J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 26: Compact Spaces

Definition

Let \(X\) be a set and \(\mathcal{A} \subseteq \mathcal{P} X\). Then \(\mathcal{A}\) covers \(X\), or is a covering of \(X\), iff \(\bigcup \mathcal{A} = X\). Given \(Y \subseteq X\), it covers \(Y\) iff \(Y \subseteq \bigcup \mathcal{A}\).

Definition

Let \(X\) be a topological space and \(\mathcal{A} \subseteq \mathcal{P} X\). Then \(\mathcal{A}\) is an open covering of \(X\) iff it covers \(X\) and every element of \(\mathcal{A}\) is open.

Definition

A topological space \(X\) is compact iff every open covering has a finite subset that also covers \(X\).

Lemma 26.1

Let \(X\) be a topological space. Let \(Y\) be a subspace of \(X\). Then \(Y\) is compact if and only if every covering of \(Y\) by sets open in \(X\) has a finite subset that covers \(Y\).

Proof

• If \(Y\) is compact, then every covering of \(Y\) by sets open in \(X\) has a finite subset that covers \(Y\).
• Assume \(Y\) is compact.
• Let \(\mathcal{U}\) be a covering of \(Y\) by sets open in \(X\).
• \( \{ U \cap Y \mid U \in \mathcal{U} \}\) is an open covering of \(Y\).
• Pick a finite subcovering \(\{U_1 \cap Y, \ldots, U_n \cap Y\}\), say.
• \(\{U_1, \ldots, U_n\}\) is a finite subset of \(\mathcal{U}\) that covers \(Y\).
• If every covering of \(Y\) by sets open in \(X\) has a finite subset that covers \(Y\), then \(Y\) is compact.
• Assume every covering of \(Y\) by sets open in \(X\) has a finite subset that covers \(Y\).
• Let \(\mathcal{U}\) be an open covering of \(Y\).
• \(\{U \text{ open in } X \mid U \cap Y \in \mathcal{U} \}\) is a covering of \(Y\) by sets open in \(X\).
• Pick a finite subset \(\{U_1, \ldots, U_n\}\) that covers \(Y\).
• \(\{U_1 \cap Y, \ldots, U_n \cap Y\}\) is a finite subset of \(\mathcal{U}\) that covers \(Y\).
• \(\Box\)

Theorem 26.2

Every closed subspace of a compact space is compact.

Proof

• Let \(X\) be a compact space.
• Let \(Y\) be a closed subspace of \(X\).
• Let \(\mathcal{U}\) be a covering of \(Y\) by sets open in \(X\).
• \(\mathcal{U} \cup \{X-Y\}\) is an open covering of \(X\).
• Pick a finite subcovering \(\{U_1, \ldots, U_n\}\) or \(\{U_1, \ldots, U_n, X - Y\}\) where \(U_1, \ldots, U_n \in \mathcal{U}\).
• \(\{U_1, \ldots, U_n\}\) is a finite subset of \(\mathcal{U}\) that covers \(Y\).
• \(\Box\)

Lemma 26.4

Let \(X\) be a Hausdorff space. Let \(Y\) be a compact subspace of \(X\). Let \(x_0 \in X - Y\). Then there exist disjoint open sets \(U\) and \(V\) with \(x_0 \in U\) and \(Y \subseteq V\).

Proof

• For all \(y \in Y\), there exist disjoint neighbourhoods \(U\) of \(x_0\) and \(V\) of \(y\).
• \(\{V \text{ open in } X \mid \exists U \text{ open in } X. x_0 \in U \wedge U \cap V = \emptyset \}\) is a covering of \(Y\) by sets open in \(X\).
• Pick a finite subcovering \(\{V_1, \ldots, V_n\}\).
• Let \(V = V_1 \cup \cdots \cup V_n\)
• For \(i = 1, \ldots, n\), pick a neighbourhood \( U_i \) of \(x\) such that \(U_i \cap V_i = \emptyset\).
• Let \(U = U_1 \cap \cdots \cap U_n\)
• \(U \cap V = \emptyset\)
• \(x_0 \in U\)
• \(Y \subseteq V\)
• \(\Box\)

Theorem 26.3

Every compact subspace of a Hausdorff space is closed.

