Let \(\mathcal T,\mathcal T'\) be two topologies on \(X\). Show that if \((X,\mathcal T)\) is compact and Hausdorff, \(\mathcal T\subseteq\mathcal T'\), and \(\mathcal T\ne\mathcal T'\), then \((X,\mathcal T')\) is Hausdorff but not compact.
Solution
1. \((X,\mathcal T')\) is Hausdorff.
If \(x\neq y\), there are disjoint \(U,V\in\mathcal T\) with \(x\in U\) and \(y\in V\), and \(U,V\in\mathcal T'\) because \(\mathcal T\subseteq\mathcal T'\).
2. The identity \(i\colon(X,\mathcal T')\to(X,\mathcal T)\) is a continuous bijection.
For \(U\in\mathcal T\), \(i^{-1}(U)=U\in\mathcal T'\).
3. If \((X,\mathcal T')\) is compact, then \(\mathcal T'=\mathcal T\).
Let \(F\) be \(\mathcal T'\)-closed. \(F\) is compact as a closed subset of a compact space, so \(i(F)=F\) is compact in \((X,\mathcal T)\) by step 2, and hence \(\mathcal T\)-closed because \((X,\mathcal T)\) is Hausdorff. Thus every \(\mathcal T'\)-closed set is \(\mathcal T\)-closed, so \(\mathcal T'\subseteq\mathcal T\), and with \(\mathcal T\subseteq\mathcal T'\) we get equality.
4. Q.E.D.
Step 1 gives the Hausdorff property, and since \(\mathcal T\neq\mathcal T'\), step 3 shows that \((X,\mathcal T')\) is not compact.