(Jan ’04 # A3) Let \((X,\tau)\) be a Hausdorff space and let \[\tau'=\{U\subseteq X\mid X\setminus U\text{ is compact}\}\cup\{\varnothing\}.\] Show that \(\tau'\) is a topology on \(X\) and is coarser than \(\tau\). Show that, in general, they need not be equal.
Solution
We first verify the topology axioms for \[ \tau'=\{U\subseteq X:X\setminus U\text{ is compact}\}\cup\{\varnothing\}. \] Clearly \(\varnothing,X\in\tau'\). If \(U_i\in\tau'\), then \[ X\setminus\bigcup_iU_i=\bigcap_i(X\setminus U_i). \] Choose one nonempty \(U_j\) if there is one. The set on the right is closed in the compact set \(X\setminus U_j\), because each compact subset of a Hausdorff space is closed. Hence it is compact. Thus arbitrary unions lie in \(\tau'\). If \(U_1,\dots,U_n\in\tau'\), then \[ X\setminus\bigcap_{k=1}^nU_k=\bigcup_{k=1}^n(X\setminus U_k), \] a finite union of compact sets, hence compact. Thus \(\tau'\) is a topology.
Since \(X\) is Hausdorff, every compact subset of \(X\) is \(\tau\)-closed. Therefore every nonempty \(U\in\tau'\) is \(\tau\)-open, so \(\tau'\subseteq\tau\).
Equality need not hold. Take \(X=\mathbb R\) with its usual topology. Then \((-1,1)\in\tau\), but \[ \mathbb R\setminus(-1,1)=(-\infty,-1]\cup[1,\infty) \] is not compact. Hence \((-1,1)\notin\tau'\), so \(\tau'\ne\tau\).