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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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How do you define "weaker"? –  Asaf Karagila Dec 31 '12 at 16:45
    
@Asaf: In this case it means strictly weaker: $S\subsetneqq T\subsetneqq U$. You might prefer the term strictly coarser. –  Brian M. Scott Dec 31 '12 at 17:27
    
@Asaf: Yes, what Brian gave is the definition I'm using. Should have included that. –  Alex Petzke Dec 31 '12 at 20:26
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up vote 2 down vote accepted

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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I can morally justify the first sentence to myself, but what's the proof? Also, I understand by the last sentence that there is a strictly finer topology in the case of X infinite, but how do we prove that ANY finer topology (in case it's not discrete) is not compact? –  Alex Petzke Dec 31 '12 at 21:27
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@Alex: If $\langle X,\tau\rangle$ is Hausdorff, it’s certainly $T_1$, so every finite $F\subseteq X$ is closed. If $X$ is finite, then every subset of $X$ is closed, and hence every subset of $X$ is also open. The rest wasn’t part of your original question, so I didn’t realize that you were asking about it as well. The identity function from $\langle X,U\rangle$ to $\langle X,T\rangle$ is continuous, so every $U$-compact set is $T$-compact. Let $K\subseteq X$ be $U$-closed but not $T$-closed. Then $K$ is not $T$-compact, hence not $U$-compact, and $X$ can’t be $U$-compact. –  Brian M. Scott Dec 31 '12 at 23:52
    
I understand the first part. To make sure I'm getting the rest... Are we applying the result "Let $X$ be a compact space, $Y$ and space and $f:X \to Y$ a continuous function from $X$ onto $Y$. Then $Y$ is compact." to subsets of $(X,U)$ and $(X,T)$ then we know that $K \subseteq U$ exists since $(X,U)$ has open sets that $(X,T)$ does not, then applying (since $(X,T)$ is compact Hausdorff) "Let $X$ be a compact Hausdorff space. A subset $A$ of $X$ is compact if and only if it is closed.", then it follows from there? Hope you can follow that. –  Alex Petzke Jan 1 '13 at 3:48
    
@Alex: Yes, that’s the idea. –  Brian M. Scott Jan 1 '13 at 4:12
    
Good. It's that kind of multistep proof that I always have difficulty seeing. Thanks for the help. –  Alex Petzke Jan 1 '13 at 17:09
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