What property on a space guarantees that compact sets are closed
What property on a space guarantees that closed sets are compact?
Show that a continuous bijection from a compact space to a Hausdorff space is necessarily a homeomorphism.
Solution
1. In a Hausdorff space every compact subset is closed.
For a compact \(K\) and \(x\notin K\), separate \(x\) from each point of \(K\) and use compactness to obtain one neighborhood of \(x\) disjoint from \(K\).
2. In a compact space every closed subset is compact.
If \(F\) is closed and \(\mathcal U\) covers \(F\), then \(\mathcal U\cup\{X\setminus F\}\) covers \(X\); a finite subcover restricts to one of \(F\).
3. A continuous bijection \(f:X\to Y\) from compact \(X\) to Hausdorff \(Y\) is a homeomorphism.
For closed \(F\subseteq X\), \(F\) is compact, so \(f(F)\) is compact and therefore closed in \(Y\). Thus \(f\) is a closed bijection, hence \(f^{-1}\) is continuous.
4. For arbitrary subsets \(A\subseteq X\) and \(B\subseteq Y\),\[\boxed{\overline{A\times B}^{\,X\times Y}=\overline A^{\,X}\times\overline B^{\,Y}.}\]
If \((x,y)\) is in the left closure, projections show \(x\in\overline A\) and \(y\in\overline B\). Conversely, if \(x\in\overline A\) and \(y\in\overline B\), every basic neighborhood \(U\times V\) of \((x,y)\) meets \(A\times B\) because \(U\cap A\) and \(V\cap B\) are both nonempty.