Prove that a closed subset of a compact topological space is compact. Prove that in a Hausdorff topological space a compact subset is closed.
Solution
1. A closed subset of a compact topological space is compact.
Let \(X\) be compact and let \(F\subseteq X\) be closed. Suppose \[ F\subseteq\bigcup_{i\in I}U_i, \] where every \(U_i\) is open in the subspace \(F\). For each \(i\), choose an open set \(V_i\subseteq X\) such that \[ U_i=F\cap V_i. \] Then \[ \{V_i:i\in I\}\cup\{X\setminus F\} \] is an open cover of \(X\). By compactness of \(X\), finitely many members cover \(X\). Discarding \(X\setminus F\) if it occurs, the corresponding finitely many \(V_i\) still cover \(F\). Intersecting with \(F\), the corresponding finitely many \(U_i\) cover \(F\). Thus every open cover of \(F\) has a finite subcover, so \(F\) is compact.
2. Let \(X\) be Hausdorff, let \(K\subseteq X\) be compact, and let\[ x\in X\setminus K. \]For each \(y\in K\), choose disjoint open sets \(U_y,V_y\subseteq X\) with\[ y\in U_y, \qquad x\in V_y. \]
For every \(y\in K\), we have \(x\ne y\). Since \(X\) is Hausdorff, the two distinct points \(x\) and \(y\) have disjoint open neighborhoods. Name the neighborhood of \(y\) by \(U_y\) and the neighborhood of \(x\) by \(V_y\).
3. The point \(x\) has an open neighborhood disjoint from \(K\).
The family \[ \{U_y:y\in K\} \] is an open cover of \(K\). Compactness gives points \[ y_1,\ldots,y_n\in K \] such that \[ K\subseteq U_{y_1}\cup\cdots\cup U_{y_n}. \] Set \[ V=V_{y_1}\cap\cdots\cap V_{y_n}. \] Then \(V\) is an open neighborhood of \(x\). If \(z\in V\cap K\), then some \(i\) satisfies \(z\in U_{y_i}\), while \(z\in V\) implies \(z\in V_{y_i}\). This contradicts \[ U_{y_i}\cap V_{y_i}=\varnothing. \] Hence \[ V\cap K=\varnothing. \]
4. Every compact subset of a Hausdorff space is closed.
By 3, every point of \(X\setminus K\) has an open neighborhood contained in \(X\setminus K\). Therefore \(X\setminus K\) is open, and hence \(K\) is closed.