Chapter 1 (cont.) · 2020. 8. 26. · Chapter 1 (cont.) Daniel H. Luecking Aug 26, 2020 Daniel H. Luecking Chapter 1 (cont.) Aug 26, 20201/22

Chapter 1 (cont.)

Daniel H. Luecking

Aug 26, 2020

Daniel H. Luecking Chapter 1 (cont.) Aug 26, 2020 1 / 22

Outline

1 Metric spaces

2 Open and closed sets

3 Compact sets


Distance functions

Let X be a set and d : X × X → [0,∞). We call d a metric if

1 ∀x , y ∈ X , d(x , y) = 0 if and only if x = y .

2 ∀x , y ∈ X , d(x , y) = d(y , x).

3 ∀x , y , z ∈ X , d(x , z) ≤ d(x , y) + d(y , z).

A set X with a metric d is called a metric space.

In a metric space (X , d) if x0 ∈ X and r > 0 denoteB(x0, r) = {x ∈ X : d(x , x0) < r}. We call this the open ball centered atx0 with radius r . It is also referred to as a neighborhood (abbr. nbd) ofx0. We also have the closed balls: B(x0, r) = {x ∈ X : d(x , x0) ≤ r}.


Open sets

A set U in any metric space (X , d) is called open iff

∀x ∈ U, ∃δ > 0, B(x , δ) ⊆ U .

That is, U contains a nbd of each of its points. For example, an open ballB(x0, r) is open: if x ∈ B(x0, r) then B(x , δ) ⊆ B(x0, r) ifδ = r − d(x , x0).

In general, if∃δ > 0, B(x , δ) ⊆ A

we call x an interior point of A. So U is open iff all its points are interiorpoints.

If A is a set and x ∈ X satisfies ∀δ > 0, B(x , δ) ∩ A 6= ∅(stated another way: every nbd of x meets A) we call x a point of closureof A. We say A is closed iff it contains all of its points of closure. Forexample, closed balls B(x0, r) are closed.


Open sets


∀x ∈ U, ∃δ > 0, B(x , δ) ⊆ U .






Open sets


∀x ∈ U, ∃δ > 0, B(x , δ) ⊆ U .






Relationship between open and closed

By negating the definitions we see that x is a point of closure of A if andonly if it is not an interior point of Ac . Thus, A is closed if and only if Ac

is open.

Let IntA (or Ao) denote the set of interior points of A and A (or Cl(A))the set of points of closure. From their definitions, IntA ⊆ A ⊆ A. Also,IntA is always open and A is always closed.

In fact, IntA is the largest open set contained in A and A is the smallestclosed set that contains A.




is open.






is open.




Properties of open and closed sets

For any metric space X open sets satisfy:

1 X and ∅ are open.

2 If Aj , j = 1, 2, 3, . . . , n are open, so is⋂n

j=1 Aj .

3 If A is any family of open sets then⋃A is open.

And closed sets satisfy:

1 X and ∅ are closed.

2 If Bj , j = 1, 2, 3, . . . , n are closed, so is⋃n

j=1 Bj .

3 If B is any family of closed sets then⋂B is closed.


Properties of open and closed sets

For any metric space X open sets satisfy:

1 X and ∅ are open.

2 If Aj , j = 1, 2, 3, . . . , n are open, so is⋂n

j=1 Aj .

3 If A is any family of open sets then⋃A is open.

And closed sets satisfy:

1 X and ∅ are closed.

2 If Bj , j = 1, 2, 3, . . . , n are closed, so is⋃n

j=1 Bj .

3 If B is any family of closed sets then⋂B is closed.


Open sets in R

A set U ⊆ R is open iff for all x ∈ U there exists δ > 0 such that(x − δ, x + δ) ⊆ U. Any open interval in R is open.

Theorem

If a set U ⊆ R is open if and only if U is a union of countably manydisjoint open intervals.

Proof: Let U be open. Define a relation on U by x ∼ y iff there is an openinterval (a, b) ⊆ U with x , y ∈ (a, b). I leave it as an exercise to prove that∼ is an equivalence relation and the equivalence classes are open intervals.

There are countably many, because each contains a different rationalnumber.

For the converse: any union of open sets is open.


Open sets in R


Theorem






Open sets in R


Theorem






Open sets in R


Theorem






Open sets in R


Theorem






Closed sets in R

Closed intervals are closed as are [a,∞) and (−∞, b]. Finite unions ofclosed intervals are closed.

The Cantor middle thirds set (defined below) is closed, being theintersection closed sets.

