- Show that a closed subset \(A\) of a compact space \(X\) is compact.
Solution
1. Let \(\mathcal U\) be an open cover of the closed subset \(A\subseteq X\).
We prove that \(\mathcal U\) has a finite subcover.
2. Since \(A\) is closed, \(X\setminus A\) is open, and\[\mathcal U\cup\{X\setminus A\}\]is an open cover of \(X\).
The members of \(\mathcal U\) cover \(A\), while \(X\setminus A\) covers every point outside \(A\).
3. Compactness of \(X\) gives a finite subcover of this enlarged cover.
This is the definition of compactness.
4. Removing \(X\setminus A\) if it occurs leaves finitely many members of \(\mathcal U\) that cover \(A\). Hence \(A\) is compact.
The removed set contains no point of \(A\).