Compactness

Question

• Why are infinite products of compact sets not necessarily compact in the box topology?
• Clarify why this is the case: To show $f^{-1}$ is continuous, we want to show that if $Z\subseteq X$ is closed, then $f(Z)$ is also closed.

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Overview §

Compactness is a topological generalization of closed and bounded sets in $\mathbb{R}$, and specifically the Heine-Borel property.

In analysis, and specifically Modern Analysis I, one first defines compactness of a set $A$ using open coverings and then proves that compactness implies both closed and bounded.

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Limit point and sequential compactness §

Definition: Limit point compact

A metric space $(X,d)$ is limit point compact if every infinite subset of $X$ has a limit point.

Definition: Sequentially compact metric space

A metric space $(X,d)$ is called sequentially compact if every sequence has a convergent subsequence.

Theorem (Munkres 28.2): Equivalent forms of compactness in metric spaces

Let $(X,d)$ be a metrizable space. Then the following are equivalent:

• (i) $X$ is compact;
• (ii) $X$ is limit point compact;
• (iii) $X$ is sequentially compact

Note that (i) $⟹$ (ii) in any topological space.

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In metric spaces §

Definition: Open covers in metric spaces

Let $(X,d)$ be a metric space and $A\subseteq Z$. A family of sets $\{G_{\alpha }{\}}_{\alpha \in J}$ is called an open cover of $A$ if all $G_{\alpha }$ are open and if $A\subseteq {\bigcup }_{\alpha \in J}{G}_{\alpha }$.

Definition: Compact sets in metric spaces

Let $(X,d)$ be a metric space. We say that $A\subseteq X$ is compact if every open cover ${\bigcup }_{\alpha \in J}{G}_{\alpha }$ of $A$ has a finite subcover. That is, there exists some other indexing set $K$ such that $J\subseteq K$ and $|K|<\infty$.

Lemma: Hausdorff property for metric spaces

Let $(X,d)$ be a metric space. Given $x,y\in X$ where $x\ne y$, there exist open sets $A,B$ such that $x\in A$ and $y\in B$ and $A\cap B=\varnothing$.

Proposition (Rudin 2.34, 2.35): Compact sets are closed and bounded; closed subsets are also compact

Let $(X,d)$ be a metric space and let $A\subset X$ be compact. Then:

• (i) $A$ is bounded;
• (ii) If $B\subseteq A$ is a closed subset of $A$, then $B$ is also compact;
• (iii) $A$ is closed.

Theorem: Limit point compactness implies subsequential compactness in metric spaces

If $(X,d)$ is a metric space that is limit point compact, then it is also “subsequentially” compact.

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In topological spaces §

Definition: Open cover, compact topological space

An open covering of a topological space $X$ is a collection of subsets $\mathcal{C}\subseteq \mathcal{P}(X)$ such that each $U\in \mathcal{C}$ is open in $X$ and ${\bigcup }_{U\in \mathcal{C}}=X$.

The space $X$ is compact if for every open cover $\mathcal{C}$ of $X$, there exists a sub-collection $\mathcal{C}’\subseteq \mathcal{C}$ such that $\mathcal{C}’$ is finite and also covers $X$.

Lemma (Munkres 26.1):

A subspace $Y\subseteq X$ is compact if and only if every covering of $Y$ by open sets in $X$ contains a finite subcover (more precisely, every collection of open sets in $X$ whose union contains $Y$ has a finite sub-collection whose union also contains $Y$).

Lemma (Munkres 26.2 & 26.3):

• (i) If $X$ is compact and $Y\subseteq X$ is closed, then $Y$ is compact.
• (ii) If $Y\subseteq X$ is compact and $X$ is Hausdorff, then $Y$ is closed in $X$.

Transclude of (Theorem)-Heine-Borel#^282006

Products and compactness §

Transclude of (Theorem)-Tube-lemma#^thm-tube-lemma

Theorem (Munkres 26.7): The product of finitely many compact spaces is compact.

