, define the set

Ux,y = {z ∈ X |hx,y(z) < f(z) + ε}

The continuity of hx,y and f make Ux,y an open set, as given any z ∈ Ux,y, oneonly needs to adjust z by some δ in order to ensure that z + δ is also in Ux,y. Byconstruction of hx,y, x and y are both in Ux,y. Therefore, the set of all Ux,y wherey varies in X \ {x}, or {Ux,y | y ∈ X \ {x}}, is an open cover of X. Because X iscompact, there must be a finite subcover {y1, y2, . . . , yn}.

Define the function gx = min{hx,y1 , hx,y2 , . . . , hx,yn}. By the above corollary,

gx ∈ A. Now, we have that gx(x) = f(x) + ε2 which is greater than f(x). Also,

given any z ∈ X, it must belong to some Ux,yi , so gx(z) ≤ hx,yi(z) < f(z) + ε,completing the proof. �

62


Proof of Stone-Weierstrass. For this proof, we will show that for every f ∈ C(X,R)and ε > 0, there exists a function g ∈ A such that f(z) < g(z) < f(z) + ε. Thisis sufficient because this condition is equivalent to A being dense in C(X,R); moreinformally, we are showing that for any function in C(X,R), there is a function inA that is arbitrarily close to it.

First, for every x ∈ X, define the set

Vx = {z ∈ X | gx(z) > f(z)}where gx is defined as in the previous proof. Again, because gx and f are bothcontinuous, Vx is an open set. Furthermore, gx(x) = f(x)+ ε

2 which is greater thanf(x), so x ∈ Vx. Therefore, {Vx |x ∈ X} is an open cover of X. Now, because X iscompact, we can extract a finite subcover {x1, x2, . . . , xn}.

Define the function g = max{gx1, gx2

, . . . , gxn}. From the corollary above, weknow that g ∈ A. Now, given any z ∈ X, it is clear from the construction of eachgx that g(z) < f(z) + ε. Furthermore, every z is in some Vxi , so g(z) > f(z).Therefore, we have that f(z) < g(z) < f(z) + ε, and we are done! �

63


24. 04/19/2011: Integration Part One

24.1. A Review of Stone-Weierstrass.

The theorem states that given C(X,R), the supremum norm, and a subalgebraA (namely 1 ∈ A and A separates the points of X), A is dense in C(X,R), orA = C(X,R).Example: The polynomials are dense in C([a, b],R), as the function x separatespoints. Likewise, the polynomials in two variables are dense in C([a, b]× [c, d],R),because given (p1, q1) and (p2, q2), the function f(x, y) = (x − p1)2 + (y − p2)2

separates the two points. It is easy to see that this can be extrapolated to Rn.

24.2. Darboux Integration.

Definition. Let [a, b] be an interval in R. A partition of [a, b] is a subset of [a, b],{t0, t1, . . . , tn}, such that t1 = a, tn = b, and for all i where 0 ≤ i ≤ n−1, ti < ti+1.

Definition. Let f : [a, b] → R be a bounded function. Given S ⊆ [a, b], we willdefine the following values:

M(f, S) = sups∈S

f(s)

m(f, S) = infs∈S

f(s)

Definition. Let P be a partition. The upper Darboux sum is defined as

U(f, P ) =

n∑i=1

M(f, [ti−1, ti])(ti − ti−1)

The lower Darboux sum is defined as

L(f, P ) =

n∑i=1

m(f, [ti−1, ti])(ti − ti−1)

Clearly, it is the case that

m(f, [a, b])(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤M(f, [a, b])(b− a)

for any P . This motivates our next definition.

Definition.

U(f) = infPU(f, P )

L(f) = supPL(f, P )

In essence, U(f) is the infimum of U(f, P ) as you vary the partition you are taking.Now, it is clear what our definition of the integral will be.

Definition. f : [a, b]→ R is Darboux integrable if L(f) = U(f), and we denote

the integral as∫ baf(x)dx = U(f).

64


Example: Consider a function f : [0, 1]→ R such that f(x) = 0 if x is irrational andf(x) = 1 if x is rational. Then, clearly M(f, [ti−1, ti]) = 1 and m(f, [ti−1, ti]) = 0for any partition P . Therefore, we can compute the following:

U(f, P ) =

n∑i=1

1(ti − ti−1) = 1

L(f, P ) = 0

This shows that f is not Darboux integrable.

