Question

Let (X,T) be a Hausdorff space and K = {A ⊂ X| A∩K is open ∀K ⊂ X compact}. Show that K is a finer topology than K, and that (X,K) is compact.

Answer 1

The topology is finer because, if $U\subseteq X$ is open then, by definition it is open in subspace topology, infact open for evry compact subset of $X,$ i.e. $U\cap A$ is open in $A.$ Hence $U\in K.$ Hence $K$ has a finer topology. But its not compact if $X$ is not.

Counterexample : Take the real line and endow it with the the standard topology. Then it is Hausdorff, but real line is non-compact. In particular take, $\mathbb{R}=\cup_{n=1}^{\infty} (-n,n)$ . This is a cover in $K$ since it has finer topology. But it has no finite cover. Infact take finitely many and consider the largest $n.$ Then $n+1$ remains uncovered in that collection.