Let \(Y\) be a subset of the topological space \(X\text{.}\) A collection \(\mathcal{A}=\{A_i\}_{i\in I}\) is a covering of \(Y\) if \(Y\subseteq \bigcup_{i\in I} A_i\text{.}\) As a shorthand, we will often say in this case that \(Y=\bigcup_{i\in I}A_i\) is a covering.
Given a property of subsets (e.g., open, closed, etc.), a covering \(\mathcal{A}=\{A_i\}_{i\in I}\) is said to satisfy this property if \(A_i\) does for all \(i\in I\text{.}\)
Definition1.14.2.Compact space.
A topological space \(X\) is compact if every open covering of \(X\) has a finite subcovering: i.e., if \(X=\bigcup_{i\in I}U_i\) is an open covering, then there is a finite set \(\{i_1,i_2,\dots,i_n\}\subseteq I\) such that \(X=\bigcup_{j=1}^n U_{i_j}\text{.}\)
A subset \(Y\subseteq X\) is compact if it is compact with respect to the subspace topology. Equivalently, \(Y\) is compact if any covering \(Y\subseteq \bigcup_{i\in I} U_i\) by open sets of \(X\) has a finite subcovering.
Remark1.14.3.Compactness is a topological property.
If \(f\colon X\rightarrow Y\) is a homeomorphism, then \(X\) is compact if and only if \(Y\) is compact. Indeed \(f\) defines a bijection
between open covers \(\mathcal{U}=\{U_i\}_{i\in I}\) of \(X\) and open covers of \(Y\text{;}\) and furthermore, this maps finite subcovers of \(\mathcal{U}\) to finite subcovers of \(f(\mathcal{U})\text{.}\)
Example1.14.4.Elementary examples.
Any finite set is compact.
Any space with the trivial topology is compact.
A discrete space is compact if and only if it is finite.
\(\R\) is not compact. Consequently, since compactness is a topological property (i.e., preserved by homeomorphisms), no open interval is compact.
Example1.14.5.\(K\cup\{0\}\) is compact.
Show that \(Y=K\cup\{0\}=\{1/n\colon n\in \Z_+\}\cup\{0\}\) is a compact subset of \(\R\text{.}\)
Theorem1.14.6.Finite closed intervals are compact.
Finite closed intervals \([a,b]\) are compact in \(\R\text{.}\)
Proof.
Let \(\mathcal{U}=\{U_i\}_{i\in I}\) be an open covering of \([a,b]\text{.}\) Define \(A\) to be the set of all \(x\in [a,b]\) such that \([a, x]\subseteq \bigcup_{i\in I}U_i\) has a finite subcovering. Note that \(A\) is nonempty, since clearly \(a\in A\text{.}\) I will show further that \(A\) is open and closed in \([a,b]\text{.}\) Since \([a,b]\) is connected, it will follows that \(A=[a,b]\text{,}\) and hence that \([a,b]\subseteq \bigcup_{i\in I}U_i\) has a finite subcovering, as desired.
\(A\) is open.
Suppose \(c\in A\text{.}\) By definition of \(A\) there is a finite subcover \([a,c]\subseteq\bigcup_{j=1}^nU_{i_j}\text{.}\) Without loss of generality, we may assume that \(c\in U_{i_1}\text{.}\) Since \(U_{i_1}\) is open, we can find an \(\epsilon > 0\) such that \(U=(c-\epsilon, c+\epsilon)\cap (a,b)\subseteq U_{i_1}\text{.}\) But then clearly \([a,x]\subseteq \bigcup_{j=1}^nU_{i_j}\) for all \(x\in U\text{,}\) showing that \([a,x]\) admits a finite subcovering for all such \(x\text{,}\) and hence that \(U\subseteq A\text{.}\) This proves \(A\) is open.
\(A\) is closed.
Take \(c\in [a,b]-A\text{.}\) Since \(\{U_i\}_{i\in I}\) is a cover of \([a,b]\text{,}\) we have \(c\in U_{i_0}\) for some \(i_0\in I\text{.}\) Since \(U_{i_0}\) is open, we can find an \(\epsilon > 0\) such that \(U=(c-\epsilon, c+\epsilon)\cap (a,b)\subseteq U_{i_0}\text{.}\) But then for all \(x\in U\) we have \(x\notin A\text{:}\) indeed, if for such an \(x\) we had a finite subcovering \([a,x]\subseteq\bigcup_{j=1}^nU_{i_j}\text{,}\) then \([a,c]\subseteq U_{i_0}\cup \bigcup_{j=1}^nU_{i_j}\) would be a finite subcovering of \([a,c]\text{,}\) contradicting the fact that \(c\notin A\text{.}\)
Theorem1.14.7.Compact implies closed in Hausdorff spaces.