If $X$ and $Y$ are compact, then their product $X\times Y$ is also compact. By induction, the product of finitely many compact spaces is compact.

This holds for arbitrarily many spaces in the product topology, but not for infinite products in the box topology.

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Continuous functions and compactness §

Theorem (Munkres 26.5 & 26.6): Continuous functions and compactness

• (i) If $f:X\to Y$ is continuous and $X$ is compact, then $f(X)$ is compact.
• (ii) Let $f:X\to Y$ be a continuous bijection. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

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Theorem: Extreme value theorem on a metric space

Let $(X,d_X)$ be a metric space and $K\subseteq X$ be a compact subset, and suppose $f:X\to \mathbb{R}$ is a continuous function. Then $f$ attains its maximum and minimum within $K$; that is, there exists $x_0\in K$ such that

$$f(x_0)=\underset{x\in X}{\sup }\{f(x)\}.$$

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Examples §

#wip compact and non-compact intervals of R

The finite complement topology on any set is compact §

If $X$ is a set equipped with the finite complement topology, then each open set in an open covering of $X$ will cover all but finitely many points of $X$. If there are $n$ points that are not covered by a set $U$, then we will need at most $n$ neighborhoods to cover the remaining points, and so we can construct a finite cover with $n+1$ elements.

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Review §

Topology

• When can compactness be used to verify if a continuous bijection is a homeomorphism (i.e., under what conditions do you not need to check that the inverse image is continuous)?
• Are countably infinite products of compact sets also compact?

Modern Analysis I

• Prove the Hausdorff property holds for metric spaces. (Hint: what open sets arise naturally from $x,y$?)

Honors Mathematics B

• Show that $(0,1]$ is not compact. Is $[a,\infty )$ closed and/or compact?

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Proof appendix §

Lemma: Hausdorff property for metric spaces

Let $(X,d)$ be a metric space. Given $x,y\in X$ where $x\ne y$, there exist open sets $A,B$ such that $x\in A$ and $y\in B$ and $A\cap B=\varnothing$.

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Proof from Modern Analysis I.

Let $r=d(x,y)>0$, the distance between $x$ and $y$. Set $A=B_{r/4}(x)$ and $B=B_{r/4}(y)$. Then we have $x\in A,y\in B$, and $A\cap B=\varnothing$ as desired.

Proposition (Rudin 2.34, 2.35): Compact sets are closed and bounded; closed subsets are also compact

Let $(X,d)$ be a metric space and let $A\subset X$ be compact. Then:

• (i) $A$ is bounded;
• (ii) If $B\subseteq A$ is a closed subset of $A$, then $B$ is also compact;
• (iii) $A$ is closed.

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Sketch from Modern Analysis I.

• (iii) Show $A^C$ is open by showing that for all $x\in A^C$, there exists $B_{ϵ}(x)\subseteq A^C$.

Theorem: Limit point compactness implies subsequential compactness in metric spaces

If $(X,d)$ is a metric space that is limit point compact, then it is also “subsequentially” compact.

Link to original

Sketch from Modern Analysis I.

Let $E:=\{x_n|n\in \mathbb{N}\}$ be the set of all points in the sequence.

• Case 1: $E$ is finite.
  • Trick: We can make an (infinite) subsequence from a finite sequence by taking the final value over and over again
  • There exists $a\in E$ such that $x_n=a$ for infinitely many $n$
  • Make a convergent subsequence by choosing all the points that equal $a$
• Case 2: $E$ is infinite.
  • Limit point compactness $⟹$ $E$ has limit point $x\in X$.
  • For all $k\in \mathbb{N}$, exists $x_{n(k)}\ne x$ in $E$ where $d(x_{n(k)},x)<1/k$.
  • Make a convergent subsequence by removing elements from $\{n(k)|k\in \mathbb{N}$} so that $n(k_i)>n(k_j)$ for $i>j$