K. Jänich, Topology. Chapter I: Fundamental Concepts. 8: Compactness

Definition 1 An open cover of a topological space \(X\) is a set \(\mathcal{U}\) of open sets such that \(\bigcup \mathcal{U} = X\).

Definition 2 A topological space is compact iff every open cover has a finite subcover.

Definition 3 Let \(X\) be a topological space and \(f : X \rightarrow \mathbb{R}\). Then \(f\) is locally bounded iff, for all \(x \in X\), there exists a neighbourhood \(N\) of \(x\) such that \(f\) is bounded on \(N\).

Proposition 4 Let \(X\) be a topological space and \(f : X \rightarrow \mathbb{R}\). If \(f\) is continuous then \(f\) is locally bounded.

Proof Assume \(f\) is continuous. Let \(x \in X\). Then \(f^{-1}(f(x) - 1, f(x)+1)\) is a neighbourhood of \(x\)

Proposition 4 Let \(X\) be a compact space and \(f : X \rightarrow \mathbb{R}\). If \(f\) is locally bounded then \(f\) is bounded.

Proof Assume \(f\) is locally bounded. Then \( \( U \text{ open in \} X : f \text{ is bounded on } U \} \) is an open cover of \(X\). Pick a finite subcover \( \{ U_1, \ldots, U_n \} \). For \( i = 1, \ldots, n \), pick \(m_i, M_i \in \mathbb{R}\) such that \(f(U_i) \subseteq (m_i, M_i)\). Let \(M = \max(M_1, \ldots, M_n) \) and \(m = \min(m_1, \ldots, m_n)\). Then \(f(X) \subseteq (m,M)\). \(\Box\)

Definition 5 Let \(X\) be a topological space and \(Y\) a metric space. Let \((f_n : X \rightarrow Y)\) be a sequence of functions. Then \((f_n)\) converges uniformly to \(f : X \rightarrow Y\) iff, for all \( \epsilon > 0 \), there exists \(N\) such that, for all \(x \in X\) and \(n \geq N\), we have \(d(f_n(x),f(x)) < \epsilon\).

Definition 6 Let \(X\) be a topological space and \(Y\) a metric space. Let \((f_n : X \rightarrow Y)\) be a sequence of functions. Then \((f_n)\) is locally uniformly convergent with limit \(f : X \rightarrow Y\) iff, for all \(x \in X\), there exists a neighbourhood \(N\) of \(x\) such that \((f_n \restriction N)\) converges uniformly to \(f \restriction N\).

Proposition 7 Let \(X\) be a topological space and \(Y\) a metric space. Let \((f_n : X \rightarrow Y)\) be a sequence of functions and \(f : X \rightarrow Y\). If \(X\) is compact and \((f_n)\) is locally uniformly convergent with limit \(f\), then \((f_n)\) converges uniformly to \(f\).

Proof We have \(\{ U \text{ open in } X : (f_n \restriction U) \text{ converges uniformly to } f \restriction U \}\) is an open cover of \(X\). Pick an open subcover \(\{U_1, \ldots, U_k\}\). Let \(\epsilon > 0\). Pick \(N_1\), ..., \(N_k\) such that, for all \(x \in U_i\) and \(n \geq N_i\), we have \(d(f_n(x),f(x)) < \epsilon\). Let \(N = \max(N_1, \ldots, N_k)\). Then, for all \(x \in X\) and \(n \geq N\), we have \(d(f_n(x),f(x))<\epsilon\). \(\Box\)

Definition 8Let \(X\) be a topological space and \(\{A_i\}_{i \in I}\) a family of subsets. Then \(\{A_i\}_{i \in I}\) is locally finite iff, for all \(x \in X\), there exists a neighbourhood \(N\) of \(x\) that intersects \(A_i\) for only finitely many \(i\).

Proposition 9 Let \(X\) be a topological space and \(\{A_i\}_{i \in I}\) a family of subsets. If \(X\) is compact and \(\{A_i\}_{i \in I}\) is locally finite then \(\{i \in I : A_i \neq \emptyset\}\) is finite.