Example: Now, consider the more familiar function f(x) = x on [0, b]. For easeof working with the partitions, we will assume that our partition divides the in-terval into segments of equal width, or in other words, let ti = ib

n . Then, clearly

M(f, [ti−1, ti]) = ibn and m(f, [ti−1, ti]) = (i−1)b

n . Therefore,

U(f, P ) =

n∑i=1

ib

n· bn

=b2

n2· n(n+ 1)

2

L(f, P ) =

n∑i=1

(i− 1)b

n· bn

=b2

n2· n(n− 1)

2

Both b2n(n+1)2n2 and b2n(n−1)

2n2 tend to b2

2 , so inf U(f, P ) ≤ b2

2 and supL(f, P ) ≥ b2

2 .However, we cannot conclude that the function is Darboux integrable until we knowfor sure that L(f) is always less than or equal to U(f). In fact, this is our nexttheorem.

Theorem. Let f : [a, b]→ R be bounded. L(f) ≤ U(f).

Proof. Let P and Q be partitions of [a, b] such that P ⊆ Q. First, we will showthat

(11) L(f, P ) ≤ L(f,Q) ≤ U(f,Q) ≤ U(f, P )

We will assume that |Q| = |P | + 1, as the full argument can be shown throughinduction. Therefore, we have

P = {t0, t1, . . . , ti−1, ti, . . . , tn}Q = {t0, t1, . . . , ti−1, u, ti, . . . , tn}

The middle inequality is trivial, and the first and third inequalities can be provedsimilarly, so we will only prove the first inequality, namely L(f, P ) ≤ L(f,Q).Notice that

L(f,Q)− L(f, P ) = m(f, [ti−1, u])(u− ti) +m(f, [u, ti])(ti − u)

−m(f, [ti−1, ti])(ti − ti−1)

≥ m(f, [ti−1, ti])(u− ti−1) +m(f, [ti−1, ti])(ti − u)

−m(f, [ti−1, ti])(ti − ti−1) = 0

Therefore, for any P,Q such that P ⊆ Q, inequality (1) holds. Now, consider P ∪Q,where P and Q are arbitrary partitions. Applying (1) twice, we obtain

L(f, P ) ≤ L(f, P ∪Q) ≤ U(f, P ∪Q) ≤ U(f,Q)

Thus, we now know that L(f, P ) ≤ U(f,Q) for any P and Q. If we fix any partition65


P , we know that L(f, P ) ≤ U(f), because L(f, P ) is less than or equal to U(f,Q)for any Q. However, this also implies that U(f) is greater than or equal to L(f, P )for any P , because our fixed P was arbitrary. Therefore, L(f) ≤ U(f), and we aredone. �

66


25. 04/21/11: Integration Part Two

25.1. More Darboux Integration.

Recall that last time, we discussed the Darboux integral. Given a partition P anda bounded function f on [a, b], we have:

U(f, P ) =∑

M(f, [ti−1, ti])(ti − ti−1)

L(f, P ) =∑

m(f, [ti−1, ti])(ti − ti−1)

U(f) = infPU(f, P )

L(f) = supPL(f, P )∫ b

a

f(x)dx = U(f) = L(f)

where the final equation only makes sense when U(f) = L(f).Also, last time, we proved that L(f, P ) ≤ U(f, P ) and L(f) ≤ U(f), where the

crucial lemma was that if we have two partitions P and Q where P ⊆ Q, then

L(f, P ) ≤ L(f,Q) ≤ U(f,Q) ≤ U(f, P )

Theorem. Let f be bounded on [a, b]. f is Darboux integrable if and only if for allε > 0, there exists a partition P such that L(f, P ) > U(f, P )− ε.


(⇐) Due to the theorem in the previous lecture, in order to show that f isDarboux integrable, we only need to show that L(f) ≥ U(f). Let ε > 0be given, and let P be a partition such that the inequality in the theoremholds. Then, notice that

L(f) ≥ L(f, P ) > U(f, P )− ε ≥ U(f)− εBecause ε was arbitrary, L(f) ≥ U(f), so f is Darboux integrable.

(⇒) Let ε > 0 be given. Then, because f is Darboux integrable, there existsa partition P such that U(f, P ) < U(f) + ε

2 and a partition Q such thatL(f,Q) > L(f)− ε

2 . To see this, assume that this was not the case. Then,there would be some ε

2 such that for all partitions P , U(f, P ) ≥ U(f) + ε2 .