Let \(Y\) be a compact subset of the topological space \(X\text{.}\)
\(Y\) is closed.
Given any \(x\in X-Y\) there are disjoint open sets \(U, V\) such that \(x\in U\) and \(Y\subseteq V\text{.}\)
Proof.
(1) follows directly from (2). To prove the latter, note that for each \(y\in Y\) we can find disjoint open neighborhoods \(U_{x,y}, V_y\) of \(x\) and \(y\text{,}\) respectively. Since \(Y\subseteq \bigcup_{y\in Y}V_y\) is an open covering of \(Y\text{,}\) there is a finite subcovering \(Y\subseteq \bigcup_{i=1}^nV_{y_i}\text{.}\) The open sets
\begin{align*}
U \amp = \bigcap_{i=1}^nU_{x,y_i}\\
V \amp =\bigcup_{i=1}^nV_{y_i}
\end{align*}
satisfy the desired conditions.
Theorem1.14.8.Compactness inherited by closed subspaces.
Let \(Y\) be a closed subset of the topological space \(X\text{.}\) If \(X\) is compact, then \(Y\) is compact.
Proof.
Let \(Y\subseteq \bigcup_{i\in I}U_i\) be an open covering of \(Y\) by open sets of \(X\text{.}\) Since \(Y\) is closed, the collection
Theorem1.14.9.Compactness preserved under continuous image.
Let \(f\colon X\rightarrow Y\) be a continuous function. If \(X\) is compact, then \(f(X)\) is compact.
Proof.
If \(f(X)\subseteq \bigcup_{i\in I}U_i\) is an open covering of \(f(X)\text{,}\) then \(X=\bigcup_{i\in I}f^{-1}(U_i)\) is an open covering of \(X\text{.}\) Since \(X\) is compact, there is a finite subcover \(X=\bigcup_{j=1}^nf^{-1}(U_{i_j})\text{.}\) It follows that \(f(X)\subseteq \bigcup_{j=1}^nU_{i_j}\) is a finite subcover of our original covering \(f(X)\subseteq \bigcup_{i\in I}U_i\text{.}\)
Theorem1.14.10.Compactness and closed maps.
Let \(f\colon X\rightarrow Y\) be continuous. If \(X\) is compact and \(Y\) is Hausdorff, then \(f\) is a closed map.
Proof.
Let \(C\subseteq X\) be closed. Since \(X\) is compact, \(C\) is compact. By Theorem 1.14.9, \(f(C)\) is compact. By Theorem 1.14.7, since \(Y\) is Hausdorff, \(f(C)\) is closed.
Corollary1.14.11.Compactness and homeomorphisms.
Let \(f\colon X\rightarrow Y\) be continuous and bijective. If \(X\) is compact and \(Y\) is Hausdorff, then \(f\) is a homeomorphism.
Proof.
By Theorem 1.14.10, the map \(f\) is closed, hence a homeomorphism.
Definition1.14.12.Finite intersection property.
A collection \(\mathcal{C}=\{C_i\}_{i\in I}\) of subsets of \(X\) satisfies the finite intersection property if every finite subcollection \(\{C_{i_1}, C_{i_2}, \dots, C_{i_n}\}\) has non trivial intersection: i.e., \(\bigcap_{j=1}^n C_{i_j}\ne\emptyset\text{.}\)
Theorem1.14.13.Closed formulation of compactness.
Let \(X\) be a topological space. The following statements are equivalent.
\(X\) is compact.
If \(\mathcal{C}=\{C_i\}_{i\in I}\) is a collection of closed sets that satisfies the finite intersection property, then \(\bigcap_{i\in I}C_i\ne \emptyset\text{.}\)
Corollary1.14.14.Nested closed sets in compact space.
Let \(X\) be a compact space. Given a nested sequence of nonempty closed subsets \(C_1\supseteq C_2\supseteq C_3\supseteq \dots\text{,}\) we have \(\bigcap_{n=1}^\infty C_n\ne \emptyset\text{.}\)
