Assume that $(\text{X}, T)$ is compact and Hausdorff. Prove that a comparable but different topological space $(\text{X},T')$ is not.

Question

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...

Answer 1

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 $T = T'$.