However, this would imply that there is some U2(f), namely U(f) + ε2 ,

that is greater than U(f) which is still a lower bound for the set of allU(f, P ). This is a contradiction, because U(f) is the greatest lower boundof {U(f, P )}.

Now, consider P ∪ Q. By a result proved in the last lecture, we knowthat U(f, P ) ≥ U(f, P ∪Q) and L(f,Q) ≤ L(f, P ∪Q). Therefore, we havethat

U(f, P ∪Q)−L(f, P ∪Q) ≤ U(f, P )− L(f,Q)

<(U(f) +

ε

2

)−(L(f)− ε

2

)= U(f)− L(f) + ε

However, f is Darboux integrable, so U(f) = L(f), and rearranging theinequality, we obtain L(f, P ∪ Q) > U(f, P ∪ Q) − ε, which is our desiredresult.

�67


25.2. Riemann Integration.

Now we will go over Riemann’s definition of an integral, which we will see isactually equivalent to Darboux integration.

Definition. We will define the mesh of a partition P as such:

mesh(P ) = max{tk − tk−1 | k ≥ 1}

In other words, the mesh P is the length of the largest subinterval of P .The next theorem states another equivalent condition for integrability, but is not

the Riemann integral.

Theorem. A bounded function f : [a, b]→ R is Darboux integrable if and only if forall ε > 0, there exists a δ > 0 such that for all partitions P of [a, b], if mesh(P ) < δ,then U(f, P )− L(f, P ) < ε.


(⇐) If we assume that mesh(P ) < δ implies U(f, P ) − L(f, P ) < ε, then wecan simply take the partition P that divides [a, b] into subintervals of equallength less than δ, and using the previous theorem, we have our result.

(⇒) Suppose f is integrable on [a, b], and let ε > 0 be given. By the previoustheorem, we can choose a partition P such that U(f, P ) − L(f, P ) < ε

2 .Now, because f is bounded, there exists a B > 0 such that |f(x)| ≤ B forall x ∈ [a, b].

Let Q be any partition such that mesh(Q) < δ, and let R = P ∪ Q. Ifwe assume that |R| = |Q|+ 1, then, repeating an argument from the proofof L(f) ≤ U(f), we have

L(f,R)− L(f,Q) ≤ B ·mesh(Q)− (−B) ·mesh(Q) = 2B ·mesh(Q)

Let |P | = m. Thus, R can only have, at most, m elements that are not inQ. By induction, we can see that, if this is the case, then

L(f,Q)− L(f, P ) ≤ 2mB ·mesh(P )

Thus, we can set δ = ε8mB , so that 2mB · mesh(P ) < 2mBδ = ε

4 . Now,because |P | < |R|, we have L(f, P ) − L(f,Q) < ε

4 , and repeating thesame steps for the Upper Darboux sums, we get U(f,Q) − U(f, P ) < ε

4 .Combining these two facts, we obtain

U(f,Q)− L(f,Q) < U(f, P )− L(f, P ) +ε

2<ε

2+ε

2= ε

completing the proof.

�

Definition. Let f : [a, b] → R be a bounded function, and let P be a partition of[a, b]. f is Riemann integrable if there exists an r ∈ R such that for all ε > 0, thereis a δ > 0 such that if mesh(P ) < δ, then∣∣∣∣∣r −

n∑k=1

f(xk)(tk − tk−1)

∣∣∣∣∣ < ε

r is known as the Riemann integral of f on [a, b], and is denoted∫ baf .

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Note that the symbol for the Riemann integral is the same as the one for theDarboux integral. This will be explained by the next theorem. Note, however,that for the sake of establishing these two integrals’ equivalence, we will denote the

Riemann integral as R∫ baf .

Theorem. A bounded function f : [a, b]→ R is Darboux integrable if and only if itis Riemann integrable, and the values for each integral are equal.


(⇒) Suppose f is Darboux integrable on [a, b], and let ε > 0 be given. Fromthe previous theorem, we can choose δ > 0 such that mesh(P ) < δ impliesU(f, P ) − L(f, P ) < ε. We want to show that the above condition forRiemann integrability is satisfied in this case. So, let P be a partition suchthat mesh(P ) < δ. We will also denote

∑nk=1 f(xk)(tk − tk−1) as S.

