# Point-Set

## Summary and Topics

Some key high-level topics:

• Connectedness
• Compactness
• Metric spaces
• Hausdorff spaces

The following properties are “pushed forward” through continuous maps, in the sense that if property $$P$$ holds for $$X$$ and $$f:X\to Y$$, then $$f(X)$$ also satisfies $$P$$:

• Compactness
• Separability
• If $$f$$ is surjective:
• Connectedness
• Density

The following are not preserved:

• Openness
• Closedness

## Metric Spaces and Analysis

A decreasing collection of nested nonempty compact sets $$C_1 \supset C_2 \supset \cdots$$ in a topological space $$X$$ is always nonempty.

If $$\left\{{[a_n, b_n] {~\mathrel{\Big\vert}~}n\in {\mathbf{Z}}^{\geq 0}}\right\}$$ is a nested sequence of compact intervals in a topological space $$X$$, then their intersection is nonempty.

If $$X$$ is a complete metric space and the diameters $${\operatorname{diam}}([a_n, b_n]) \overset{n\to\infty}\longrightarrow 0$$, then their intersection contains exactly one point.

A continuous function on a compact set is uniformly continuous.

Take $$\left\{{B_{{\varepsilon}\over 2}(y) {~\mathrel{\Big\vert}~}y\in Y}\right\}\rightrightarrows Y$$, pull back to an open cover of $$X$$, has Lebesgue number $$\delta_L > 0$$, then $$x' \in B_{\delta_L}(x) \implies f(x), f(x') \in B_{{\varepsilon}\over 2}(y)$$ for some $$y$$.

Lipschitz continuity implies uniform continuity (take $$\delta = {\varepsilon}/C$$)

Counterexample to the converse: $$f(x) = \sqrt x$$ on $$[0, 1]$$ has unbounded derivative.

For $$f:X \to Y$$ continuous with $$X$$ compact and $$Y$$ ordered in the order topology, there exist points $$c, d\in X$$ such that $$f(x) \in [f(c), f(d)]$$ for every $$x$$.

A metric space $$X$$ is sequentially compact iff it is complete and totally bounded.

A metric space is totally bounded iff every sequence has a Cauchy subsequence.

A metric space is compact iff it is complete and totally bounded.

If $$X$$ is a complete metric space, $$X$$ is a Baire space: the intersection of countably many dense open sets in $$X$$ is again dense in $$X$$.

## Compactness

$$U\subset X$$ a Hausdorff spaces is closed $$\iff$$ it is compact.

A closed subset $$A$$ of a compact set $$B$$ is compact.

• Let $$\left\{{A_i}\right\} \rightrightarrows A$$ be a covering of $$A$$ by sets open in $$A$$.
• Each $$A_i = B_i \cap A$$ for some $$B_i$$ open in $$B$$ (definition of subspace topology)
• Define $$V = \left\{{B_i}\right\}$$, then $$V \rightrightarrows A$$ is an open cover.
• Since $$A$$ is closed, $$W\coloneqq B\setminus A$$ is open
• Then $$V\cup W$$ is an open cover of $$B$$, and has a finite subcover $$\left\{{V_i}\right\}$$
• Then $$\left\{{V_i \cap A}\right\}$$ is a finite open cover of $$A$$.

The continuous image of a compact set is compact.

Let $$f:X\to f(X)$$ be continuous. Take an open covering $$\mathcal{U} \rightrightarrows f(X)$$, then $$f^{-1}(\mathcal{U}) \rightrightarrows X$$, which is cover by opens since $$f$$ is continuous. Take a finite subcover by compactness of $$X$$, then they push forward to a finite subcover of $$f(X)$$.

A closed subset of a Hausdorff space is compact.

## Separability

A retract of a Hausdorff/connected/compact space is closed/connected/compact respectively.

Points are closed in $$T_1$$ spaces.

## Maps and Homeomorphism

A continuous bijection $$f: X\to Y$$ with $$X$$ is compact and $$Y$$ is Hausdorff is a homeomorphism.

Show that $$f^{-1}$$ is continuous by showing $$f$$ is a closed map. If $$A\subseteq X$$ is closed in a compact space, $$A$$ is compact. The continuous image of a compact set is compact, so $$f(A)$$ is compact. A compact set in a Hausdorff space is closed, so $$f(A)$$ is closed in $$Y$$.