Proof \(\{U \text{ open in } X : U \text{ intersects } A_i \text{ for only finitely many } i\}\) is an open cover of \(X\). Pick a finite subcover \(\{U_1, \ldots, U_n\}\). For \(j = 1, \ldots, n\), let \(I_j = \{ i \in I : U_j \text{ intersects } A_i \}\), which is finite. Then \(\{ i \in I : A_i \neq \emptyset\} = I_1 \cup \cdots \cup I_n\). \(\Box\)

Definition 10 Let \(X\) be a topological space and \(A \subseteq X\). Then \(A\) is locally finite iff every point in \(X\) has a neighbourhood whose intersection with \(A\) is finite.

Proposition 11 Let \(X\) be a topological space and \(A \subseteq X\). If \(X\) is compact and \(A\) is locally finite then \(A\) is finite.

Proof \(\{U \text{ open in \} X : U \cap A \text{ is finite} \}\) is an open cover of \(X\). Pick a finite subcover \(\{U_1, \ldots, U_n\}\). Then \(U_i \cap A\) is finite for each \(i\) and so \(A = (U_1 \cap A) \cup \cdots \cup (U_n \cap A)\) is finite. \(\Box\)

Example 5 in this section I cannot understand - it involves differentiable vector fields on manifolds, which I do not have enough topology to understand yet.

Proposition 12 For every open cover \(\mathcal{U}\) of \([0,1]\), there exists \(\delta > 0\), called a Lebesgue number for \(\mathcal{U}\), such that every subinterval of length \(\delta\) is included in some element of \(\mathcal{U}\).

Proof Suppose for a contradiction this is not true. For every positive integer \(n\), pick a subinterval \(I_n\) of length \(1/n\) which is not included in any element of \(\mathcal{U}\). Let \((m_n)\) be the sequence consisting of the midpoints of the intervals \(I_n\). Pick a convergent subsequence \((m_{r_n})\) (by the Bolzano-Weierstrass Theorem). Let \(l\) be its limit. Pick \(U \in \mathcal{U}\) such that \(l \in U\). Pick \( \epsilon > 0 \) such that \((l - \epsilon, l + \epsilon) \subseteq U\). Pick \(N\) large enough that \(d(m_{r_N},l) < \epsilon / 2\) and \(1/2N < \epsilon / 2\). Then \(I_{r_N} = (m_{r_N} - 1/2r_N, m_{r_N} + 1/2r_N) \subseteq (m_{r_N} - 1/2N, m_{r_N} + 1/2N) \subseteq (l-\epsilon,l+\epsilon) \subseteq U\), contradicting the fact that \(I_{r_N}\) is not included in any element of \(\mathcal{U}\). \(\Box\)

Proposition 13 The interval \([0,1]\) is compact.

Proof Let \(\mathcal{U}\) be any open covering of \([0,1]\). Pick a Lebesgue number \(\delta\) for \(\mathcal{U}\). Pick a finite covering \(\{I_1, \ldots, I_n\}\) of \([0,1]\) by intervals of length \(\delta\). For \(i=1, \ldots, n\), pick \(U_i \in \mathcal{U}\) such that \(I_i \subseteq U_i\). Then \(\{U_1, \ldots, U_n\}\) covers \([0,1]\). \(\Box\)

Proposition 14 The continuous image of a compact space is compact.

Proof Let \(X\) be a compact space. Let \(Y\) be a topological space and \(f : X \rightarrow Y\) a surjective continuous function. Let \(\mathcal{U}\) be an open cover of \(Y\). Then \(\{ f^{-1}(U) : U \in \mathcal{U} \}\) is an open cover of \(X\). Pick a finite subcover \(\{f^{-1}(U_1), \ldots, f^{-1}(U_n)\}\). Then \(\{U_1, \ldots, U_n\}\) is a finite subcover of \(Y\). \(\Box\)

Proposition 15 Let \(X\) be a topological space and \(Y\) a subspace of \(X\). Then \(Y\) is compact if and only if, for every set \(\mathcal{U}\) of open sets in \(X\) that covers \(Y\) (i.e. \(Y \subseteq \bigcup \mathcal{U}\)), there is a finite subset of \(\mathcal{U}\) that covers \(Y\).

