Math 360: Other axioms equivalent to the fundamental axiom

Math 360: Other axioms equivalent to thefundamental axiom

D. DeTurck

University of Pennsylvania

September 18, 2017

D. DeTurck Math 360 001 2017C: Suprema 1 / 25

IVT equivalent to the FA

Theorem

Let F be an ordered field for which the intermediate value theoremholds. Then the fundamental axiom holds.

Let a1 ≤ a2 ≤ a3 ≤ · · · and an < A for all n be a nondecreasingsequence that is bounded from above. We need to show that thesequence has a limit.

Define f (x) via

f (x) =

{−1 if x < an for some n+1 otherwise

If a < a1 and b > A then by the IVT there must be a pointc ∈ (a, b) where f (x) is discontinuous.


Conclusion of proof

We must have c ≥ an for all n, because if x < aN for some N thenno matter what ε is, we can choose δ < 1

2aN − x . Then|x − y | < δ implies that y < aN − 1

2(aN − x), so thatf (x)− f (y) = (−1)− (−1) = 0.

Next, we show that for all ε > 0, there is an N so that c − aN < ε(and so c − an < ε for all n ≥ N, since the sequence is increasingand c ≥ an for all n. If not, then f (c) = +1 and in fact f (x) = +1for all x > c − ε, so f would be continuous at c .

Therefore, c is the limit of the sequence a1, a2, . . . which provesthe existence of the limit and hence the fundamental axiom.


Maximum and supremum

The integers are “well-ordered”

This means that a nonempty set S of integers that is boundedabove has a maximum, that is an integer n such that n ∈ S andm ≤ n for all m ∈ S .

The real numbers are not well-ordered. For instance the openinterval (0, 1) does not have a maximum. But it does have asupremum.

Definition of the supremum of a set of real numbers

Let S be a non-empty set of real numbers. Say that s is thesupremum of S (or the least upper bound of S) if s ≥ t for allt ∈ S and, if it is true that r ≥ t for all t ∈ S then r ≥ s.

Lemma

If the supremum exists, then it is unique.


A characterization of the supremum

Lemma

Let S be a non-empty set of real numbers. Then s is thesupremum of S if and only if both s ≥ t for all t ∈ S and, givenε > 0 there is a t ∈ S such that t ≥ s − ε.

By the way, the notation is supS or supt∈S

t for the supremum of S .

All of the above works in any ordered field, but note that there arebounded sets of rational numbers (for instance {x ∈ Q | x2 < 2})that do not have suprema. However:

The supremum principle

Let S be a non-empty set of real numbers that is bounded above.Then S has a supremum.


Proof of the supremum principle

Since the principle is false in Q, we must use the fundamentalaxiom. Time for some lion-hunting!

Choose a0 to be any point of S , and b0 to be a number so thatb0 > s for every s ∈ S . Then [a0, b0] ∩ S 6= ∅.

Define the sequences a0 ≤ a1 ≤ a2 ≤ · · · and b0 ≥ b1 ≥ b2 ≥ · · ·as follows: For each n, Let c be the midpoint of the interval[an−1, bn−1]. If [c, bn−1] ∩ S 6=∅ then set an = c and bn = bn−1,otherwise set an = an−1 and bn = c . Observe that bn ≥ s for everys ∈ S , [an, bn] ∩ S 6= ∅, and bn − an = 1

2(bn−1 − an−1).

Therefore both the sequences {an} and {bn} converge to acommon limit L, which must be the supremum of S .


Equivalence

We proved the supremum principle based on the fundamentalaxiom. But if you assume the supremum principle as the axiom,then you can prove the fundamental axiom from it: If you have abounded increasing sequence a1 ≤ a2 ≤ · · · , then the limit is justthe supremum of the set {a1, a2, . . .}.

The supremum is the least upper bound of a set of real numbers.Correspondingly, the greatest lower bound of the set S is called theinfimum of S , denoted inf S .


IVT via sup

IVT: If f : [a, b]→ R is continuous and f (a) ≤ 0 ≤ f (b) then thereis a c ∈ [a, b] with f (c) = 0.