Proof

• Let \(X\) be a Hausdorff space.
• Let \(Y\) be a compact subspace of \(X\).
• Let \(x \in X - Y\).
• Pick disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(Y \subseteq V\).
• \(x \in U \subseteq X - Y\)
• \(\Box\)

Example 6

We cannot remove the hypothesis that the space is Hausdorff. Give \(\mathbb{R}\) the finite complement topology. Then every subspace of \(\mathbb{R}\) is compact, but only the finite subsets are closed.

Theorem 26.5

The continuous image of a compact space is compact.

Proof

• Let \(X\) be a compact space and \(Y\) a topological space. Let \(f : X \rightarrow Y\) be a continuous surjection. Prove: \(Y\) is compact.
• Let \(\mathcal{U}\) be an open covering of \(Y\).
• \(\{ f^{-1}(U) \mid U \in \mathcal{U} \}\) is an open covering of \(X\).
• Pick a finite subcovering \(\{f^{-1}(U_1), \ldots, f^{-1}(U_n)\}\).
• \(\{U_1, \ldots, U_n\}\) is a finite subset of \(\mathcal{U}\) that covers \(Y\).
• \(\Box\)

Theorem 26.6

Let \(f : X \rightarrow Y\) be a bijective continuous function. If \(X\) is compact and \(Y\) is Hausdorff then \(f\) is a homeomorphism.

Proof

• Let \(A\) be closed in \(X\). Prove: \(f(A)\) is closed in \(Y\).
• \(A\) is compact.
• Theorem 26.2
• \(f(A)\) is compact.
• Theorem 26.5
• \(f(A)\) is closed.
• Theorem 26.3
• \(\Box\)

Lemma 26.8 (The Tube Lemma)

Let \(X\) be a topological space and \(Y\) a compact space. Let \(x_0 \in X\). Let \(N\) be an open set in \(X \times Y\) that includes \(\{x_0\} \times Y\). Then there exists a neighbourhood \(W\) of \(x_0\) such that \(W \times Y \subseteq N\).

Proof

• For all \(y \in Y\), there exist neighbourhoods \(U\) of \(x_0\) and \(V\) of \(y\) such that \(U \times V \subseteq N\).
• \(\{ V \text{ open in } Y \mid \exists U \text{ open in } X. x_0 \in U \wedge U \times V \subseteq N \}\) is an open covering of \(Y\).
• Pick a finite subcovering \(\{V_1, \ldots, V_n\}\).
• For \(i =1, \ldots, n\), pick a neighbourhood \(U_i\) of \(x_0\) such that \(U_1 \times V_1 \subseteq N\).
• Let \(W = U_1 \cap \cdots \cap U_n\).
• \(W \times Y \subseteq N\).
• \(\Box\)

Theorem 26.7

The product of two compact spaces is compact.

Proof

• Let \(X\) and \(Y\) be compact spaces.
• Let \(\mathcal{A}\) be an open covering of \(X \times Y\).
• \(\{ W \text{ open in } X \mid W \times Y \text{ can be covered by finitely many elements of } \mathcal{A} \}\) is an open covering of \(X\).
• Let \(x_0 \in X\).
• \(\{x_0\} \times Y\) is compact.
• It is homeomorphic to \(Y\).
• Pick finitely many elements \(A_1, \ldots, A_n \in \mathcal{A}\) that cover \(\{x_0\} \times Y\).
• Let \(N = A_1 \cup \cdots \cup A_n\).
• \(\{x_0\} \times Y \subseteq N\)
• Pick a neighbourhood \(W\) of \{x_0\} such that \(W \times Y \subseteq N\).
• Tube Lemma
• \(W \times Y\) is covered by \(A_1\), \ldots, \(A_n\).
• Pick a finite subcovering \(\{W_1, \ldots, W_n\}\).
• \(X \times Y = (W_1 \times Y) \cup \cdots \cup (W_n \times Y)\)
• \(X \times Y\) can be covered by finitely many elements of \(\mathcal{A}\).
• \(\Box\)

Definition

Let \(X\) be a set and \(\mathcal{C} \subseteq \mathcal{P} X\). Then \(\mathcal{C}\) has the finite intersection property iff every nonempty finite subset of \(\mathcal{C}\) has nonempty intersection.