Every space has at least one retraction - for example, the constant map $$r:X \to\left\{{x_0}\right\}$$ for any $$x\_0 \in X$$.

A continuous bijective open map is a homeomorphism.

For $$f:X\to Y$$, TFAE:

• $$f$$ is continuous
• $$A\subset X \implies f({ \operatorname{cl}}_X(A)) \subset { \operatorname{cl}}_X(f(A))$$
• $$B$$ closed in $$Y \implies f^{-1}(B)$$ closed in $$X$$.
• For each $$x\in X$$ and each neighborhood $$V \ni f(x)$$, there is a neighborhood $$U\ni x$$ such that $$f(U) \subset V$$.

See Munkres page 104.

If $$f:X\to Y$$ is continuous where $$X$$ is compact and $$Y$$ is Hausdorff, then

• $$f$$ is a closed map.
• If $$f$$ is surjective, $$f$$ is a quotient map.
• If $$f$$ is injective, $$f$$ is a topological embedding.
• If $$f$$ is bijective, it is a homeomorphism.

## The Tube Lemma

Let $$X, Y$$ be spaces with $$Y$$ compact, and let $$x_0\in X$$. Let $$N\subseteq X\times Y$$ be an open set containing the slice $$x_0 \times Y$$, then there is a neighborhood $$W\ni x$$ in $$X$$ such that $$N \supset W\times Y$$:

Compactness in one factor is a necessary condition. For a counterexample, $${\mathbf{R}}^2$$ and let $$N$$ be the set contained between a Gaussian and its reflection across the $$x{\hbox{-}}$$axis. Then no tube about $$y=0$$ is entirely contained within $$N$$:

• For each $$y\in Y$$ choose neighborhoods $$A_y, B_y \subseteq Y$$ such that \begin{align*} (x, y) \in A_y \times B_y \subseteq U .\end{align*}
• By compactness of $$Y$$, reduce this to finitely many $$B_y \rightrightarrows Y$$ so $$Y = \bigcup_{j=1}^n B_{y_j}$$
• Set $$O\coloneqq\cap_{j=1}^n B_{y_j}$$; this works.
\todo[inline]{Check this proof!}

• Let $$\left\{{U_j\times V_j {~\mathrel{\Big\vert}~}j\in J}\right\} \rightrightarrows X\times Y$$.
• Fix a point $$x_0\in X$$, then $$\left\{{x_0}\right\}\times Y \subset N$$ for some open set $$N$$.
• By the tube lemma, there is a $$U^x \subset X$$ such that the tube $$U^x \times Y \subset N$$.
• Since $$\left\{{x_0}\right\}\times Y \cong Y$$ which is compact, there is a finite subcover $$\left\{{U_j \times V_j {~\mathrel{\Big\vert}~}j\leq n}\right\} \rightrightarrows\left\{{x_0}\right\}\times Y$$.
• “Integrate the $$X$$”: write \begin{align*}W = \cap_{j=1}^n U_j,\end{align*} then $$x_0 \in W$$ and $$W$$ is a finite intersection of open sets and thus open.
• Claim: $$\left\{{U_j \times V_j {~\mathrel{\Big\vert}~}j\leq n}\right\}\rightrightarrows W\times Y$$
• Let $$(x, y) \in W\times Y$$; want to show $$(x, y)\in U_j \times V_j$$ for some $$j\leq n$$.
• Then $$(x_0, y) \in \left\{{x_0}\right\}\times Y$$ is on the same horizontal line
• $$(x_0, y)\in U_j \times V_j$$ for some $$j$$ by construction
• So $$y\in V_j$$ for this $$j$$
• Since $$x\in W$$, $$x\in U_j$$ for every $$j$$, thus $$x\in U_j$$.
• So $$(x, y) \in U_j \times V_j$$

## “Analysis”-esque Results in Topology

$${\mathbf{Q}}\subset {\mathbf{R}}$$ is not open and not closed.

\

This follows because every neighborhood of $$q\in {\mathbf{Q}}$$ contains an irrational and every neighborhood of $$q' \in {\mathbf{R}}\setminus{\mathbf{Q}}$$ contains a rational.

## Exercises

Show that any map between compact Hausdorff spaces must be proper.

If $$Y$$ is compact Hausdorff, $$U$$ is compact iff $$U$$ is closed, and by continuity closed sets pull back.