Let E be the subset of [a, b] where f (x) ≤ 0. Then a ∈ E and E isbounded above by b, so E has a supremum, call it c . We’ll showthat |f (c)| ≤ ε for all ε > 0, which implies that f (c) = 0.

There are three possibilities: either c = a or c = b or a < c < b. Ifc = a then f (c) = f (a) ≤ 0 but for all x > c we have f (x) > 0.Given ε > 0, there is a δ such that |f (c)− f (x)| < ε if x − c < δ.But then −ε < f (c)− f (x) < ε so −ε < f (x)− ε < f (c) ≤ 0, so|f (c)| < ε.

If c = b then for any δ > 0 there is an x > c − δ with x ∈ E , sof (x) ≤ 0. Given ε > 0, we have δ so that |f (c)− f (y)| < ε ifc − y < δ. Choosing x so that c − x is less than this δ gives us−ε < f (c)− f (x) < ε, or 0 ≤ f (c) < f (x) + ε ≤ ε (since c = b),so |f (c)| < ε again.


End of proof of IVT via sup

Finally, if a < c < b and we are given ε > 0, we know there is aδ > 0 so that if |c − x | < δ then |f (c)− f (x)| < ε. Choose x+ tobe any number between c and c + δ, then x+ 6∈ E so0 < f (x+) < ε. And we can choose an x− ∈ E between c − δ andc so that −ε < f (x−) ≤ 0.

We know f (c)− f (x+) > −ε, so f (c) > f (x+)− ε > −ε and weknow f (c)− f (x−) < ε, so f (c) < f (x−) + ε ≤ ε. Therefore|f (c)| < ε as claimed.

So now we know that |f (c)| < ε for every ε > 0, so f (c) = 0,proving the IVT.

We could also prove the mean value inequality using a supremumargument.


The Bolzano-Weierstrass theorem

Bolzano-Weierstrass theorem

If {xn} is a bounded sequence of real numbers (so there is an M sothat |xn| < M for all n), then there is a subsequence of {xn} (say,xn(1), xn(2), . . .) that is convergent.

There might be more than one convergent subsequence — forexample take xn = (−1)n.

Prove this theorem by lion-hunting (that’s how Bolzano did it) —choose the half of the interval that has infinitely many terms in it.

This theorem is (not surprisingly) equivalent to the fundamentalaxiom.


Closed sets in R

These are generalizations of open and closed intervals.

Definition: Closed sets

A set F ⊂ R is closed if, for any sequence {xn} where xn ∈ F andxn → x as n→∞, we have x ∈ F

Sometimes we say that a closed set “contains all of its limitpoints”.

Theorem (Bolzano-Weierstrass-ish)

(i) If F is a closed and bounded set in R, then every sequence in Fhas a subsequence converging to a point of F .(ii) Conversely, if F is a subset of R such that every sequence in Fhas a subsequence converging to a point of F , then F is a closedbounded set.


Proofs

(i) If {xn} is a sequence in F it is by definition bounded, so byBolzano-Weierstrass it has a convergent subsequence. But if F isclosed, then the subsequence converges to a point of F .

(ii) If F is not closed, then we can find xn ∈ F and x 6∈ F suchthat xn → x as n→∞. Since any subsequence of a convergentsequence converges to the same limit, no subsequence of the {xn}can converge to a point of F .

If F s not bounded, we can find xn ∈ F so that |xn| > n. If x ∈ Rthen the inequality

|xn − x | ≥ |xn| − |x | > n − |x |

shows that no subsequence of the xn can converge.


Open sets

A set U ⊂ R is open if, whenever x ∈ U, there exists an ε > 0such that if |y − x | < ε then y ∈ U.

So any x ∈ U lies in the “interior” of the set. It is true that[a, b] ⊂ R is closed, (a, b) ⊂ R is open, (a, b] is neither open norclosed, and R is both open and closed.

U is open if and only if each point of U is the center of an openinterval lying entirely within U.

Lemma

U is open if and only if its complement is closed.

(Think about sequences in U or in its complement.)


