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Say that a topological space is CH if it is both compact and Hausdorff. Let $T$ and $T'$ be two topologies on the same set X that are comparable but different, i.e., $T$ is either strictly smaller or strictly larger than $T'$. Assume that $(\text{X}; T )$ is CH. Prove that $(\text{X}; T')$ is not CH.

I just can seem to get started, so I was wondering if anybody could give me a hint...

If $T' \subsetneqq T$, then the identity $id: (\text{X},T) \rightarrow (\text{X}, T')$ is continuous and surjective. This implies that $id(\text{X},T) = (\text{X},T')$ is also compact. So in order to show that $(\text{X},T')$ is not CH, I need to show that it is not Hausdorff, right? But I'm stuck...

Martin Sleziak
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Artus
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1 Answers1

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It's (IMO) easier to prove the equivalent statement:

If $(X,T)$ and $(X,T')$ are both CH and comparable, then $T = T'$.

You should use the fact that a continuous bijection $K \to X$ where $K$, is CH and $X$ is Hausdorff, is a homeomorphism. If eg. $T \subset T'$, then $\operatorname{id} : (X,T') \to (X,T)$ is continuous. Thus it's a homeomorphism, from which you conclude $T = T'$.

Najib Idrissi
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  • Thanks a lot. We know that $id: (\text{X}, T) \rightarrow (\text{X},T')$ is bijective because it is the identity. It is continuous if we assume $T' \subseteqq T$. There is a corollary in my notes which says that if $f:X \rightarrow Y$ is a continuous function with $X$ compact and $Y$ Hausdorff, then $f$ is closed. So if I use this corollary, I won't have to use the fact that $K$ is $CH$, right? It is sufficient for $K$ to be compact. Is that correct? – Artus May 18 '14 at 18:51
  • That's correct, yes. You still need CH for when they are comparable the other way, though. – Najib Idrissi May 18 '14 at 19:19