Theorem 26.9

Let \(X\) be a topological space. Then \(X\) is compact if and only if, for every nonempty set \(\mathcal{C}\) of closed sets in \(X\) having the finite intersection property, the intersection \(\bigcap \mathcal{C}\) is nonempty.

Proof

This is the dual of the definition of compactness.

Exercises

Exercise 1

(a)

Let \(\mathcal{T}\) and \(\mathcal{T}'\) be two topologies on the set \(X\); suppose that \(\mathcal{T}' \supseteq \mathcal{T}\). What does compactness of \(X\) under one of these topologies imply about compactness under the other?

Solution

If \(\mathcal{T}'\) is compact then \(\mathcal{T}\) is compact. For if \(\mathcal{U} \subseteq \mathcal{T}\) covers \(X\), then since we have \(\mathcal{U} \subseteq \mathcal{T}'\), we can deduce that some finite subset of \(\mathcal{U}\) covers \(X\).

It is not necessarily true that if \(\mathcal{T}\) is compact then \(\mathcal{T}'\) is compact. Take \(\mathcal{T}\) to be the indiscrete topology on \(\mathbb{Z}_+\) and \(\mathcal{T}'\) the discrete topology. Then \(\mathcal{T}\) is compact but \(\mathcal{T}'\) is not.

(b)

Show that if \(X\) is compact Hausdorff under both \(\mathcal{T}\) and \(\mathcal{T}'\), then either \(\mathcal{T}\) and \(\mathcal{T}'\) are equal or they are incomparable.

Solution

If \(\mathcal{T} \subseteq \mathcal{T}'\), then the identity function \(\mathrm{id}_X\) is a continuous bijection from \((X,\mathcal{T}')\) to \((X, \mathcal{T})\), hence a homeomorphism by Theorem 26.6. This means \(\mathcal{T} = \mathcal{T}'\).

Exercise 2

(a)

Show that in the finite complement topology on \(\mathbb{R}\), every subspace is compact.

Solution

• Let \(Y \subseteq \mathbb{R}\).
• Let \(\mathcal{U}\) be a set of open sets in \(\mathbb{R}\) that cover \(Y\).
• Assume w.l.o.g. \(Y\) is nonempty.
• Pick \(y_0 \in Y\).
• Pick \(U_0 \in \mathcal{U}\) such that \(y_0 \in U_0\).
• Let \(Y - U_0 = \{y_1, \ldots, y_n\}\)
• \(Y - U_0\) is finite because \(X - U_0\) is finite.
• For \(i =1, \ldots, n\), pick \(U_i \in \mathcal{U}\) such that \(y_i \in U_i\).
• \(\{U_0, U_1, \ldots, U_n\}\) covers \(Y\).
• \(\Box\)

(b)

If \(\mathbb{R}\) has the topology consisting of all sets \(A\) such that \(\mathbb{R} - A\) is either countable or all of \(\mathbb{R}\), is \([0,1]\) a compact subspace?

Solution

It is not. For \(n \in \mathbb{Z}_+\), let \(A_n = [0,1] - \{1/n, 1/(n+1), 1/(n+2), \ldots\}\). Then \(\{A_n \mid n \in mathbb{Z}_+\}\) is an open covering of \([0,1]\) but it has no finite subcovering.

Exercise 3

Show that a finite union of compact subspaces of \(X\) is compact.

Solution

• Let \(X\) be a topological space and \(Y_1\), \ldots, \(Y_n\) be compact subspaces. Let \(Y = Y_1 \cup \cdots \cup Y_n\).
• Let \(\mathcal{U}\) be a covering of \(Y\) by sets open in \(X\).
• For \(i = 1,\ldots, n\), pick a finite subset \(\mathcal{U}_i \subseteq \mathcal{U}\) that covers \(Y_i\).
• \(\mathcal{U}_1 \cup \cdots \cup \mathcal{U}_n\) is a finite subset of \(\mathcal{U}\) that covers \(Y\).
• \(\Box\)

Exercise 4

Show that every compact subspace of a metric space is bounded. Find a metric space in which not every closed bounded subspace is compact.