Topology

Proposition

Let τ be the collection of open sets in R. Then

• ∅ ∈ τ and R ∈ τ .

• If Uα ∈ τ for all α ∈ A then ∪α∈AUα ∈ τ .

• If U1, . . .Un ∈ τ then ∩ni=1Ui ∈ τ .

Proposition

Let F be the collection of closed sets in R. Then

• ∅ ∈ F and R ∈ F .

• If Fα ∈ F for all α ∈ A, then ∩α∈FFα ∈ F .

• If F1, . . .Fn ∈ F then ∪ni=1Fi ∈ F .


Another characterization of continuity

Theorem

The function f : R→ R is continuous if and only if f −1(U) is open(in the domain) whenever U is open (in the codomain).

Recall that f −1(U) = {x ∈ R | f (x) ∈ U}.

This is almost immediate from the definitions. If f is continuousand U is open in the target R, we have that if x ∈ f −1(U) thenf (x) ∈ U. Since U is open there is an ε > 0 so that(f (x)− ε , f (x) + ε) ⊂ U. Since f is continuous, there is a δ > 0so that |f (x)− f (y)| < ε whenever |x − y | < δ, so the interval(x − δ , x + δ) ⊂ f −1(U), so f −1(U) is open (it contains an openinterval around each of its points).


Another characterization of continuity

Theorem

The function f : R→ R is continuous if and only if f −1(U) is open(in the domain) whenever U is open (in the codomain).

Conversely, if f −1(U) is open whenever U is open, then choosex ∈ R and ε > 0. Since the interval of “radius” ε centered at f (x)is open, we have that f −1 of this interval is also open containing x .But then there is a δ so that (x − δ , x + δ) in in f −1 of thisinterval, i.e., |f (x)− f (y)|ε whenever |x − y | < δ. so f iscontinuous.

Corollary

If g is continuous and f is continuous on the range of g then thecomposition f ◦ g is continuous.


The extreme value theorem

Theorem

Let f : [a, b]→ R be continuous, where [a, b] is a closed, boundedinterval. Then the image of f is a closed, bounded subset of R (inparticular there are points of [a, b] where f attains its minimumand maximum values).

Proof that the image is bounded: Suppose there were a sequenceof points xn ∈ [a, b] such that |f (xn)| > n for all n. A subsequenceof the xn’s converges to a point x ∈ [a, b]. But then f (xn)→ f (x)and a convergent sequence is bounded, a contradiction.

Proof that the image is closed: Let yn be a convergent sequence ofpoints in the image of f . Then yn = f (xn) for each n, wherexn ∈ [a, b]. But then there is a convergent subsequence of the xnto a point x ∈ [a, b], and f (x) is the limit of the yn and is in theimage. Therefore the image is closed.


Limits and derivatives

We hadn’t yet actually defined the notion of the limit of a functionas x → a ∈ R. So here we go:

Definition (limit)

Let E ⊂ R, let a ∈ E and L ∈ R. Let f be a functionf : (E − {a})→ R (or f : E → R). Say that f (x)→ L as x → a(through values of x ∈ E ) or lim

x→af (x) = L if, given ε > 0 we can

find a δ(ε, a) > 0 such that whenever x ∈ E and |x − a| < δ thenwe have

|f (x)− L| < ε.

Note that f doesn’t have to be defined at x = a, and even if it is,it doesn’t matter what the value of f (a) is, in order for the limit toexist. But from our previous definition of continuity, we see that fis continuous at a if and only if lim

x→af (x) exists and is equal to f (a).


The deriviative (redux)

Definition (derivative)

Let U ⊂ R be open, and x ∈ U. Then f : U → R is differentiableat x with derivative f ′(t) if and only if

f (x + h)− f (x)

h→ f ′(x) as h→ 0

This is, of course, just a restatement of our earlier definition.

If f (x) = c then f ′(x) = 0, and if g(x) = x then g ′(x) = 1,because

f (x + h)− f (x)

h= 0 and

g(x + h)− g(x)

h= 1.

Proposition

If f is differentiable at x , then f is continuous at x .


