J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 29: Local Compactness

Definition

A space \(X\) is said to be locally compact at \(x\) if there is some compact subspace \(C\) of \(X\) that contains a neighbourhood of \(x\). If \(X\) is locally compact at each of its points, \(X\) is said to be locally compact.

Example 1

\(\mathbb{R}\) is locally compact. For any point \(x\), the compact subspace \([x-1,x+1]\) includes the neighbourhood \((x-1,x+1)\).

\(\mathbb{Q}\) is not locally compact.

• Assume for a contradiction \(C\) is a compact subspace of \(\mathbb{Q}\) that includes \((-\epsilon,\epsilon) \cap \mathbb{Q}\) for some \(\epsilon > 0\).
• Pick an irrational \(\xi \in (-\epsilon, \epsilon)\).
• Pick a strictly increasing sequence of rationals \((q_n)\) in \((-\epsilon, \epsilon)\) that converges to \(\xi\).
• \(\{(-\infty, q_n) \cap \mathbb{Q} \mid n \in \mathbb{Z}_+\} \cup \{(\xi, +\infty) \cap \mathbb{Q}\}\) is a sequence of open sets in \(\mathbb{Q}\) that covers \(C\).
• No finite subset covers \(C\).
• This is a contradiction.
• \(\Box\)

Example 2

\(\mathbb{R}^n\) is locally compact: for any point \(\vec{x}\), the compact subspace \(\prod_{i=1}^n [x_i - 1, x_1 + 1]\) includes the neighbourhood \(\prod_{i=1}^n (x_i -1, x_1 + 1)\).

\(\mathbb{R}^\omega\) is not locally compact.

• Assume for a contradiction there is a compact subspace of \(\mathbb{R}^\omega\) that includes \(U = \prod_{i=1}^n (-\epsilon, \epsilon) \times \mathbb{R} \times \cdots\).
• \(\overline{U} = \prod_{i=1}^n [-\epsilon, \epsilon] \times \mathbb{R} \times \cdots\) is compact.
• This is a contradiction.
• \(\Box\)

Example 3

Any linearly ordered set \(X\) with the least upper bound property is locally compact in the order topology. Every element has a neighbourhood that is included in a closed interval, which is compact.

Theorem 29.1

Let \(X\) be a topological space. Then \(X\) is locally compact Hausdorff if and only if there exists a space \(Y\) such that:

1. \(X\) is a subspace of \(Y\).
2. The set \(Y - X\) consists of a single point.
3. \(Y\) is a compact Hausdorff space.

If \(Y\) and \(Y'\) are two spaces satisfying these conditions, then there is a unique homeomorphism of \(Y\) with \(Y'\) that is the identity on \(X\).

Proof

• If \(X\) is locally compact Hausdorff, then there exists a space \(Y\) satisfying 1-3.
• Assume \(X\) is locally compact Hausdorff.
• Let \(Y = X + \{\infty\}\) under the topology \(\mathcal{U} = \{U \mid U \text{ is open in } X \} \cup \{ Y - C \\mid C \subseteq X \text{ is compact} \}\).
• \(X\) is a subspace of \(Y\).
• For all \(U\) open in \(X\), there exists \(V \in \mathcal{U}\) such that \(U = V \cap X\).
• Take \(V = U\)
• For all \(V \in \mathcal{U}\), we have \(V \cap X\) is open in \(X\).
• For all \(U\) open in \(X\) we have \(U \cap X\) is open in \(X\).
• Since \(U \cap X = U\)
• For all \(C \subseteq X\) compact we have \((Y - C) \cap X\) is open in \(X\).
• Since \(C\) is closed in \(X\) (Theorem 26.3) and \((Y - C) \cap X = X - C\).
• \(Y - X = \{ \infty \}\)
• \(Y\) is compact.
• Let \(\mathcal{V} \subseteq \mathcal{U}\) be any open covering of \(Y\).
• Pick \(C \subseteq X\) compact such that \(Y - C \in \mathcal{V}\).
• There must be such an element of \(\mathcal{V}\) since \(\infty\) is in some element of \(\mathcal{V}\).
• Pick a finite subset \(\{V_1, \ldots, V_n\}\) of \(\mathcal{V}\) that covers \(C\).
• \(\{V_1, \ldots, V_n, Y - C\} is a finite subset of \(\mathcal{V}\) that covers \(Y\).
• \(Y\) is Hausdorff.
• Let \(x,y \in Y) with \(x \neq y\).
• Case \(x,y \in X\).
• Pick \(U,V\) disjoint and open in \(X\) such that \(x \in U\) and \(y \in V\).
• \(U\) and \(V\) are disjoint open sets in \(Y\) with \(x \in U\) and \(y \in V\).
• Case \(x \in X\) and \(y = \infty\).
• Pick a compact \(C \subseteq X\) that includes a neighbourhood \(U\) of \(x\).
• \(U\) and \(Y - C\) are disjoint open sets in \(Y\) with \(x \in U\) and \(\infty \in Y - C\).
• Case \(x = \infty\) and \(y \in X\).
• Similar.
• If there exists a space \(Y\) satisfying 1-3, then \(X\) is locally compact Hausdorff.
• Assume there exists \(Y\) satisfying 1-3.
• Let \(Y - X = \{\infty\}\).
• \(X\) is locally compact.
• Let \(x \in X\).
• Pick disjoint open sets \(U\) and \(V\) in \(Y\) such that \(x \in U\) and \(\infty \in V\).
• \(Y - V \subseteq X\) is compact and includes the neighbourhood \(U\) of \(x\).
• \(X\) is Hausdorff.
• Theorem 17.1
• If \(Y\) and \(Y'\) are two spaces satisfying 1-3, then there is a unique homeomorphism of \(Y\) with \(Y'\) that is the identity on \(X\).
• Let \(Y - X = \{\infty\}\) and \(Y' - X = \{\infty'\}\).
• Define \(f : Y \rightarrow Y'\) by \(f(x) = x\) for \(x \in X\) and \(f(\infty) = \infty'\).
• \(f\) is continuous.
• Let \(V \subseteq Y'\) be open.
• Case \(\infty' \in Y'\)
• \(Y' - V\) is closed in \(X\)
• \(f^{-1}(V) = Y - (Y' - V)\) is open in \(Y\)
• Case \(\infty' \notin Y'\)
• \(f^{-1}(V) = V\) is open in \(X\), hence in \(Y\).
• \(f^{-1}\) is continuous.
• Similar.
• \(\Box\)