Solution

• Let \(X\) be a metric space. Let \(Y\) be a compact subspace of \(X\).
• Assume w.l.o.g. \(Y\) is nonempty.
• Pick \(y_0 \in Y\).
• \(\{B(y_0,n) \mid n \in \mathbb{Z}_+\}\) is a covering of \(Y\) by sets open in \(X\).
• Pick a finite subset \(\{B(y_0,n_1), \ldots, B(y_0,n_k)\}\) that covers \(Y\)
• Let \(N = \max(n_1, \ldots, n_k)\)
• \(Y \subseteq B(y_0,N)\)
• \(Y\) is bounded with diameter \(\leq 2N\)
• \(\Box\)

In \(\mathbb{R}\) under the standard bounded metric, the set \(\mathbb{R}\) is closed and bounded but not compact.

Exercise 5

Let \(A\) and \(B\) be disjoint compact subspaces of the Hausdorff space \(X\). Show that there exist disjoint open sets \(U\) and \(V\) including \(A\) and \(B\) respectively.

Solution

• For all \(a \in A\), there exist disjoint neighbourhoods \(U\) of \(a\) and \(V\) of \(B\).
• Lemma 26.4
• \(\{U \text{ open in } X \mid \exists V \text{ open in } V. B \subseteq V \wedge U \cap V = \emptyset\}\) is a covering of \(A\) by sets open in \(X\).
• Pick a finite subset \(\{U_1, \ldots, U_n\}\) that covers \(A\).
• Let \(U = U_1 \cup \cdots \cup U_n\).
• For \(i = 1, \ldots, n\), pick a neighbourhood \(V_i\) of \(B\) disjoint from \(U_i\).
• Let \(V = V_1 \cap \cdots \cap V_n\).
• \(U \cap V = \emptyset\)
• \(A \subseteq U\)
• \(B \subseteq V\)
• \(\Box\)

Exercise 6

Show that if \(f : X \rightarrow Y\) is continuous, where \(X\) is compact and \(Y\) is Hausdorff, then \(f\) is a closed map.

Proof

• Let \(C \subseteq X\) be closed.
• \(C\) is compact.
• Theorem 26.2
• \(f(C)\) is compact.
• Theorem 26.5
• \(f(C)\) is closed.
• Theorem 26.3
• \(\Box\)

Exercise 7

Show that if \(Y\) is compact, then the projection \(\pi_1 : X \times Y \rightarrow X\) is a closed map.

Solution

• Let \(C \subseteq X \times Y\) be closed.
• Let \(a \in X - \pi_1(C)\). Prove: there exists a neighbourhood \(U\) of \(a\) such that \(U \subseteq X - \pi_1(C)\).
• \(\{a\} \times Y \subseteq (X \times Y) - C\)
• Pick a neighbourhood \(U\) of \(a\) such that \(U \times Y \subseteq (X \times Y) - C\)
• Tube Lemma
• \(U \subseteq X - \pi_1(C)\)
• \(\Box\)

Exercise 8

Theorem. Let \(f : X \rightarrow Y\); let \(Y\) be compact Hausdorff. Then \(f\) is continuous if and only if the graph of \(f\),

\[ G_f = \{(x, f(x)) \mid x \in X\} \enspace , \]

is closed in \(X \times Y\).

Solution

• If \(f\) is continuous then \(G_f\) is closed.
• Assume \(f\) is continuous.
• Let \((x,y) \in (X \times Y) - G_f\)
• \(y \neq f(x)\)
• Pick disjoint neighbourhoods \(U\) of \(y\) and \(V\) of \(f(x)\)
• \((x,y) \in f^{-1}(V) \times U \subseteq (X \times Y) - G_f\)
• If \(G_f\) is closed then \(f\) is continuous.
• Assume \(G_f\) is closed.
• Let \(V\) be open in \(Y\).
• \(G_f \cap (X \times (Y - V))\) is closed.
• \(\pi_1(G_f \cap (X \times (Y - V))) = \{x \in X \mid f(x) \notin V\}\) is closed.
• Exercise 7
• \(f^{-1}(V)\) is open.
• \(\Box\)