Start with the interval C0 = [0, 1] and remove the open middle third,(1/3, 2/3) leaving two closed intervals C1 = [0, 1/3] ∪ [2/3, 1]. Repeat thiswith those two intervals to getC2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1]. Repeat this indefinitelyand we get C =

⋂∞n=1 Cn.

C is nonempty because it contains all the endpoints of the intervals ineach Cn. Thus it is infinite.


Closed sets in R




⋂∞n=1 Cn.



Closed sets in R



Start with the interval C0 = [0, 1] and remove the open middle third,(1/3, 2/3) leaving two closed intervals C1 = [0, 1/3] ∪ [2/3, 1].

Repeat thiswith those two intervals to getC2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1]. Repeat this indefinitelyand we get C =

⋂∞n=1 Cn.



Closed sets in R



Start with the interval C0 = [0, 1] and remove the open middle third,(1/3, 2/3) leaving two closed intervals C1 = [0, 1/3] ∪ [2/3, 1]. Repeat thiswith those two intervals to getC2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1].

Repeat this indefinitelyand we get C =

⋂∞n=1 Cn.



Closed sets in R




⋂∞n=1 Cn.



Closed sets in R




⋂∞n=1 Cn.



An uncountable set containing no nontrivial interval

C contains no nontrivial intervals and yet. . .

C is in fact uncountable: there is a one-to-one correspondence between Cand the set S of sequences of 0’s and 1’s. We can associate with x ∈ C asequence of left and right moves from the interval in Cn−1 that contains xto one of the two intervals it splits into in Cn. Using 0 for left and 1 forright, we get an element of S .

It is an exercise to prove this correspondence is bijective.

The Cantor set is also bounded. Closed and bounded sets play a specialrole in real analysis. They have a property known as compactness.




C is in fact uncountable: there is a one-to-one correspondence between Cand the set S of sequences of 0’s and 1’s.

We can associate with x ∈ C asequence of left and right moves from the interval in Cn−1 that contains xto one of the two intervals it splits into in Cn. Using 0 for left and 1 forright, we get an element of S .
















Compact sets

If U is a family of sets in a metric space X and A ⊆ X we will say U is acover of A if A ⊆

⋃U . If the sets in U are all open we call it an open

cover of A. A subcover of U is just a subfamily that also covers A. Afinite (sub)cover is just one that contains only a finite number of sets.

We say A is compact iff every open cover of A has a finite subcover.

Every compact set A is contained in some ball B(x0, r): if not then theopen cover U = {B(x0, n) : n ∈ N} has no finite subcover.

Every compact set A is closed: If not, let x0 be a point of closure of Athat is not in A, then U = {B(x0, r)c : r > 0} is an open cover with nofinite subcover.


Compact sets








Compact sets








Compact sets








The Heine-Borel Theorem

Theorem (Heine-Borel)

A set A in Rn is compact if and only if it is closed and bounded.

A useful consequence of H-B is the following

Theorem (Cantor)

If Fn, n ∈ N are closed bounded nonempty sets in Rn and nested (i.e.,Fn+1 ⊆ Fn for all n) then

⋂∞n=1 Fn 6= ∅.

Note: if the sets lie in R and Fn ⊆ [an, bn] with bn − an → 0, then⋂∞n=1 Fn is a singleton.

Proof: If the intersection is empty then by de Morgan’s laws, F cn , n > 1,

is an open cover of F1 with no finite subcover. For if F cnj

, j = 1, 2, . . . ,m

covers F1 with nm the largest index. Then F1 ∼⋃m

j=1 Fnj = Fnm 6= ∅.






Theorem (Cantor)


⋂∞n=1 Fn 6= ∅.




, j = 1, 2, . . . ,m


j=1 Fnj = Fnm 6= ∅.






Theorem (Cantor)


⋂∞n=1 Fn 6= ∅.




, j = 1, 2, . . . ,m


j=1 Fnj = Fnm 6= ∅.






Theorem (Cantor)


⋂∞n=1 Fn 6= ∅.




, j = 1, 2, . . . ,m


j=1 Fnj = Fnm 6= ∅.


Bolzano-Weierstrass Theorem

Let us say the a set satisfies the Bolzano-Weierstrass property if everysequence in A has a convergent subsequence with its limit in A. (This is avalid concept in any metric space once we define convergence.)

Theorem (Bolzano-Weierstrass)

A set A in Rn satisfies the Bolzano-Weierstrass property if and only it isclosed and bounded.

If A is not bounded it contains a sequence xn with d(xn, 0) > n for everyn. No subsequence of this can converge. If A is not closed and x is a pointof closure not in A then we can consruct a sequence {xn} ⊂ A withxn → x . This has convergent subsequences but they all have limit x /∈ A.