Derivative rules – sum and difference

Suppose f (x) and g(x) are differentiable functions on (a, b), withderivatives f ′(x) and g ′(x).

Sum rule: (f + g)′(x) = f ′(x) + g ′(x)

(f + g)(x + h)− (f + g)(x)

h=

f (x + h)− f (x)

h+

g(x + h)− g(x)

h→ f ′(x) + g ′(x).

Constant multiple rule: (cf )′(x) = cf ′(x) if c ∈ R

(cf )(x + h)− (cf )(x)

g= c

f (x + h)− f (x)

h→ cf ′(x).

Difference rule: (f − g)′(x) = f ′(x)− g ′(x)

Since f − g = f + (−1)g , this follows from the two rules above.


Derivative rules – products

Product rule: (f × g)′(x) = f (x)g ′(x) + f ′(x)g(x)

(f × g)(x + h)− (f × g)(x)

h

=f (x + h)g(x + h)− f (x + h)g(x) + f (x + h)g(x)− f (x)g(x)

h

= f (x + h)g(x + h)− g(x)

h+

f (x + h)− f (x)

hg(x)

→ f (x)g ′(x) + f ′(x)g(x)

Using the product rule, we can (by induction) prove that iff (x) = xn then f ′(x) = nxn−1 for non-negative integers n.

So from the rules so far, we now know that polynomials aredifferentiable.


Derivative rules – quotients

Quotient rule:

(f

g

)′(x) =

f ′(x)g(x)− f (x)g ′(x)

(g(x))2, if g(x) 6= 0

Let q(x) =

(f

g

)(x). Then q(x)g(x) = f (x) and we can use the

product rule to write

q(x)g ′(x) + q′(x)g(x) = f ′(x)

Solve for q′ and remember what q is to write:

q′(x) =f ′(x)− q(x)g ′(x)

g(x)=

f ′(x)g(x)− f (x)g ′(x)

(g(x))2.

Use this to show that the derivative of1

xn= − n

xn+1. So we can

now differentiate all integer powers, and all rational functions(quotients of polynomials).


Derivative rules – compositions

Chain rule: (f ◦ g)′(x) = f ′(g(x))g ′(x).

This one is a little bit tricky. We have to handle the casesg ′(x) 6= 0 and g ′(x) = 0 separately.

If g ′(x) 6= 0 then g(x + h) 6= g(x) for sufficiently small h, eventhough it is true that g(x + h)− g(x)→ 0 as h→ 0. So we have

(f ◦ g)(x + h)− (f ◦ g)(x)

h=

f (g(x + h))− f (g(x))

h

=f (g(x + h))− f (g(x))

g(x + h)− g(x)· g(x + h)− g(x)

h

→ f ′(g(x))g ′(x)


Proof of chain rule when g ′(x) = 0

If g ′(x) = 0, then for any ε > 0 we can find δ > 0 so that δ < εand |g(x + h)− g(x)| < εh < ε2 provided |h| < δ. and also so that∣∣∣∣ f (x + k)− f (x)

k− f ′(x)

∣∣∣∣ < ε if |k | < δ.

We can rewrite this as

f ′(x)k − ε|k| < f (x + k)− f (x) < f ′(x)k + ε|k |.

Substitute g(x) for x and g(x + h)− g(x) for k and get

f ′(g(x))(g(x + h)− g(x))− ε|g(x + h)− g(x)|< f (g(x + h))− f (g(x))

< f ′(g(x))(g(x + h)− g(x)) + ε|g(x + h)− g(x)|


Proof cont’d

Now divide everything by h and take absolute values and get∣∣∣∣ f (g(x + h))− f (g(x))

h

∣∣∣∣ < |f ′(g(x)) + ε|∣∣∣∣g(x + h)− g(x)

h

∣∣∣∣< |f ′(g(x)) + ε|ε

which shows, since this is true for all ε > 0, that the derivative off (g(x)) is zero at a point where g ′ = 0.

Use this to find the derivative of x1/q and xp/q for p, q ∈ Z(q 6= 0), so now we can differentiate all rational powers of x .


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Math 360: Other axioms equivalent to the fundamental axiom