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?

  • $\begingroup$ You might want to modify in but not necessarily Hausdorff/compact. $\endgroup$ – Pece May 12 '13 at 15:53
  • $\begingroup$ @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. $\endgroup$ – Martin May 12 '13 at 15:54
  • $\begingroup$ @Martin Ok then, any strictly weaker/stronger. $\endgroup$ – Pece May 12 '13 at 15:58
  • $\begingroup$ @Pece: Okay, we agree :-) $\endgroup$ – Martin May 12 '13 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.

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks, that cleared things up nicely. I got bogged down trying to utilise results from lectures to show that weaker implies compact and stronger implies Hausdorff rather than just going back to the definitions. For weaker implies not Hausdorff, (correct me if I'm wrong) you're showing that Hausdorff implies topology can't be weaker, yes? $\endgroup$ – lokodiz May 12 '13 at 16:17
  • $\begingroup$ I assume that the topology is weaker (not strictly) and so it's automatically compact; suppose it's hausdorff too and you get that the two topologies are the same (observe that it's not a proof by contradiction, but by contraposition) so that if it's strictly weaker than you can't have hausdorff property. $\endgroup$ – Edoardo Lanari May 12 '13 at 16:25

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.

| cite | improve this answer | |
  • $\begingroup$ Hmm, I knew this already, and I recall working out why this implies that a set with a weaker topology is compact (as the image of a compact set under a continuous function); I just didn't write it down! $\endgroup$ – lokodiz May 12 '13 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.