Exercise 9

Generalize the Tube Lemma as follows:

Theorem. Let \(A\) and \(B\) be subspaces of \(X\) and \(Y\), respectively. Let \(N\) be an open set in \(X \times Y\) containing \(A \times B\). If \(A\) and \(B\) are compact, then there exist open sets \(U\) and \(V\) in \(X\) and \(Y\), respectively, such that

\[ A \times B \subseteq U \times V \subseteq N \]

Solution

• For all \(a \in A\), there exist neighbourhoods \(U\) of \(a\) and \(V\) of \(B\) such that \(U \times V \subseteq N\).
• For all \(b \in B\), there exist neighbourhoods \(U\) of \(a\) and \(V\) of \(b\) such that \(U \times V \subseteq N\).
• \(\{V \text{ open in } Y \mid \exists U \text{ open in } X. a \in U \wedge U \times V \subseteq N \}\) is a covering of \(B\) by sets open in \(Y\).
• Pick a finite subcovering \(\{V_1, \ldots, V_n\}\).
• For \(i = 1, \ldots, n\), pick a neighbourhood \(U_i\) of \(a\) such that \(U_i \times V_i \subseteq N\).
• Let \(U = U_1 \cap \cdots \cap U_n\)
• Let \(V = V_1 \cup \cdots \cup V_n\)
• \(a \in U\)
• \(B \subseteq V\)
• \(U \times V \subseteq N\)
• \(\{U \text{ open in } X \mid \exists V \text{ open in } Y. B \subseteq V \wedge U \times V \subseteq N \}\) is a covering of \(A\) by sets open in \(X\).
• Pick a finite subcovering \(\{U_1, \ldots, U_n\}\).
• For \(i=1, \ldots, n\), pick a neighbourhood \(V_i\) of \(B\) such that \(U_i \times V_i \subseteq N\).
• Let \(U = U_1 \cup \cdots \cup U_n\).
• Let \(V = V_1 \cap \cdots \cap V_n\).
• \(A \subseteq U\)
• \(B \subseteq V\)
• \(U \times V \subseteq N\)
• \(\Box\)

Exercise 10

(a)

Prove the following partial converse to the Uniform Limit Theorem:

Theorem. Let \(f_n : X \rightarrow \mathbb{R}\) be a sequence of continuous functions, with \(f_n(x) \rightarrow f(x)\) as \(n \rightarrow \infty\) for all \(x \in X\). If \(f\) is continuous, and the sequence \((f_n)\) is monotone increasing, and if \(X\) is compact, then the convergence is uniform.

Solution

• Let \(\epsilon > 0\).
• For all \(x \in X\), there exists a neighbourhood \(U\) of \(x\) such that there exists \(N\) such that \(\forall y \in U. f(y) - f_N(y) < \epsilon\)
• Pick \(N\) such that \(\forall n \geq N. f(x) - f_n(x) < \epsilon / 3\)
• Pick a neighbourhood \(U_1\) of \(x\) such that \(\forall y \in U_1. |f(x) - f(y)| < \epsilon / 3\)
• Pick a neighbourhood \(U_2\) of \(x\) such that \(\forall y \in U_2. |f_N(x) - f_N(y)| < \epsilon / 3\)
• Let \(U = U_1 \cap U_2\)
• \( \forall y \in U. f(y) - f_N(y) < \epsilon\)
\[ \begin{align} f(y) - f_N(y) & \leq |f(y) - f(x)| + (f(x) - f_N(x)) + |f_N(x) - f_N(y)| & (\text{Triangle inequality}) \\ & < \epsilon /3 + \epsilon / 3 + \epsilon / 3 \\ & = \epsilon \end{align} \]
• \(\{U \text{ open in } X \mid \exists N. \forall x \in U. f(x) - f_N(x) < \epsilon \}\) is an open covering of \(X\).
• Pick a finite subcover \(\{U_1, \ldots, U_n\}\)
• For \(i = 1, \ldots, n\), pick \(N_i\) such that \(\forall x \in U_i. f(x) - f_{N_i}(x) < \epsilon\).
• Let \(N = \max(N_1, \ldots, N_n)\)
• For all \(x \in X\) and \(n \geq N\) we have \(f(x) - f_n(x) < \epsilon\)
\[ \begin{align} f(x) - f_n(x) & \leq f(x) - f_N(x) \\ & \leq f(x) - f_{N_i}(x) & (x \in U_i) \\ & < \epsilon \end{align} \]

