2.2–Compactnessmath.furman.edu/~tlewis/math41/Pugh/chap2/sec2.pdfOutline 1 Bolzano-Weierstrass and Heine-Borel Theorems 2 Some examples of compact sets 3 Nested compact sets 4 Continuity

§2.2–Compactness

Tom Lewis

Fall Term 2006

Tom Lewis () §2.2–Compactness Fall Term 2006 1 / 20

Outline

1 Bolzano-Weierstrass and Heine-Borel Theorems

2 Some examples of compact sets

3 Nested compact sets

4 Continuity and compactness

5 Uniform continuity and compactness


Bolzano-Weierstrass and Heine-Borel Theorems

Problem

Construct a sequence of points (pn) in [0, 1] that does not converge.

Construct a sequence of points (pn) in [0, 1] that does not have aconvergent subsequence.

Definition

Let (M, d) be a metric space. A set K ⊂ M is said to be compact orsequentially compact if every sequence of points (pn) in K has asubsequence (pnk

) that converges to a point in K .



Problem



Definition





Problem



Definition





Problem



Definition





Theorem

Every compact set is closed and bounded.

Proof.

Let K be compact.

Let (pn) be a sequence of points in K and suppose that pn → p.Show that p ∈ K .

Fix a point p ∈ M and consider the sets

M1p ⊂ M2p ⊂ M3p ⊂ · · ·

Either one of these sets contains K or we can extract a sequence (pn)satisfying pn ∈ K ∩ (Mnp)c . This sequence cannot have a convergentsubsequence.



Theorem


Proof.

Let K be compact.



M1p ⊂ M2p ⊂ M3p ⊂ · · ·




Theorem


Proof.

Let K be compact.



M1p ⊂ M2p ⊂ M3p ⊂ · · ·




Theorem


Proof.

Let K be compact.



M1p ⊂ M2p ⊂ M3p ⊂ · · ·




Theorem

A closed subset of a compact set is compact.

Proof.

Let K ⊂ M be a compact set in a metric space and let F ⊂ K be closed.We need to show that F is compact.

Let (pn) be a sequence of points in F . Then (pn) is a sequence ofpoints in K .

Thus there exists a subsequence (pnk) converging to a point p in K .

Since F is closed, p ∈ F as well.



Theorem


Proof.







Theorem


Proof.







Theorem


Proof.







Theorem


Proof.







Theorem

Let (xn) be a sequence in the metric space (M, d). If p ∈ M has theproperty that for every ε > 0 the set

{k : xk ∈ Mεp}

is infinite, then there exists a subsequence of (xn) converging to p.

Proof.

Select n1 such that xn1 ∈ M1p. In general, select nk > nk−1 such that

xnk∈ M1/kp.

Then d(xnk, p) < 1/k and therefore xnk

→ p.



Theorem

Let (xn) be a sequence in the metric space (M, d). If p ∈ M has theproperty that for every ε > 0 the set

{k : xk ∈ Mεp}

is infinite, then there exists a subsequence of (xn) converging to p.

Proof.

Select n1 such that xn1 ∈ M1p. In general, select nk > nk−1 such that

xnk∈ M1/kp.

Then d(xnk, p) < 1/k and therefore xnk

→ p.



Theorem

The closed interval [a, b] ⊂ R is compact

Proof.

Let (xn) be a sequence of points in [a, b]. Let

C = {x ∈ [a, b] : xn < x for finitely many n}.

Show that l.u.b.C exists. Let c = l.u.b.C .

Show, by contradiction, that there exists a subsequence of (xn)converging to c .



Theorem


Proof.







Theorem


Proof.







Theorem


Proof.







Theorem

The Cartesian product of two compact sets is compact.

Proof.

Let M and N be metric spaces and let A ⊂ M and B ⊂ N becompact sets. We must show that A× B is compact.

Let xn = (an, bn) be a sequence of points in A× B.

Now carefully analyze the component sequences. Since A is compact,there exist natural numbers 1 ≤ n1 < n2 < n3 < · · · such that (ank

)converges to a point a ∈ A.

Since B is compact, there exist natural numbers1 ≤ m1 < m2 < m3 < · · · taken from the set {n1, n2, n3, · · · } suchthat (bmk

) converges to a point b ∈ B.

Finally (amk, bmk

) converges to (a, b) in A× B.



Theorem


Proof.







Finally (amk, bmk




Theorem


Proof.







Finally (amk, bmk




Theorem


Proof.







Finally (amk, bmk




Theorem


Proof.







Finally (amk, bmk




Theorem


Proof.







Finally (amk, bmk




Theorem


Proof.







