# Weaker/Stronger Topologies and Compact/Hausdorff Spaces

In my topology lecture notes, I have written:

"By considering the identity map between different spaces with the same underlying set, it follows that for a compact, Hausdorff space:

$\bullet$ any weaker topology is compact, but not Hausdorff

$\bullet$ any stronger topology is Hausdorff, but not compact "

However, I'm struggling to see why this is. Can anyone shed some light on this?

• You might want to modify in but not necessarily Hausdorff/compact.
– Pece
Commented May 12, 2013 at 15:53
• @Pece: The statements are correct as written. Recall the basic fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Commented May 12, 2013 at 15:54
• @Martin Ok then, any strictly weaker/stronger.
– Pece
Commented May 12, 2013 at 15:58
• @Pece: Okay, we agree :-) Commented May 12, 2013 at 16:00

Weaker implies compact and stronger implies Hausdorff is a general fact (an open cover in the weaker one is an open cover in the original too, pick there a finite subcover and you conclude. If you can separate two point in the original topology you can do it in the richer one because you can use the same open sets).Then suppose $(X, \tau)$ is the topological space from which we start, if $\sigma$ is a weaker topology on $X$ and $(X,\sigma)$ is compact-Hausdorff, then: $Id: (X,\tau) \rightarrow (X,\sigma)$ is a continous bijection between compact-Hausdorff spaces, thus an homeomorphism. As it's the identity we get $\sigma=\tau$; similarly for the other case.
Hint. $X$ being a set, a topology $\tau$ is weaker than a topology $\sigma$ on $X$ if and only if the application $$(X, \sigma) \to (X,\tau), x \mapsto x$$ is continuous.