It is obvious that L(f, P ) ≤ S ≤ U(f, P ). Now, notice that by theprevious theorem, we have

U(f, P ) < L(f, P ) + ε ≤ L(f) + ε =

∫ b

a

f + ε

L(f, P ) > U(f, P )− ε ≥ U(f)− ε =

∫ b

a

f − ε

Therefore,∫ baf−ε ≤ S ≤

∫ baf+ε, so because ε was arbitrary, f is Riemann

integrable, and R∫ ba

=∫ ba

.(⇐) Suppose f is Riemann integrable, and let ε > 0 be given. Choose δ and r as

given in the definition of Riemann integrability. Now, let P be a partitionsuch that mesh(P ) < δ such that |P | = n. For each tk ∈ P , there is anxk ∈ [tk−1, tk] such that f(xk) < m(f, [tk−1, tk]) + ε. Again, such xks existby the definition of infimum. Now, if we consider these xk, it is the casethat

n∑k=1

f(xk)(tk − tk−1) ≤ L(f, P ) + ε(b− a)

Furthermore, by assumption, |r −∑nk=1 f(xk)(tk − tk−1)| < ε. Therefore,

substituting S for the summation, we have

L(f) ≥ L(f, P ) ≥ S − ε(b− a) > r − ε− ε(b− a)

Thus, L(f) ≥ r by the arbitrary nature of ε, and similarly, U(f) ≤ r.Therefore, because we know L(f) ≤ U(f), L(f) = U(f) = r, f is Darboux

integrable and∫ baf = R

∫ baf .

�

Now that we have proved this theorem, we can remove the R from the Riemann

integral and use one symbol for both, namely∫ baf .

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26. 4/28/11: The Fundamental Theorem of Calculus

26.1. Integration Facts.

Recall the symbol∫ baf(x)dx where f is a bounded function from [a, b] to R. We

have learned two equivalent ways of calculating the integral:

• The Darboux integral:∫ baf(x)dx = U(f) where

U(f) = inf U(f, P ) =∑

M(f, [ti−1, ti])(ti − ti−1)

L(f) = supL(f, P ) =∑

m(f, [ti−1, ti])(ti − ti−1)

and the integral is defined if U(f) = L(f).• The Riemann integral: Take xi ∈ [ti−1, ti] for all i. Then,∫ b

a

f(x)dx =∑

f(xi)(ti − ti−1)

The integral is defined if for all ε there exists a δ such that there exists anr that satisfies |

∑f(xi)(ti − ti−1)− r| < ε if mesh(P ) < δ.

Our goal today is to prove the Fundamental Theorem of Calculus.

Theorem. Any monotonic function is integrable.


Theorem. Any continuous function is integrable.

Proof. We will use the previous theorem that states that a function f is integrableif and only if for all ε > 0, there is a partition P such that U(f, P ) ≤ L(f, P ) + ε.Let ε > 0 be given. f is uniformly continuous because [a, b] is compact. Thus, thereexists a d > 0 such that |x− y| < δ implies |f(x)− f(y)| < ε

b−a for all x, y ∈ [a, b].

So, choose a partition P such that max{ti − ti−1 | i = 1, . . . , n} < δ. We can seethat

M(f, [ti−1, ti]) < m(f, [ti−1, ti]) +ε

b− abecause each subinterval has width less than δ, so the function does not differ bymore than ε

b−a in each subinterval. Therefore, we now have

n∑i=1

M(f, [ti−1, ti])(ti−ti−1) <

n∑i=1

m(f, [ti−1, ti])(ti−ti−1)+

(n∑i=1

(ti − ti−1)

)(

ε

b− a)

which implies U(f, P ) < L(f, P ) + ε. �

26.2. Properties of the Integral.

Proposition (Linearity of∫

). Let f, g : [a, b] → R. If∫ bafdx and

∫ bagdx exist,

then∫ ba

(f + g)dx exists, and∫ ba

(f + g)dx =∫ bafdx+

∫ bagdx.