Compactness and the Bolzano-Weierstrass property are actually equivalentin every metric space. However, they are not always equivalent to beingclosed and bounded.






If A is not bounded it contains a sequence xn with d(xn, 0) > n for everyn. No subsequence of this can converge.

If A is not closed and x is a pointof closure not in A then we can consruct a sequence {xn} ⊂ A withxn → x . This has convergent subsequences but they all have limit x /∈ A.

















Cauchy sequences

A sequence {xn} in a metric space converges to x iff

∀ε > 0, ∃N ∈ N, ∀n ≥ N, d(xn, x) < ε

(This is equivalent to d(xn, x)→ 0.)

{xn} is said to be Cauchy iff

∀ε > 0, ∃N ∈ N, ∀n, k ≥ N, d(xn, xk) < ε

The difference between these two is that the points of the Cauchysequence become close to each other rather than some fixed limit. Everyconvergent sequence is Cauchy, but the reverse may not true be true.

A metric space is said to be complete if every Caucy sequence converges.One of the axioms of the real number system is that R is complete.

Except some axiom systems for R have a different ‘completeness axiom’:

Order completeness: Every nonempty set in R that has an upper boundhas a least upper bound.


Cauchy sequences


∀ε > 0, ∃N ∈ N, ∀n ≥ N, d(xn, x) < ε

(This is equivalent to d(xn, x)→ 0.) {xn} is said to be Cauchy iff

∀ε > 0, ∃N ∈ N, ∀n, k ≥ N, d(xn, xk) < ε






Cauchy sequences


∀ε > 0, ∃N ∈ N, ∀n ≥ N, d(xn, x) < ε


∀ε > 0, ∃N ∈ N, ∀n, k ≥ N, d(xn, xk) < ε






Cauchy sequences


∀ε > 0, ∃N ∈ N, ∀n ≥ N, d(xn, x) < ε


∀ε > 0, ∃N ∈ N, ∀n, k ≥ N, d(xn, xk) < ε






Cauchy sequences


∀ε > 0, ∃N ∈ N, ∀n ≥ N, d(xn, x) < ε


∀ε > 0, ∃N ∈ N, ∀n, k ≥ N, d(xn, xk) < ε






Order completeness: the supremum of a set

A real number u is called an upper bound for a nonenpty set E if ∀x ∈ E ,x ≤ u. The order completeness axiom says that the set of upper boundsfor E has a smallest element a. We call a the supremum of E , writinga = supE .

How to prove a value a is the supremum of a set E : prove

1 that a is an upper bound: ∀x ∈ E , x ≤ a, and

2 that no smaller value is an upper bound: ∀b < a, ∃x ∈ E , x > b.

Sometimes it is easier to prove the first statement using the Archmedeanprincipal: x ≤ a if and only if ∀ε > 0, x < a + ε. The above becomes:

The value a = supE iff for all ε > 0

1 ∀x ∈ E , x < a + ε, and

2 ∃x ∈ E , x > a− ε.

Note: the second statement implies that supE ∈ E .









1 ∀x ∈ E , x < a + ε, and

2 ∃x ∈ E , x > a− ε.










1 ∀x ∈ E , x < a + ε, and

2 ∃x ∈ E , x > a− ε.



Order completeness: the infimum of a set

If we reverse all inequalities we get: every set E that has a lower bound(obvious definition) has a greatest lower bound called the infimum of E :inf E .

Similar to the case for sup we get the proof scheme:

The value b = inf E iff for all ε > 0

1 ∀x ∈ E , x > b − ε, and

2 ∃x ∈ E , x < b + ε.

If E has no upper bound, we write supE =∞.If E has no lower bound, we write inf E = −∞If E ⊂ F then supE ≤ supF and inf E ≥ inf F .






1 ∀x ∈ E , x > b − ε, and

2 ∃x ∈ E , x < b + ε.







1 ∀x ∈ E , x > b − ε, and

2 ∃x ∈ E , x < b + ε.

If E has no upper bound, we write supE =∞.If E has no lower bound, we write inf E = −∞

If E ⊂ F then supE ≤ supF and inf E ≥ inf F .






1 ∀x ∈ E , x > b − ε, and

2 ∃x ∈ E , x < b + ε.



The lim sup and lim inf of a sequence of reals

Given a sequence {xn} in R let En = {xn, xn+1, xn+2, . . . } and gn = supEn.Because En+1 ⊂ En, {gn} is monotone decreasing.

So a = limn→∞

gn exists

and we call it the superior limit of {xn} writing a = lim supn→∞

xn.