Proof Assume \(Y\) is compact. Let \(\mathcal{U}\) be a set of open sets in \(X\) that covers \(Y\). Then \(\{U \cap Y : U \in \mathcal{U}\}\) is an open cover of \(Y\) and so has a finite subcover \(\{U_1 \cap Y, \ldots, U_n \cap Y\}\), say. Then \(\{U_1, \ldots, U_n\}\) is a finite subset of \(\mathcal{U}\) that covers \(Y\).

Conversely, assume that every set of open sets in \(X\) that covers \(Y\) has a finite subset that covers \(Y\). Let \(\mathcal{U}\) be an open cover of \(Y\). Then \(\{U \text{ open in } X : U \cap Y \in \mathcal{U}\}\) is a set of open sets in \(X\) that covers \(Y\). Pick a finite subset \(\{U_1, \ldots, U_n\}\) that covers \(Y\). Then \(\{U_1 \cap Y, \ldots, U_n \cap Y\}\) is a finite subcover of \(\mathcal{U}\). \(\Box\)

Proposition 16 A closed subspace of a compact space is compact.

Proof Let \(X\) be a compact space and \(Y\) be closed in \(X\). Let \(\mathcal{U}\) be a set of open sets in \(X\) that covers \(Y\). Then \(\mathcal{U} \cup \{X - Y\}\) is an open cover of \(X\). Pick a finite subcover, which may have the form \(\{U_1, \ldots, U_n\}\) or \(\{U_1, \ldots, U_n, X-Y\}\) where \(U_1, \ldots, U_n \in \mathcal{U}\). Then \(\{U_1, \ldots, U_n\}\) is a finite subset of \(\mathcal{U}\) that covers \(Y\). \(\Box\)

Proposition 17 Let \(X\) and \(Y\) be nonempty topological spaces. Then the following are equivalent.

1. \(X\) and \(Y\) are compact.
2. \(X+Y\) is compact.
3. \(X \times Y\) is compact.

Proof

\(1 \Rightarrow 2\) Assume \(X\) and \(Y\) are compact. Let \(\mathcal{U}\) be an open cover of \(X + Y\). Then \(\{U : \exists V. U + V \in \mathcal{U} \}\) is an open cover of \(X\), and \(\{V : \exists U. U + V \in \mathcal{U}\}\) is an open cover of \(Y\). Pick finite subcovers \(\{U_1, \ldots, U_n\}\) of the first and \(\{V'_1, \ldots, V'_m\}\) of the second. Pick sets \(V_1\), ..., \(V_n\) such that \(U_1 + V_1, \ldots, U_n + V_n \in \mathcal{U}\) and sets \(U'_1\), ..., \(U'_m\) such that \(U'_1 + V'_1, \ldots, U'_m + V'_m \in \mathcal{U}\). Then \(\{U_1 + V_1, \ldots, U_n + V_n, U'_1 + V'_1, \ldots, U'_m + V'_m \}\) is a finite subcover of \(\mathcal{U}\).

\(2 \Rightarrow 1\) This holds since \(U + \emptyset\) and \(\emptyset + V\) are closed subspaces of \(X + Y\).

\(1 \Rightarrow 3\) Assume \(X\) and \(Y\) are compact. Let \(\mathcal{W}\) be an open cover of \(X \times Y\).

Step 1For every \(x \in X\), there exists a neighbourhood \(U\) of \(x\) such that \(U \times Y\) is covered by a finite subset of \(\mathcal{W}\).

Proof of Step 1Let \(x \in X\). For all \(y \in Y\), there exists \(W \in \mathcal{W}\) such that there exist neighbourhoods \(U\) of \(x\) and \(V\) of \(y\) such that \(U \times V \subseteq W\). Hence \(\{V \text{ open in } Y : \exists U \text{ a neighbourhood of } x. \exists W \in \mathcal{W}. U \times V \subseteq W \} \) is an open cover of \(Y\).

