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I have been working on this problem from Principles of Topology by Croom:

"Let $X$ be a set with three different topologies $S$, $T$, $U$ for which $S$ is weaker than $T$, $T$ is weaker than $U$, and $(X,T)$ is compact and Hausdorff. Show that $(X,S)$ is compact but not Hausdorff, and $(X,U)$ is Hausdorff but not compact."

I managed to show that $(X,T)$ compact implies $(X,S)$ compact, and that $(X,T)$ Hausdorff implies $(X,U)$ Hausdorff.

However, I realized that there must be an error (right?) when it comes to $(X,T)$ compact implying that $(X,U)$ is noncompact, since if $X$ is finite (which it could be, as the problem text doesn't specify), then it is compact regardless of the topology on it. What are the minimal conditions that we need in order for there to be a counterexample to $(X,T)$ compact implying $(X,U)$ compact? Just that $X$ is not finite?

Does this also impact showing $(X,T)$ Hausdorff implies $(X,S)$ is not Hausdorff?

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    @Asaf: Yes, what Brian gave is the definition I'm using. Should have included that.2012-12-31

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If $X$ is finite, the only compact Hausdorff topology on $X$ is the discrete topology, so $T=\wp(X)$. In this case there is no strictly finer topology $U$. If $X$ is infinite, the discrete topology on $X$ is not compact, so if $T$ is a compact Hausdorff topology on $X$, there is always a strictly finer topology.

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    @Alex: You’re welcome.2013-01-01