Some special cases: If {xn} is not bounded above then supEn = +∞ forall n and so lim sup

n→∞xn = +∞.

If xn → −∞ then lim supn→∞

xn = −∞In all other cases, lim sup xn is a real number.

The inferior limit is defined by the same process, reversing the order:

lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })

If {xn} is not bounded below then lim infn→∞

xn = −∞.

If xn → +∞ then lim infn→∞

xn = +∞In all other cases, lim sup xn is a real number.



Given a sequence {xn} in R let En = {xn, xn+1, xn+2, . . . } and gn = supEn.Because En+1 ⊂ En, {gn} is monotone decreasing. So a = lim

n→∞gn exists


xn.


n→∞xn = +∞.




lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })


xn = −∞.






n→∞gn exists


xn.


n→∞xn = +∞.




lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })


xn = −∞.






n→∞gn exists


xn.


n→∞xn = +∞.


xn = −∞

In all other cases, lim sup xn is a real number.


lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })


xn = −∞.






n→∞gn exists


xn.


n→∞xn = +∞.




lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })


xn = −∞.






n→∞gn exists


xn.


n→∞xn = +∞.




lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })


xn = −∞.






n→∞gn exists


xn.


n→∞xn = +∞.




lim infn→∞

xn = limn→0

(inf{xn, xn+1, xn+2, . . . })


xn = −∞.




Proving a = lim sup xn

How to prove a finite value a equals lim sup xn:

If a ∈ R then a = lim sup xn if and only if for every ε > 0

1 the number of n with xn > a + ε is finite, and

2 the number of n with xn > a− ε is infinite

The first of these says that eventually gn ≤ a + ε and the second says thatall gn > a− ε, so together they say |gn − a| → 0.

Similarly, if b ∈ R then b = lim inf xn if and only if for every ε > 0

1 the number of n with xn < b − ε is finite, and

2 the number of n with xn < b + ε is infinite

It can be shown that a = lim sup xn is the largest value that is the limit ofsome subsequence of {xn}, and b = lim inf xn is the smallest value that isa limit fo some subsequence of {xn}.
























Properties of lim sup

Example: If xn = (−1)n(1 + 1/n) then lim sup xn = 1 and lim inf xn = −1.

1 lim inf xn ≤ lim sup xn.

2 {xn} converges if and only if lim sup xn = lim inf xn and lim xn is theircommon value.

3 lim sup−xn = − lim inf xn and lim inf −xn = − lim sup xn4 lim sup(xn + yn) ≤ lim sup xn + lim sup yn and

lim inf(xn + yn) ≥ lim inf xn + lim inf yn5 If lim xn = a then lim sup(xn + yn) = a + lim sup yn and

lim inf(xn + yn) = a + lim inf yn.

6 If f is continuous and increasing then lim sup f (xn) = f (lim sup xn).

Number 4 is based on the simple fact that if a ≥ xn and b ≥ yn thena + b ≥ xn + yn. Number 6 is based on the fact that thesup(a + E ) = a + supE for any real a and any set E .
















3 lim sup−xn = − lim inf xn and lim inf −xn = − lim sup xn

4 lim sup(xn + yn) ≤ lim sup xn + lim sup yn andlim inf(xn + yn) ≥ lim inf xn + lim inf yn

5 If lim xn = a then lim sup(xn + yn) = a + lim sup yn andlim inf(xn + yn) = a + lim inf yn.









lim inf(xn + yn) ≥ lim inf xn + lim inf yn

5 If lim xn = a then lim sup(xn + yn) = a + lim sup yn andlim inf(xn + yn) = a + lim inf yn.
































Number 4 is based on the simple fact that if a ≥ xn and b ≥ yn thena + b ≥ xn + yn.

Number 6 is based on the fact that thesup(a + E ) = a + supE for any real a and any set E .












Convergence and closure

The definition of convergence in metric spaces can be rewritten in terms ofnbds: {xn} converges to x if and only if

∀ε > 0,∃N ∈ N, ∀n ≥ N, xn ∈ B(x , ε).

In particular, if {xn} ⊂ A then the limit belongs to A.

This proves the ‘if’ portion of the following

Theorem

x ∈ A if and only if there is a sequence {xn} ⊂ A with xn → x

Proof: Let x ∈ A. Choose any points xn in A ∩ B(x , 1/n). Then0 ≤ d(xn, x) < 1/n→ x .


Convergence and closure

The definition of convergence in metric spaces can be rewritten in terms ofnbds: {xn} converges to x if and only if

∀ε > 0,∃N ∈ N, ∀n ≥ N, xn ∈ B(x , ε).