Pick a finite subcover \(\{V_1, \ldots, V_n\}\). For \(i = 1, \ldots, n\), pick a neighbourhood \(U_i\) of \(x\) and \(W_i \in \mathcal{W}\) such that \(U_i \times V_i \subseteq W_i\). Let \(U = U_1 \cap \cdots \cap U_n\). Then \(U\) is a neighbourhood of \(x\) and \(U \times Y\) is covered by \(\{W_1, \ldots, W_n\}\). \(\Box\)

So \(\{U \text{ open in } X : U \times Y \text{ is covered by a finite subset of } \mathcal{W} \}\) is an open cover of \(X\). Pick a finite subcover \(\{U_1, \ldots, U_n\}\). Then \(X \times Y = (U_1 \times Y) \cup \cdots \cup (U_n \times Y)\) is covered by a finite subset of \(\mathcal{W}\).

\(3 \Rightarrow 1\) By Proposition 14. \(\Box\)

Proposition 18 A compact subspace of a Hausdorff space is closed.

Proof Let \(X\) be a Hausdorff space and \(Y\) a compact subspace of \(X\). Let \(x \in X - Y\); we shall prove \(x\) is an interior point of \(X - Y\).

For all \(y \in Y\), there exist disjoint open neighbourhoods \(U\) of \(x\) and \(V\) of \(y\). So \(\{V \text{ open in \} X : \exists U \text{ open neighbourhood of } x. U \cap V = \emptyset\}\) is a set of open sets that covers \(Y\). Pick a finite subset \(\{V_1, \ldots, V_n\}\) that covers \(Y\). Pick open neighbourhoods \(U_1, \ldots, U_n\) of \(x\) such that, for each \(i\), we have \(U_i \cap V_i = \emptyset\). Then \(U_1 \cap \cdots \cap U_n\) is an open set with \(x \in U_1 \cap \cdots \cap U_n \subseteq X - Y\) as required. \(\Box\)

Definition 19 Let \(X\) be a metric space and \(A \subseteq X\). Then \(A\) is bounded if and only if there exists \(B > 0\) such that, for all \(x,y \in A\), we have \(d(x,y) < B\).

Proposition 20 A compact subspace of a metric space is bounded.

Proof Let \(C\) be a compact subspace of a metric space \(X\). If \(C\) is empty, it is bounded vacuously. Otherwise, pick \(a \in C\). Then \(\{B_N(a) : N \in \mathbb{Z}^+\}\) is a set of open sets that cover \(C\), and so some finite subset covers \(C\). Therefore, there exists some positive integer \(N\) such that \(C \subseteq B_N(a)\). Hence, for all \(x,y \in C\), we have \(d(x,y) < 2N\). \(\Box\)

Heine-Borel Theorem Let \(A\) be a subspace of \(\mathbb{R}^n\). Then \(A\) is compact if and only if it is closed and bounded.

Proof If \(A\) is compact then \(A\) is closed (Proposition 18) and bounded (Proposition 20). Conversely, assume \(A\) is closed and bounded. Pick \(K > 0\) such that \(A \subseteq [-K,K]^n\). We have \([-K,K]\) is compact (because it is homeomorphic to [0,1]), hence \([-K,K]^n\) is compact (Proposition 17) and so \(A\) is compact (Proposition 16). \(\Box\)

Proposition 21 Let \(X\) be a compact space, \(Y\) a Hausdorff space, and \(f : X \rightarrow Y\). If \(f\) is continuous and bijective, then it is a homeomorphism.

Proof We must prove \(f^{-1}\) is continuous. Let \(C \subseteq X\) be closed. Then \(C\) is compact (Proposition 16), hence \(f(C)\) is compact (Proposition 14), and so \(f(C)\) is closed (Proposition 18). \(\Box\)

Example 22 It is not true in general that a continuous bijection is a homeomorphism. Let \(f : [0, 2 \pi) \rightarrow \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \}\) be the function \(f(\theta) = (\cos \theta, \sin \theta) \). (This function takes the line \([0, 2 \pi)\) and wraps it around the unit circle once.) Then \(f\) is a continuous bijection, but is not a homeomorphism, because the image of \(f\) is compact but its domain is not.