# Comparable topologies are equal if underlying spaces are Compact and Hausdorff.

Since it considers the possibility of the topologies being same , I assume the underlying set is the same(Correct me if I'm wrong). So let $X$ be our set and $\tau_1$ and $\tau_2$ are the two topologies that are comparable. So without loss of generality I can assume that $\tau_1 \subseteq \tau_2$ and under both these topologies the space $X$ is both compact and hausdorff. So now I have to prove that $\tau_1 = \tau_2$. It suffices to show that $\tau_2 \subseteq \tau_1.$

Choose any arbitrary open set $U$ from $\tau_2.$ To show this is also open in $\tau_1.$ By compactness of $\tau_1$, $U$ has an open cover satisfying $$U\subset \bigcup_{k=1}^{k=n}V_k$$ where $V_k's$ are open sets in $\tau_1$.If not a proper subset then $U$ is already open in $\tau_1.$ So let's assume it is proper subset.

Then there exists a point $v_1$ in the RHS set that is not in $U$.By Hausdorff compactness we can find open sets $$A_1\cap B_1=\Phi$$ such that $$U\subset A_1 \\\text{and}\\ v_1\in B_1.$$ Now $$\bigcup_{k=1}^{k=n}(V_k\cap A_1)=W_1$$ is the new finite open cover of $U.$ If this is also not equal then we will find anothe $v_2$ with similar situation and this process goes on. Since the $W_i's$ always contain the $U$ and cannot be infinitely small so this process has to stop at one point and that is when we have arrived at an equation like $$U=\bigcup_{k=1}^{k=n}C_k$$ and $U$ is proved to be an open set in $\tau_1.$

Is this proof correct $?$ Please point out if there are mistakes and also tell how to correct those. I think I need to show that the process should stop after finite steps because infinite intersection of open sets may not be open but I do not know how to.

Thanks.

• Right at the beginning you're misusing compactness. Compactness is not the statement "Every set has a finite cover," but rather "Every open cover of the whole space has a finite subcover." – Noah Schweber Aug 26 '16 at 17:28
• @NoahSchweber : I know that. From any open cover I can find a finite subcover so there we have a finite open cover . Doesn't work? – user118494 Aug 26 '16 at 17:34
• It's not that the statement is false, it's that it's useless. Any set in any space has a finite open cover - namely, the whole space itself! $X$ is open in any topology on $X$. So that's not going to get you far in the proof. – Noah Schweber Aug 26 '16 at 17:38
• Also, you haven't justified how the $A_1$ and $B_1$ exist, and you haven't explained why "this process has to stop at one point" (why can't the $W_i$s be infinitely small? and what exactly does that mean?). – Noah Schweber Aug 26 '16 at 17:40
• Searching for compact Hausdorff finer gives a few useful links: math.stackexchange.com/questions/828324/…, math.stackexchange.com/questions/328923/…, proofwiki.org/wiki/… etc. – Martin Sleziak Aug 27 '16 at 7:48

This is not correct. There is no reason that the sets $A_1$ and $B_1$ that you claim to exist must exist. For instance, imagine $X=[0,1]$, $\tau_1$ is the usual topology, and $U=X\setminus \{v_1\}$. I also don't at all follow why your process would need to end after finitely many steps.
As a hint that you're barking up the wrong tree, you haven't actually used compactness anywhere. For any topology and any $U$, you could just take $n=1$ and $V_1=X$. The power of compactness is not that there exists some finite open cover (which is trivial, you can always just take the whole space), it's that no matter what open cover you start with you can find finitely many of the sets which still cover.
I find this question easier to think about in terms of closed sets instead of open sets. Try proving that any $\tau_2$-closed set must be $\tau_1$-compact, and then use the fact that a compact subset of a Hausdorff space is closed.