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

See more here.

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 \(X\).


\(U\subset X\) 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.


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.


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.