Finally (amk, bmk




Corollary

The Cartesian product of m compact sets is compact.

Proof.

Use induction and the previous result.

Corollary

A box[a1, b1]× [a2, b2]× · · · × [am, bm]

is compact in Rm

Proof.

This result is a special case of the previous corollary, since each closedinterval [ai , bi ] is compact in R.



Corollary


Proof.


Corollary

A box[a1, b1]× [a2, b2]× · · · × [am, bm]

is compact in Rm

Proof.




Corollary


Proof.


Corollary

A box[a1, b1]× [a2, b2]× · · · × [am, bm]

is compact in Rm

Proof.




Corollary


Proof.


Corollary

A box[a1, b1]× [a2, b2]× · · · × [am, bm]

is compact in Rm

Proof.




Theorem (Bolzano-Weierstrass)

Any bounded sequence in Rm has a convergent subsequence.

Proof.

A bounded sequence is contained within a box.

Theorem (Heine-Borel)

Every closed and bounded subset of Rm is compact.

Proof.

Let A be a closed and bounded subset of Rm.

Since A is bounded, it is contained in a box, which is a compact setin Rm.

Since A is closed, it is a closed subset of a compact set and thereforecompact.





Proof.




Proof.








Proof.




Proof.








Proof.




Proof.








Proof.




Proof.








Proof.




Proof.






Definition

Let `∞ denote the set of all bounded sequences of real numbers.

Example

Thus (1/n) and (sin(n)) are elements of `∞, being bounded, but (n) is notin `∞, being unbounded.

Definition

Given two sequences a = (an) and b = (bn) in `∞, let

d(a, b) = sup{|ak − bk | : k ≥ 1}

Problem

Let a = (1/n) and b = (2− 1/n). Show d(a, b) = 2.



Definition


Example


Definition


d(a, b) = sup{|ak − bk | : k ≥ 1}

Problem




Definition


Example


Definition


d(a, b) = sup{|ak − bk | : k ≥ 1}

Problem




Definition


Example


Definition


d(a, b) = sup{|ak − bk | : k ≥ 1}

Problem




Theorem

d : `∞ × `∞ → R is a metric.

Remark

Our next problem shows that the Heine-Borel theorem is special to Rm

and therefore has a limited scope. In general closed and bounded sets arenot compact.

Problem

Let 0 ∈ `∞ denote the sequence of zeros: 0, 0, 0, · · · . Show that the set{a ∈ `∞ : d(a, 0) ≤ 1} is closed and bounded but is not compact.



Theorem

d : `∞ × `∞ → R is a metric.

Remark



Problem




Theorem

d : `∞ × `∞ → R is a metric.

Remark



Problem



Some examples of compact sets

Some compact sets

Any finite subset of a metric space.

The union of finitely many compact sets. (Why?)

Any closed subset of a compact set.

The Cartesian product of finitely many compact sets.

The intersection of arbitrarily many compact sets. (Why?)

The unit ball in Rm.

The boundary of a compact set.

The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.



Some compact sets








The set {1/n : n ≥ 1} ∪ {0} ⊂ R.


Nested compact sets

Definition

A collection of sets A1,A2,A3, · · · is said to be nested provided that

A1 ⊃ A2 ⊃ A3 ⊃ · · ·

Problem

For n ≥ 1, let An = (0, 1/n).

Show that the sets are nested.

What is ∩n≥1An?


Nested compact sets

Definition


A1 ⊃ A2 ⊃ A3 ⊃ · · ·

Problem

For n ≥ 1, let An = (0, 1/n).


What is ∩n≥1An?


Nested compact sets

Definition


A1 ⊃ A2 ⊃ A3 ⊃ · · ·

Problem

For n ≥ 1, let An = (0, 1/n).


What is ∩n≥1An?


Nested compact sets

Definition


A1 ⊃ A2 ⊃ A3 ⊃ · · ·

Problem

For n ≥ 1, let An = (0, 1/n).


What is ∩n≥1An?


Nested compact sets

Theorem

The intersection of a nested sequence of compact non-empty sets iscompact and non-empty.

Proof.

Let (Kn) be such a collection of compact sets and let K = ∩Kn.

K is compact, since it is a closed subset of a compact set, K ⊂ K1.

We must show that K is not empty. To see this, choose a pointan ∈ Kn.

Use compactness to extract a subsequence (ank) which converges to a

point a. This point must be in each of the sets Kn and hence in K .


Nested compact sets

Theorem


Proof.







Nested compact sets

Theorem


Proof.







Nested compact sets

Theorem


Proof.







Nested compact sets

Theorem


Proof.







