Point-Set

Summary and Topics

Some key high-level topics:

  • Connectedness
  • Compactness
  • Metric spaces
  • Hausdorff spaces
proposition (The continuous image of a...):

The following properties are “pushed forward” through continuous maps, in the sense that if property P holds for X and f:XY, 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

theorem (Cantor's Intersection Theorem):

A decreasing collection of nested nonempty compact sets C1C2 in a topological space X is always nonempty.

theorem (Cantor's Nested Intervals Theorem):

If {[an,bn] | nZ0} 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 diam([an,bn])n0, then their intersection contains exactly one point.

proposition (Continuous on compact uniformly continuous):

A continuous function on a compact set is uniformly continuous.

proof (?):

Take {Bε2(y) | yY}, 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.

corollary (Lipschitz implies uniformly continuous):

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

remark:

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

theorem (Extreme Value Theorem):

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.

theorem (Sequentially compact if and only if complete and totally bounded):

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

theorem (Totally bounded if and only if Cauchy subsequences exist):

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

theorem (Compact if and only if complete and totally bounded):

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

theorem (Baire):

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

theorem (Closed if and only if compact in Hausdorff spaces):

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

theorem (Closed subset of compact is compact):

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

proof (?):

    
  • 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.
theorem (Continuous image of compact is compact):

The continuous image of a compact set is compact.

proof (?):

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).

theorem (Closed in Hausdorff \implies compact):

A closed subset of a Hausdorff space is compact.

Separability

proposition (Properties preserved under retracts):

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

proposition (?):

Points are closed in T_1 spaces.

Maps and Homeomorphism

theorem (Continuous bijections from compact to Hausdorff are homeomorphisms):

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

proof (?):

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.

remark (On retractions):

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.

theorem (When open maps are homeomorphisms):

A continuous bijective open map is a homeomorphism.

theorem (Characterizations of continuous maps, Munkres 18.1):

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.
proof (?):

See Munkres page 104.

theorem (Maps from compact to Hausdorff spaces, Lee A.52):

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

theorem (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:

figures/image_2021-05-21-00-28-13.png

remark:

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:

figures/image_2021-05-21-01-39-26.png

proof (Sketch):

    
  • 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.
proof (Detailed proof of the Tube Lemma):
\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

proposition (The rationals are neither open nor closed):

{\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

exercise (?):

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

solution:

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