(b)

Give examples to show that this theorem fails if you delete the requirement that \(X\) be compact, or if you delete the requirement that the sequence be monotone.

Solution

Define \(f_n : [0,1) \rightarrow \mathbb{R}\) by \(f_n(x) = -x^n\). Define \(f : [0,1) \rightarrow \mathbb{R}\) by \(f(x) = 0\). Then \(f_n(x) \rightarrow f(x)\) for all \(x\), each \(f_n\) is continuous, \(f\) is continuous, and \((f_n)\) is monotone, but the convergence is not uniform.

Define \(f_n : [0,1] \rightarrow \mathbb{R}\) by

\[ f_n(x) = \frac{1}{n^3[x-(1/n)]^2 + 1} \]

Let \(f : [0,1] \rightarrow \mathbb{R}\) be the zero function. Then \(f_n(x) \rightarrow f(x)\) for all \(x\), each \(f_n\) is continuous, \(f\) is continuous, and \([0,1]\) is compact, but the convergence is not uniform.

Exercise 11

Let \(X\) be a compact Hausdorff space. Let \(\mathcal{A}\) be a nonempty set of closed connected subspaces of \(X\) that is linearly ordered by inclusion. Then

\[ Y = \bigcap \mathcal{A} \]

is connected.

Solution

• Assume for a contradiction \(C\) and \(D\) form a separation of \(Y\).
• \(C\) and \(D\) are closed in \(X\).
• They are closed in \(Y\) which is closed in \(X\).
• Pick disjoint open sets \(U\) and \(V\) with \(C \subseteq U\) and \(D \subseteq V\).
• Exercise 5.
• \(\bigcap_{A \in \mathcal{A}} (A - (U \cup V)) \) is nonempty.
• \(\{A - (U \cup V)\} \mid A \in \mathcal{A}\) has the finite intersection property.
• Let \(A_1, \ldots, A_n \in \mathcal{A}\)
• Assume w.l.o.g. \(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n\)
• \(A_1 - (U \cup V)\) is nonempty.
• Otherwise \(A_1 \cap U\) and \(A_1 \cap V\) would be a separation of \(A_1\).
• \((A_1 - (U \cup V)) \cap \cdots \cap (A_n - (U \cup V)) = A_1 - (U \cup V) \neq \emptyset\)
• QED
• Theorem 26.9
• This is a contradiction.
• Since \(\bigcap \mathcal{A} = C \cup D \subseteq U \cup V\).
• \(\Box\)

Definition

Let \(X\) and \(Y\) be topological spaces. Let \(p : X \rightarrow Y\). Then \(p\) is a perfect map iff \(p\) is a closed map, continuous, surjective, and for all \(y \in Y\) we have \(p^{-1}(y)\) is compact.

Exercise 12

Let \(p : X \rightarrow Y\) be a perfect map. If \(Y\) is compact then \(X\) is compact.