Definition

If \(Y\) is a compact Hausdorff space and \(X\) is a proper subspace of \(Y\) whose closure equals \(Y\), then \(Y\) is said to be a compactification of \(X\). If in addition \(Y-X\) equals a single point, then \(Y\) is called the one-point compactification of \(X\).

Example

\(S^1\) is the one-point compactification of \(\mathbb{R}\). \(S^2\) is the one-point compactification of \(\mathbb{R}^2\). When we think of \(S^2\) as the one-point compactification of \(\mathbb{C}\) we call it the Riemann sphere or extended complex plane.

Theorem 29.2

Let \(X\) be a Hausdorff space. Then \(X\) is locally compact if and only if given \(x \in X\), and given a neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\).

Proof

• If \(X\) is locally compact then, for all \(x \in X\) and every neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\).
• Assume \(X\) is locally compact.
• Let \(Y\) be the space from Theorem 29.1
• Let \(x \in X\).
• Let \(U\) be a neighbourhood of \(x\).
• Let \(C = Y - U\)
• \(C \subseteq X\) is compact.
• Pick disjoint open sets \(V\) and \(W\) in \(Y\) containing \(x\) and \(C\) respectively.
• Lemma 26.4
• \(\overline{V}\) is compact.
• \(\overline{V} \subseteq U\)
• \(\overline{V} \subseteq Y - W \subseteq Y - C = U\)
• \(\overline{V}\) is the closure of \(V\) in \(X\).
• If, for all \(x \in X\) and every neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\), then \(X\) is locally compact.
• Assume that, for all \(x \in X\) and every neighbourhood \(U\) of \(x\), there exists a nighbourhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\).
• Pick a neighbourhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq X\).
• \(\overline{V}\) is a compact subspace of \(X\) that includes a neighbourhood of \(x\).
• \(\Box\)

Corollary 29.3

Let \(X\) be a locally compact Hausdorff space. Let \(A\) be a subspace of \(X\). If \(A\) is closed in \(X\) or open in \(X\), then \(A\) is locally compact.

Proof

• If \{A\) is closed in \(X\) then \(A\) is locally compact.
• Assume \(A\) is closed in \(X\).
• Let \(x \in A\).
• Let \(U\) be a neighbourhood of \(x\) in \(A\).
• Pick a neighbourhood \(U'\) of \(x\) in \(X\) such that \(U = U' \cap A\).
• Pick a neighbourhood \(V\) of \(x\) in \(X\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U'\).
• \(V \cap A\) is a neighbourhood of \(x\) in \(A\).
• The closure of \(V \cap A\) in \(A\) is compact.
• It is \(\overline{V} \cap A\), which is a closed subspace of \(\overline{V}\).
• The closure of \(V \cap A\) is included in \(U\).
• If \(A\) is open in \(X\) then \(A\) is locally compact.
• Assume \(A\) is open in \(X\).
• Let \(x \in A\).
• Let \(U\) be a neighbourhood of \(x\) in \(A\).
• \(U\) is a neighbourhood of \(x\) in \(X\).
• Pick a neighbourhood \(V\) of \(x\) in \(X\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\).
• \(V\) is a neighbourhood of \(x\) in \(A\).
• The closure of \(V\) in \(A\) is \(\overline{V}\).
• The closure of \(V\) in \(A\) is compact and included in \(U\).
• \(\Box\)

Corollary 29.4

A topological space is an open subspace of a compact Hausdorff space if and only if it is locally compact Hausdorff.