Proof. We will use Riemann’s method of integration. Let ε > 0 be given. Be-cause f and g are integrable, there exists a δ > 0 such that mesh(P ) < δ implies

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|∑f(xi)(ti − ti−1) −

∫ baf(x)dx| < ε, and likewise for g, for any choice of xis in

each subinterval [ti−1, ti]. Therefore, we have∣∣∣∣∑(f + g)(xi)(ti − ti−1)−∫ b

a

f(x)dx−∫ b

a

g(x)dx

∣∣∣∣ =∣∣∣∣(∑ f(xi)(ti − ti−1)−∫ b

a

f(x)dx

)+

(∑g(xi)(ti − ti−1)−

∫ b

a

g(x)dx

)∣∣∣∣The latter expression is less than ε+ ε, so because ε was arbitrary, we are done. �

Another part of this theorem is that for any c ∈ R,∫ bacf(x)dx = c

∫ baf(x)dx. The

proof of this statement is left as an exercise.

Theorem. If f is integrable, then |f | is also integrable, and∣∣∣∣∣∫ b

a

fdx

∣∣∣∣∣ ≤∫ b

a

|f |dx∗

Proof Sketch. The crucial lemmas are that

M(|f |, [ti−1, ti])−m(|f |, [ti−1, ti]) ≤ |M(f, [ti−1, ti])−m(f, [ti−1, ti])|which follows from |a| − |b| ≤ |a − b| for any a, b ∈ R, and the fact that if f, g are

integrable and f(x) ≤ g(x) for all x ∈ [a, b), then∫ baf(x)dx ≤

∫ bag(x)dx. To see

this fact, consider g − f . g − f ≥ 0, so∫ ba

(g − f)dx ≥ 0, which implies, by the

previous theorem, that∫ bagdx ≥

∫ bafdx. �

Theorem. If a < b < c for a, b, c ∈ R and f is integrable on [a, b] and [b, c], thenf is integrable on [a, c] and∫ c

a

f(x)dx =

∫ b

a

f(x)dx+

∫ c

b

f(x)dx

Proof. By Riemann’s method; left as an exercise. �

26.3. The Fundamental Theorem of Calculus.

Theorem (The Fundamental Theorem of Calculus, Part 1). Let f be differentiableon (a, b) and continuous on [a, b]. If f ′ is integrable on [a, b], then∫ b

a

f ′(x)dx = f(b)− f(a)

Proof. We will proceed by Riemann’s method of integration. Let ε > 0 be given.Choose a partition P such that∣∣∣∣∣

∫ b

a

f ′(x)dx−∑

f ′(xi)(ti − ti−1)

∣∣∣∣∣ < ε

for all xi ∈ [ti−1, ti]. By the Mean Value Theorem, for each interval [ti−1, ti], there

exists an xi ∈ (ti−1, ti) such that f ′(x) = f(ti)−f(ti−1)ti−ti−1

. Then, we have

n∑i=1

(ti − ti−1)f ′(xi) =

n∑i=1

(f(ti)− f(ti−1)) = f(b)− f(a)

∗Note that this theorem is a version of the triangle inequality.

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Therefore, the above chain of equalities implies that |f(b)− f(a)−∫ baf ′(x)dx| < ε,

so we are done. �

Theorem 12 (The Fundamental Theorem of Calculus, Part 2). Let f be an inte-grable function on [a, b], and define the function F (x) =

∫ xaf(t)dt. Then,

(1) F (x) is continuous.(2) If f is continuous at x0, then F is differentiable at x0, and F ′(x0) = f(x0).

Proof. Not covered in class, due to time constraints. Professor Jones referred theclass to the proof in the textbook. �

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Appendix: Significant Results Proved in Homework

This is a list of all of the results that have been assigned as homework that weshould know.

(1) Let S and T be nonempty subsets of R with the property that s ≤ t for alls ∈ S and t ∈ T . Then:(a) S is bounded above and T is bounded below.(b) supS ≤ inf T .

(2) Let I be the set of real numbers that are not rational numbers, and leta, b ∈ R. If a < b, then there exists an x ∈ I such that a < x < b.

(3) Let A and B be nonempty bounded subsets of R, and let S be the set ofall sums a+ b where a ∈ A and b ∈ B. Then:(a) supS = supA+ supB.(b) inf S = inf A+ inf B.

(4) Every element of a complete ordered field is the supremum of the set ofrationals that are less than it.

(5) In a complete ordered field, a nonzero element is positive if and only if itis a square.

(6) Let F and G be complete ordered fields. There exists a unique bijectionf : F → G such that

f(x+ y) = f(x) + f(y) and f(xy) = f(x)f(y) .

