Show that if $X$ is compact Hausdorff under both $\mathcal{T}$ and $\mathcal{T'}$, then either $\mathcal{T}$ and $\mathcal{T'}$ are equal or not comparable.
My solution:
So, we have to show $\mathcal{T}\subset \mathcal{T'}$ and $\mathcal{T}\supset\mathcal{T'}$.
First of all, we know that every subspace of compact Hausdorff is closed. Suppose $\mathcal{T}\subset\mathcal{T'}$. Then, compactness under the topology $\mathcal{T'}$ (trivially) implies compactness under the topology $\mathcal{T}$.
Again, every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T'}$. Hence $\mathcal{T}\supset\mathcal{T'}$.
Is my argument ok? Thanks.
Edit:
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Sorry for your inconvenience for the statement "…compact Hausdorff is closed." I meant:- "Every compact subspace of a Hausdorff space is closed"(Munkres Topology Thrm.: 26.3)
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Sorry for the second part too (The statement: "every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T'}$"). I wanted to say "Since $\mathcal{T}\subset\mathcal{T'}$, every closed set under the topology $\mathcal{T'}$ can be written as finite union of closed sets under the topology $\mathcal{T}$".
Then let $X-A_1,X-A_2,\dots, X-A_k\in \mathcal{T}$, and $\mathcal{A}_1,\mathcal{A}_2,\dots, \mathcal{A}_k$ are the open covering of $A_1,A_2,\dots,A_k$ respectively, so that $\bigcup_{i=1}^{k}A_i=B$, where $X-B\in \mathcal{T'}$. Then the union of finite subcollections that covers $A_1,A_2,\dots, A_k$ will cover $B$. Then compactness under the topology $\mathcal{T}$ implies compactness under the topology $\mathcal{T'}$. Then $\mathcal{T}\supset\mathcal{T'}$.
Best Answer
"...every subspace of compact Hausdorff is closed." I'm not sure what you mean by this. The space $[0,1]$ is compact and Hausdorff, but its subspace $[0,1/2)$ is not closed.
"...every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T}'$." Well if $E$ is $\mathcal{T}$-closed, then $E^c$ is $\mathcal{T}$-open, and thus by $\mathcal{T}\subseteq\mathcal{T}'$, we see that $E^c$ is $\mathcal{T}'$-open and consequently $E$ is $\mathcal{T}'$-closed. Thus every $\mathcal{T}$-closed set is $\mathcal{T}'$-closed. However how does this show that $\mathcal{T}\supseteq\mathcal{T}'$?
The easiest way to do this is to assume $\mathcal{T}\subseteq\mathcal{T}'$, and consider the identity map $f:(X,\mathcal{T}')\to(X,\mathcal{T})$ defined by $f(x)=x$. This is continuous by our assumption, and it's a bijection. Since it maps from a compact space into a Hausdorff space, we can conclude that it is a homeomorphism. Therefore $f$ maps open sets to open sets, meaning that if $U$ is $\mathcal{T}'$-open, then $U=f(U)$ is $\mathcal{T}$-open. Therefore $\mathcal{T}=\mathcal{T}'$.
If you aren't familiar with the theorem I used above, you should try proving it independently. It is tremendously useful and its proof relies on the same techniques one would employ to prove your problem without it. Here it is:
I alluded that you can prove that $\mathcal{T}\subseteq\mathcal{T}'$ implies $\mathcal{T}=\mathcal{T}'$ without using this theorem, and as I suspect this is the route you intended, here's how one would do so. Suppose $U$ is $\mathcal{T}'$-open. Then $U^c$ is $\mathcal{T}'$-closed. Since closed subsets of compact sets are compact, we know that $U^c$ is $\mathcal{T}'$-compact. As you pointed out, $\mathcal{T}'$-compactness implies $\mathcal{T}$-compactness, so $U^c$ is $\mathcal{T}$-compact. Since $\mathcal{T}$ is Hausdorff and compact subsets of Hausdorff spaces are closed, we deduce $U^c$ is $\mathcal{T}$-closed. Therefore $U$ is $\mathcal{T}$-open, which completes the proof.
I find that the following reformulation of this problem is more enlightening.