Exercises

Exercise 1

Show that the rationals \(\mathbb{Q}\) are not locally compact.

Solution

See above.

Exercise 2

Let \(\{X_\alpha\}_{\alpha \in I}\) be a family of nonempty topological spaces.

(a)

Show that if \(\prod_{\alpha \in I} X_\alpha\) is locally compact, then each \(X_\alpha\) is locally compact and \(X_\alpha\) is compact for all but finitely many \(\alpha\).

Solution

Each \(X_\alpha\) is locally compact because it is homeomorphic to a subspace of \(\prod_{\alpha \in I} X_\alpha\).

• Pick a point \((x_\alpha) \in \prod_{\alpha \in I} X_\alpha\).
• Since every \(X_\alpha\) is nonempty.
• Pick a compact \(C \subseteq X_\alpha\) that includes the neighbourhood \(\prod_{\alpha \in U} U_\alpha\) of \((x_\alpha)\), where each \(U_\alpha\) is open in \(X_\alpha\), and \(U_\alpha = X_\alpha\) for all but finitely many \(\alpha\).
• If \(U_\alpha = X_\alpha\) then \(X_\alpha\) is homeomorphic to a closed subspace of \(C\).
• If \(U_\alpha = X_\alpha\) then \(X_\alpha\) is compact.
• For all but finitely many \(\alpha\), \(X_\alpha\) is compact.
• \(\Box\)

(b)

Prove the converse, assuming the Tychonoff theorem.

Solution

• Assume that every \(X_\alpha\) is locally compact, and \(X_\alpha\) is compact for all but finitely many \(\alpha\).
• For every point \((x_\alpha) \in \prod_\alpha X_\alpha\) and every neighbourhood \(U\) of \((x_\alpha)\), there exists a neighbourhood \(V\) of \((x_\alpha)\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\).
• Let \((x_\alpha) \in \prod_\alpha X_\alpha\) and \(U\) be a neighbourhood of \((x_\alpha)\).
• Assume w.l.o.g. \(U = \prod_\alpha U_\alpha\), where each \(U_\alpha\) is open in \(X_\alpha\), and \(U_\alpha = X_\alpha\) for all but finitely many \(\alpha\).
• \(\langle 2 \rangle 1 \) For all \(\alpha\) such that \(X_\alpha\) is not compact or \(U_\alpha \neq X_\alpha\), pick a neighbourhood \(V_\alpha\) of \(x_\alpha\) such that \(\overline{V_\alpha}\) is compact and \(\overline{V_\alpha} \subseteq U_\alpha\). For the values of \(\alpha\) such that \(X_\alpha\) is compact and \(X_\alpha = U_\alpha\), let \(V_\alpha = X_\alpha\).
• Theorem 29.2
• \(\prod_\alpha V_\alpha\) is a neighbourhood of \((x_\alpha)\).
• Each \(V_\alpha\) is open in \(X_\alpha\), and \(V_\alpha = X_\alpha\) for all but finitely many \(\alpha\).
• \(\langle 2 \rangle 2\) \(\overline{\prod_\alpha V_\alpha} = \prod_\alpha \overline{V_\alpha}\)
• Theorem 19.5
• \(\overline{\prod_\alpha V_\alpha}\) is compact.
• Each \(\overline{V_\alpha}\) is compact.
• If \(X_\alpha\) is not compact this is immediate from \(\langle 2 \rangle 1 \). If \(X_\alpha\) is compact then \(\overline{V_\alpha}\) is a closed subspace of a compact space, hence compact.
• QED
• Tychonoff's Theorem, \(\langle 2 \rangle 2\).
• \(\overline{\prod_\alpha V_\alpha} \subseteq \prod_\alpha U_\alpha\)
• Since each \(\overline{V_\alpha} \subseteq U_\alpha\).
• \(\prod_\alpha X_\alpha\) is locally compact.
• Theorem 29.2
• \(\Box\)

Exercise 6

Show that the one-point compactification of \(\mathbb{R}\) is homeomorphic with the circle \(S^1\).

Solution

We prove that the one-point compactification of \((0,1)\) is homeomorphic with \(S^1\). The result follows because we already know \(\mathbb{R} \cong (0,1)\).

