Details about the question between strong/weaker topologies and compact/Hausdorff

general-topology

I've just learned a corollary says

A continuous bijective map from a compact space onto a Hausdorff space is a homeomorphism.

Then I want to prove that for a compact Hausdorff topology $\omega$ on a set $X$, if $\omega_0$ is strictly stronger (weaker) than $\omega$, than is must be Hausdorff (compact) but not compact (Hausdorff).

It's obvious to see if $\omega_0$ is weaker then it is Hausdorff, since a compact subset of a Hausdorff space is closed. But I feel a bit confused to show $\omega_0$ is stronger than it is Hausdorff.

And according to Edoardo Lanari's answer, suppose $(X, \tau)$ is the topological space from which we start, if $\sigma$ is a weaker topology on $X$ and $(X,\tau)$ is compact-Hausdorff, then why the map $Id: (X,\tau) \rightarrow (X,\sigma)$ is a continuous bijection? (I mean, since $\sigma$ should be strictly weaker, so the map won't be one-one. )

May be a silly question, I'm just a newbie, can any one help me figure it out?

Edoardo Lanari, Weaker/Stronger Topologies and Compact/Hausdorff Spaces, URL (version: 2013-05-12): https://math.stackexchange.com/q/389576

Best Answer

If $(X,\tau)$ and $(X,\sigma)$ are such that $\sigma \subseteq \tau$ ($\sigma$ is weaker than $\tau$) then $\textrm{id}: (X, \tau) \to (X,\sigma)$ is continuous because a function is continuous iff every inverse image of an open set is open:

$$\forall O \in \sigma: \textrm{id}^{-1}[O]=O \in \tau \text{ as } \sigma \subseteq \tau$$

The function $\textrm{id}$ as a function from $X$ to $X$ is always a bijection (regardless of topologies on either side); this is just a set theory fact and has nothing to do with weaker/stronger topologies.

So if $(X,\tau)$ is compact and $\sigma \subseteq \tau$ is Hausdorff, then $\textrm{id}$ is continuous as I already explained and a bijection and so your very first theorem concludes from compactness of $\tau$ and Hausdorffness of $\sigma$ that $\textrm{id}: (X,\tau) \to (X, \sigma)$ is a homeomorphism. In particular it is an open map, so

$$\forall O \in \tau: \textrm{id}[O]=O \in \sigma \text{ or otherwise put: } \tau \subseteq \sigma$$

So in conclusion:

Let $(X,\tau)$ be compact Hausdorff and suppose $\sigma$ is a weaker topology than $\tau$ on $X$. Then if $\sigma$ is properly weaker, then $(X,\sigma)$ is not Hausdorff.

Similarly:

Let $(X,\tau)$ be compact Hausdorff and suppose $\sigma$ is a stronger topology than $\tau$ on $X$. Then if $\sigma$ is properly stronger, then $(X,\sigma)$ is not compact.

Proof of the latter is similar: $\textrm{id}: (X,\sigma) \to (X, \tau)$ is continuous as $\tau \subseteq \sigma$; we assumed the domain is compact and we know the codomain is Hausdorff so again we conclude this $\textrm{id}$ is also open and so $\sigma \subseteq \tau$ hence $\sigma=\tau$ contradicting *properly** stronger. So the assumption that $(X,\sigma)$ is compact must be false.

Related Question