(7) Let (sn) be a sequence in R such that sn ≥ 0 for all n ∈ N. If lim sn = 0,then lim

√sn = 0.

(8) Let (sn) be a convergent sequence. If lim sn > a, then there exists anN ∈ N such that n > N implies sn > a.

(9) Let R(x) be the field of rational functions (with real coefficients) in x. Then,

the set of elements of R(x) of the form P (x)Q(x) where P and Q are polynomials

such that the coefficient of the highest power of x is positive satisfies theproperties required to make R(x) into an ordered field.

(10) R(x), as defined above, is not Archimedean, and hence not complete.(11) Every positive element of R has a unique positive nth root, for all n ∈ N.(12) The sum of two Cauchy sequences is Cauchy.(13) The product of two Cauchy sequences is Cauchy.(14) Let (sn) and (tn) be bounded sequences in R. Then,

lim sup(sn + tn) ≤ lim sup sn + lim sup tn .

(15) A sequence (sn) is bounded if and only if lim sup |sn| <∞.(16) Addition and multiplication of sequences (in the natural way) in any or-

dered field makes the set of all sequences into a commutative ring. However,this structure does not define a ring structure on equivalence classes of se-quences, as defined in Lecture 7.

(17) Let (sn) be a sequence of nonnegative numbers, and for each n, defineσn = 1

n (s1 + s2 + · · ·+ sn). Then,(a) lim inf sn ≤ lim inf σn ≤ lim supσn ≤ lim sup sn.(b) If lim sn exists, then limσn exists, and lim sn = limσn.

(18) Let (X, d) be a metric space. Then, D(x, y) = d(x,y)1+d(x,y) also defines a metric

on X and a subset U of X is open in (X, d) if and only if it is open in (X,D).73


(19) Let d : X ×X → R be a function satisfying all of the properties of a metricexcept d(x, y) = 0 ⇒ x = y. Define ∼ on X by x ∼ y ⇐⇒ d(x, y) = 0.Then, ∼ is an equivalence relation, setting D([x], [y]) = d(x, y) is well-defined, and D makes the set of equivalence classes of X into a metricspace.

(20) Let (X, d) be a metric space. Then, |d(x, y)− d(x, z)| ≤ d(y, z).(21) Let (xn) and (yn) be two Cauchy sequences in (X, d). Then, (d(xn, yn)) is

a Cauchy sequence in R.(22) Define a function d((xn), (yn)) = lim d(xn, yn) on the set of all sequences in

a metric space (X, d). Also, define a relation ∼ on the set of all sequencesin (X, d) by (xn) ∼ (yn) ⇐⇒ d(xn, yn) → 0. d then defines a metric on

the set of equivalence classes of Cauchy sequences, which we will denote X.Furthermore, given that X can be embedded into X by defining x ∈ X tobe the equivalence class of the constant sequence of x, X is dense in X.

(23) The metrics defined by the norms || · ||1, || · ||2, and || · ||∞ are equivalent.∗

(24) Equivalent metrics define the same topology.(25) Every polynomial function p(x) = a0 + a1x + · · · + anx

n is continuous onR.

(26) The Cantor set is uncountable.(27) Let A be an infinite set and F be a finite set. Then, |A| = |A ∪ F |.(28) A uniformly continuous function maps Cauchy sequences to Cauchy se-

quences.(29) Let A be a dense subset of the metric space (X, d), and let (Y,D) be

a complete metric space. Then, given a uniformly continuous functionf : A→ Y , f admits a unique continuous extension to X.

(30) Any open subset U of R is a countable union of disjoint open intervals.(31) Let X be a compact metric space, and Y be a metric space. Then, any

continuous function from X to Y is uniformly continuous.(32) Let (X, d) be a compact metric space and let XN be the set of all sequences

in X. Then, D((xn), (yn)) =∞∑n=1

d(xn,yn)2n defines a metric on XN for which

it is sequentially compact, and hence compact.(33) Assume that any finite map in the plane can be colored with four colors

so that no adjacent countries have the same color. Then, any countablyinfinite planar map can also be colored with four colors.

(34) A compact metric space is complete.(35) The closure of a connected subset of a metric space is also connected.(36) Let E and F be connected subsets of a metric space. If E and F have empty

intersection, then their union is connected. However, their intersection(regardless of whether it is empty or not) need not be connected.