Nested compact sets

Definition

Let (M, d) be a metric space and let S ⊂ M. The extended real number

diamS = sup{d(x , y) : x , y ∈ S}

is called the diameter of S .

Problem

Let S = [0, 1]× [0, 1]× [0, 1] ⊂ R3. Find diamS.

Let S = {(x , y) : |x | ≤ 1} ⊂ R2. Find diamS.


Nested compact sets

Definition




Problem

Let S = [0, 1]× [0, 1]× [0, 1] ⊂ R3. Find diamS.



Nested compact sets

Definition




Problem

Let S = [0, 1]× [0, 1]× [0, 1] ⊂ R3. Find diamS.



Nested compact sets

Definition




Problem

Let S = [0, 1]× [0, 1]× [0, 1] ⊂ R3. Find diamS.



Nested compact sets

Corollary

Let (Kn) be a sequence of non-empty, nested, compact subsets of a metricspace (M, d). If diam(Kn) → 0 as n →∞, then ∩Kn consists of a singlepoint.

Proof.

We know that the intersection K is non-empty. We must show thatK consists of a single point.

Let us suppose that x , y ∈ K with x 6= y .

Then x , y ∈ Kn for each n ≥ 1 and therefore

diamKn ≥ d(x , y) > 0,

in contradiction to the assumption that diam(Kn) → 0 as n →∞.


Nested compact sets

Corollary


Proof.







Nested compact sets

Corollary


Proof.







Nested compact sets

Corollary


Proof.







Nested compact sets

Corollary


Proof.







Continuity and compactness

Theorem

Let (M, dm) and (N, dN) be metric spaces and let f : M → N be acontinuous function. If K ⊂ M is a compact set, then

f (K ) = {f (x) : x ∈ K}

is a compact set in N.

Proof.

Let (yn) be a sequence in f (K ).

Choose xn ∈ K such that yn = f (xn).

Since K is compact, xnk→ x , x ∈ K .

Since f is continuous, f (xnk) = ynk

→ f (x) ∈ f (K ).



Theorem


f (K ) = {f (x) : x ∈ K}


Proof.





→ f (x) ∈ f (K ).



Theorem


f (K ) = {f (x) : x ∈ K}


Proof.





→ f (x) ∈ f (K ).



Theorem


f (K ) = {f (x) : x ∈ K}


Proof.





→ f (x) ∈ f (K ).



Theorem


f (K ) = {f (x) : x ∈ K}


Proof.





→ f (x) ∈ f (K ).



Theorem


f (K ) = {f (x) : x ∈ K}


Proof.





→ f (x) ∈ f (K ).



Theorem

Let (M, d) be a metric space and let f : M → R be a continuous function.If K ⊂ M is a compact set, then there exist points u, v ∈ K such that

f (u) ≤ f (x) ≤ f (v) for all x ∈ K

Proof.

The image set f (K ) is a compact subset of R thus closed andbounded.

Thus m = g.l.b.f (K ) and M = l.u.b.f (K ) exist and are elements off (K ).

Choose u and v such that f (u) = m and f (v) = M. The conclusionfollows.



Theorem



Proof.






Theorem



Proof.






Theorem



Proof.






Theorem



Proof.





Uniform continuity and compactness

Theorem

Let (M, dM) and (N, dN) be metric spaces and let K ⊂ M be compact. Iff : K → N is continuous, then f is uniformly continuous.

Proof.

We will use proof by contradiction.

If not, there exists an ε > 0 such that for each δ > 0 there existpoints x , y ∈ K with dM(x , y) < δ and dN(f (x), f (y)) ≥ ε.

Select (xn) and (yn) in K with dM(xn, yn) < 1/n butdN(f (xn), f (yn)) ≥ ε.

(xnk) converges to some x in K . But (ynk

) must converge to x as well.

By the triangle inequality,

dN(f (xnk), f (ynk

)) ≤ dN(f (xnk), f (x)) + dN(f (x), f (ynk

)).

The right-hand side tends to 0, which provides the contradiction.



Theorem


Proof.







dN(f (xnk), f (ynk


)).




Theorem


Proof.







dN(f (xnk), f (ynk


)).




Theorem


Proof.







dN(f (xnk), f (ynk


)).




Theorem


Proof.







dN(f (xnk), f (ynk


)).




Theorem


Proof.







dN(f (xnk), f (ynk


)).



Documents

2.2–Compactnessmath.furman.edu/~tlewis/math41/Pugh/chap2/sec2.pdfOutline 1 Bolzano-Weierstrass and Heine-Borel Theorems 2 Some examples of compact sets 3 Nested compact sets 4 Continuity