Solution

• \( \langle 1 \rangle 1 \) For all \(y \in Y\) and every neighbourhood \(U\) of \(p^{-1}(y)\), there exists a neighbourhood \(W\) of \(y\) such that \(p^{-1}(W) \subseteq U\).
• Let \(y \in Y\) and let \(U\) be a neighbourhood of \(p^{-1}(y)\).
• Let \(W = Y - p(X - U)\)
• \(W\) is open
• Since \(p(X-U)\) is closed.
• \(y \in W\)
• We have \(p^{-1}(y) \subseteq U\) so \(y \notin p(X - U)\).
• \(p^{-1}(W) \subseteq U\)
• Let \(x \in p^{-1}(W)\)
• \(p(x) \in W\)
• \(p(x) \notin p(X - U)\)
• \(x \notin X - U\)
• \(x \in U\)
• Let \(\mathcal{U}\) be an open covering of \(X\).
• \(\{W \text{ open in } Y \mid \exists \mathcal{U}_0 \subseteq^{\mathrm{fin}} \mathcal{U}. p^{-1}(W) \subseteq \bigcup \mathcal{U}_0 \}\) is an open covering of \(Y\).
• Let \(y \in Y\)
• Pick \(U_1, \ldots, U_n \in \mathcal{U}\) that cover \(p^{-1}(y)\)
• There exists a neighbourhood \(W\) of \(y\) such that \(p^{-1}(W) \subseteq U_1 \cup \cdots \cup U_n\).
• By \(\langle 1 \rangle 1\).
• Pick a finite subcovering \(\{W_1, \ldots, W_n\}\).
• For \(i = 1, \ldots, n\), pick \(\mathcal{U}_i \subseteq^{\mathrm{fin}} \mathcal{U}\) such that \(p^{-1}(W_i) \subseteq \bigcup \mathcal{U}_i\).
• \(\mathcal{U}_1 \cup \cdots \cup \mathcal{U}_n\) is a finite subset of \(\mathcal{U}\) that covers \(X\).

Exercise 13

Let \(G\) be a topological group.

(a)

Let \(A\) and \(B\) be subspaces of \(G\). If \(A\) is closed and \(B\) is compact, show that \(AB\) is closed.

Solution

• Let \(c \in G - AB\)
• For all \(b \in B\) we have \(cb^{-1} \notin A\)
• For all \(b \in B\), there exist disjoint open sets \(U\) and \(V\) with \(cb^{-1} \in U\) and \(A \subseteq V\).
• For all \(b \in B\), there exist disjoint open sets \(U\) and \(V\) with \(b \in U^{-1}c\) and \(A \subseteq V\).
• \(\{U \text{ open in } G \mid \exists V \text{ open}. A \subseteq V, cU^{-1} \cap V = \emptyset\}\) is a covering of \(B\) by sets open in \(G\).
• Pick a finite subcovering \(\{U_1, \ldots, U_n\}\).
• For \(i = 1, \ldots, n\), pick \(V_i\) open with \(A \subseteq V_i\) and \(cU_i^{-1} \cap V_i = \emptyset\).
• Let \(W = cU_1^{-1} \cap \cdots \cap cU_n^{-1}\)
• \(WB^{-1} \cap A = \emptyset\)
• Let \(w \in W\), \(a \in A\) and \(b \in B\)
• Assume for a contradiction \(wb^{-1} = a\)
• Pick \(i\) such that \(b \in U_i\)
• \(w \in cU_i^{-1}\)
• Pick \(u \in U_i\) such that \(w = cu^{-1}\)
• \(a = cu^{-1}b^{-1} \in cU_i^{-1} \cap V_i\)
• This is a contradiction.
• \(c \in W \subseteq G - AB\)
• \(\Box\)

(b)

Let \(H\) be a subgroup of \(G\); let \(p : G \rightarrow G / H\) be the quotient map. If \(H\) is compact, show that \(p\) is a closed map.

Solution

• Let \(A \subseteq G\) be closed.
• \(AH\) is closed.
• By part (a)
• \(p(A)\) is closed.
• Since \(p^{-1}(p(A)) = AH\) is closed.
• \(\Box\)

(c)

Let \(H\) be a compact subgroup of \(G\). Show that if \(G/H\) is compact then \(G\) is compact.

Solution

• Let \(p : G \rightarrow G/ H\) be the quotient map.
• \(p\) is a perfect map.
• \(p\) is a closed map.
• Part (b).
• \(p\) is continuous.
• Since \(p\) is a quotient map.
• \(p\) is surjective.
• Since \(p\) is a quotient map.
• For all \(xH \in G/H\) we have \(p^{-1}(xH)\) is compact.
• Since \(p^{-1}(xH) = xH\) is homeomorphic to \(H\).
• \(G\) is compact.
• Exercise 12.
• \(\Box\)