(37) Cauchiness is not a topological property, i.e. there exists an example ofa set which can take two metrics which define the same topology, but forwhich a sequence exists which is Cauchy in one metric but not the other.

(38) Define C(v, r) ⊆ R2 to be the circle centered at v of radius r, namely ifv = (a, b), then C(v, r) = {(x, y) | (x − a)2 + (y − b2) = r2}. C(v, r) isconnected and path-connected.

∗The notion of equivalent metrics is the same as the notion of equivalent norms detailed in

Lecture 16.

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(39)⋃∞n=1 C(( 1

n , 0), 1n ), also known as the Hawaiian earring, is path-connected.

(40) The continuous image of a path-connected metric space is path-connected.(41) Let c : R→ R and s : R→ R be differentiable functions such that c(0) = 1,

s(0) = 0, c′(x) = −s(x) for all x, and s′(x) = c(x) for all x. Then,c(x) = cos(x) and s(x) = sin(x) for all x.

(42) cos(A+B) = cos(A) cos(B)− sin(A) sin(B) andsin(A+B) = sin(A) cos(B) + cos(A) sin(B).

(43) cos(t) = 0 for some unique t between 1 and 2, and cos(x) > 0 for allx ∈ (−t, t).

(44) Let π be the real number equal to 2t as defined in the item above. Then,the following are true:

cos(π

2− x)

= sin(x)

sin(π

2− x)

= cos(x)

sin(x+ π) = − sin(x)

cos(x+ π) = − cos(x)

(45) exp(x) is a bijection from R to (0,∞), and log(x) is differentiable such thatlog′(x) = 1

x .(46)

log(1− x) = −∞∑n=1

xn

n

for all x ∈ (−1, 1).(47) Let f : [0, 2]→ R be continuous and differentiable in (0, 2), and let f(0) = 0.

If −3 < f ′(x) < 4 for all x ∈ (0, 2), then −6 < f(2) < 8.(48) Let S1 := {(x, y) ∈ R2 |x2 + y2 = 1}. There is a unique θ ∈ [0, 2π) such

that x = cos θ and y = sin θ, known as the angle between the positive xaxis and the ray going from the origin through (x, y).

(49) The trigonometric polynomials, i.e.

k∑n=−k

an sinnθ + bn cosnθ

where an and bn are real numbers and k is a positive integer, are dense inC(S1,R).

(50) | cos(x)− cos(y)| ≤ |x− y| for all x, y ∈ R.(51) Let f be a differentiable on R such that a = sup{|f(x)| |x ∈ R} < 1. Then,

for any s0 ∈ R, the sequence defined by sn+1 = f(sn) converges.(52) Let f be a bounded function on [a, b], such that |f(x)| < B for all x ∈ [a, b].

Then, for all partitions P of [a, b], it is the case that

U(f2, P )− L(f2, P ) ≤ 2B(U(f, P )− L(f, P ))

Also, if f is integrable on [a, b], then f2 is integrable on [a, b].(53) Let f(x) = x sgn(sin 1

x ) for x 6= 0 and f(0) = 0. f is not piecewise contin-uous nor piecewise monotonic on [−1, 1], but it is integrable on [−1, 1].

(54) Suppose f is a continuous function on [a, b] such that f(x) ≥ 0 for all

x ∈ [a, b]. If∫ baf(x)dx = 0, then f(x) = 0 for all x ∈ [a, b].

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(55) Let f be a continuous function on [a, b] such that∫ baf(x)g(x)dx = 0 for all

continuous functions g on [a, b]. Then, f(x) = 0 for all x ∈ [a, b].(56) Assume you are standing at (0, 0), the origin in R2. Around you are circles

of equal nonzero radius on every point (a, b) where a, b ∈ Z. Then, you canneither see infinitely, nor can you see arbitrarily far. In other words, theredoes not exist a ray from (0, 0) that does not, at some point, hit a circle,and for every radius of the circles, there exists some R ∈ R such that thelength of the ray is always less than R.

Note that there are further concepts not covered in these lecture notes in the notesprovided by Professor Jones about Metric Spaces and Cardinality, as well as extraexposition and concepts in the course textbook, Elementary Analysis: The Theoryof Calculus, 1st edition, by Kenneth A. Ross, part of the Springer UTM series.

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