We identify \(S^1\) with \(\{z \in \mathbb{C} \mid |z| = 1\}\).

• Define \(f : (0,1) \rightarrow S^1\) by \(f(\theta) = e^{2 i \pi \theta}\).
• \(S^1 - f((0,1)) = \{1\}\)
• \(f\) is an embedding.
• \(f\) is injective.
• \(f : (0,1) \rightarrow S^1 - \{1\}\) is continuous.
• \(f^{-1} : S^1 - \{1\} \rightarrow (0,1)\) is continuous.
• QED
• By Theorem 29.1

Exercise 7

Show that the one-point compactification of \(\Omega\) is \(\Omega + 1\).

Solution

Since \(\Omega\) is locally compact Hausdorff, \(\Omega + 1\) is compact Hausdorff, and \((\Omega + 1) - \Omega\) is a singleton.

Exercise 8

Show that the one-point compactification of \(\mathbb{Z}_+\) is homeomorphic with the subspace \(\{0\} \cup \{1/n \mid n \in \mathbb{Z}_+\}\) of \(\mathbb{R}\).

Solution

The function that maps \(n\) to \(1/n\) is a topological embedding and \(\{0\} \cup \{1/n \mid n \in \mathbb{Z}_+\}\) is compact Hausdorff.

Exercise 9

Show that if \(G\) is a locally compact topological group and \(H\) is a subgroup, then \(G/H\) is locally compact.

Solution

By Exercise 3, since the canonical projection \(G \rightarrow G / H\) is a continuous open surjection.

Exercise 10

Show that if \(X\) is a Hausdorff space that is locally compact at the point \(x\), then for each neighbourhood \(U\) of \(x\), there is a neighbourhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq U\).

Solution

• Let \(U\) be a neighbourhood of \(x\).
• Pick a compact \(C \subseteq X\) that includes a neighbourhood of \(x\).
• Pick disjoint open subsets \(V\) and \(W'\) in \(C\) such that \(x \in V\) and \(C - U \subseteq W'\)
• Lemma 26.4
• \(\overline{V} \subseteq C\)
• Since \(C\) is closed.
• \(\overline{V}\) is compact.
• \(\overline{V} \subseteq U\)
• \(\overline{V} \subseteq C - W' \subseteq U \)
• \(\Box\)

Exercise 11

Prove the following.

(c)

Lemma. If \(p : X \rightarrow Y\) is a quotient map and \(Z\) is a locally compact Hausdorff space, then the map

\[ \pi = p \times i_Z : X \times Z \rightarrow Y \times Z \]

is a quotient map.

• \(p \times i_Z\) is surjective.
• \(p \times i_Z\) is continuous.
• \(\pi\) is strongly continuous.
• Let \(A \subseteq Y \times Z\).
• Assume \(\pi^{-1}(A)\) is open in \(X \times Z\). Prove: \(A\) is open in \(Y \times Z\).
• Let \((y,z) \in A\)
• Pick \(x \in X\) such that \(p(x) = y\)
• Pick open neighbourhoods \(U_1\) of \(x\) and \(V_1\) of \(z\) such that \(U_1 \times V_1 \subseteq \pi^{-1}(A)\).
• Pick a neighbourhood \(V\) of \(z\) such that \(\overline{V}\) is compact and \(\overline{V} \subseteq V_1\).
• Theorem 29.2
• Let \(U\) be the union of all the open sets \(U'\) in \(X\) such that \(U' \times \overline{V} \subseteq \pi^{-1}(A)\)
• \(U \times V \subseteq \pi^{-1}(A)\)
• \(U\) is saturated
• Let \(a \in U\), \(b \in X\) and assume \(p(a) = p(b)\)
• \(\{b\} \times \overline{V} \subseteq \pi^{-1}(A)\)
• For \(t \in \overline{V}\) we have \(\pi(b,t) = (p(b),t) = (p(a),t) = \pi((a,t)) \in A\).
• Pick a neighbourhood \(U'\) of \(b\) such that \(U' \times \overline{V} \subseteq \pi^{-1}(A)\)
• Tube Lemma
• \(U' \subseteq U\)
• \(b \in U\)
• \(p(U)\) is open in \(Y\)
• \(p\) maps saturated open sets to open sets.
• \((y,z) \in p(U) \times V \subseteq A\)
• \(\Box\)

(b)

Theorem. Let \(p : A \rightarrow B\) and \(q : C \rightarrow D\) be quotient maps. If \(B\) and \(C\) are locally compact Hausdorff spaces then \(p \times q : A \times C \rightarrow B \times D\) is a quotient map.

Solution

Since \(p \times q = (i_B \times q) \circ (p \times 1_C)\).