Summary and Topics
Some key highlevel 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.