In particular, if {xn} ⊂ A then the limit belongs to A.

This proves the ‘if’ portion of the following

Theorem

x ∈ A if and only if there is a sequence {xn} ⊂ A with xn → x

Proof: Let x ∈ A. Choose any points xn in A ∩ B(x , 1/n). Then0 ≤ d(xn, x) < 1/n→ x .


Continuity in metric spaces

If (X , d) and (Y , ρ) are metric spaces and f : X → Y then f is continuousat x0 ∈ X iff

∀ε > 0, ∃δ > 0, ∀x with d(x , x0) < δ, ρ(f (x), f (x0)) < ε

∀ε > 0, ∃δ > 0, ∀x ∈ B(x0, δ), f (x) ∈ B(f (x0), ε)

∀ε > 0, ∃δ > 0, f (B(x0, δ)) ⊂ B(f (x0), ε).

The last one can be restated: For every nbd B of f (x0), f −1(B) containssome nbd of x0.

If f is continuous at every point of X we say it is continuous. Making useof the last statement and the definition of open sets, f is continuous ifff −1(V ) is open (in X ) for every open set V in Y .

By taking complements, f is continuous iff for every closed set F ⊆ Y ,f −1(F ) is closed.





∀ε > 0, ∃δ > 0,∀x ∈ B(x0, δ), f (x) ∈ B(f (x0), ε)

∀ε > 0, ∃δ > 0, f (B(x0, δ)) ⊂ B(f (x0), ε).








∀ε > 0, ∃δ > 0, ∀x ∈ B(x0, δ), f (x) ∈ B(f (x0), ε)

∀ε > 0, ∃δ > 0, f (B(x0, δ)) ⊂ B(f (x0), ε).








∀ε > 0, ∃δ > 0, ∀x ∈ B(x0, δ), f (x) ∈ B(f (x0), ε)

∀ε > 0, ∃δ > 0, f (B(x0, δ)) ⊂ B(f (x0), ε).








∀ε > 0, ∃δ > 0, ∀x ∈ B(x0, δ), f (x) ∈ B(f (x0), ε)

∀ε > 0, ∃δ > 0, f (B(x0, δ)) ⊂ B(f (x0), ε).





Subspace metrics

If E ⊂ X we can restrict d to E × E and make E a metric space. Theopen sets in (E , d) are just sets of the form E ∩ U where U is open in X .

This is because balls in E are just E ∩ B(x , r) where x ∈ E and B(x , r) isa ball in X , and open sets are just unions of the open balls they contain.

Thus, a function f : E → Y can be consider for continuity and iscontinuous iff for any open V in Y , f −1(V ) = E ∩ U for some open U inX .

An important special case is when K is compact, because of the following:

Theorem

If K is compact and f : K → Y is continuous then f (K ) is compact.

Proof: Take any open cover V of f (K ). Then {f −1(V ) : V ∈ V} is anoopen cover of K . It therefore has a finite subcover f −1(V1), . . . , f −1(Vn)and then X1, . . . ,Vn is a finite cover of f (K ). .


Subspace metrics

If E ⊂ X we can restrict d to E × E and make E a metric space. Theopen sets in (E , d) are just sets of the form E ∩ U where U is open in X .This is because balls in E are just E ∩ B(x , r) where x ∈ E and B(x , r) isa ball in X , and open sets are just unions of the open balls they contain.



Theorem




Subspace metrics

If E ⊂ X we can restrict d to E × E and make E a metric space. Theopen sets in (E , d) are just sets of the form E ∩ U where U is open in X .This is because balls in E are just E ∩ B(x , r) where x ∈ E and B(x , r) isa ball in X , and open sets are just unions of the open balls they contain.



Theorem




Two important theorems

Theorem (Existence of a maximum)

Suppose K is compact and f : K → R is continuous, thensup f (K ) ∈ f (K ).

Proof: Since f (K ) ⊆ R is bounded, sup f (K ) is finite sincesup f (K ) ∈ f (K ) and f (K ) is closed we have sup f (K ) ∈ f (K ).

Theorem (Intermediate Value Theorem)

If I ⊂ R is an interval and f : I → R is continuous then f (I ) is an interval

There is a general property of sets in metric spaces called connectedness,and a general theorem the continuous functions take connected sets toconnected sets. The above is based on the fact that the connected sets inR are the intervals.

Conbining the two theorems, the continuous image of a closed interval is aclosed interval.





























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Chapter 1 (cont.) · 2020. 8. 26. · Chapter 1 (cont.) Daniel H. Luecking Aug 26, 2020 Daniel H. Luecking Chapter 1 (cont.) Aug 26, 20201/22