It turns out that not every Hausdorff topology is contained in a minimal Hausdorff topology. Let's put this question on its head: is every non-$T_2$ topology contained in a topology that is maximal with respect to the property of not being $T_2$?



Let $\tau$ be a non-Hausdorff topology on a set $X$.

Suppose first that $\tau$ is not $T_1$, so there are two points $a,b\in X$ such that every open neighbourhood of $a$ contains $b$. Then $\tau$ can be refined to the topology consisting of all subsets of $X$ except for those containing $a$ but not $b$, which is a maximal non-Hausdorff topology.

Now suppose that $\tau$ is $T_1$.

Choose two points $x$ and $y$ that are not separated by $\tau$, and consider the set of all topologies on $X$ that contain $\tau$ and do not separate $x$ and $y$. By Zorn's Lemma there is a maximal such topology $\tau'$.

The topology generated by $\tau'$ and all subsets of $X\setminus\{x,y\}$ also fails to separate $x$ and $y$, so by maximality $\tau'$ must contain all subsets of $X\setminus\{x,y\}$. Since $\tau$ is $T_1$ it follows that $\tau'$ separates every pair of points except $\{x,y\}$, and so any proper refinement of $\tau'$ is Hausdorff.

| cite | improve this answer | |
  • $\begingroup$ Thanks for this argument! - The crucial part is "By Zorn's Lemma...". Could you add a sentence or so that proves that if $\cal T$ is a chain of topologies such that for all $\tau \in \cal T$, the points $x,y$ are not separated by $\tau$, then the same holds for $\bigcup \cal T$? $\endgroup$ – Dominic van der Zypen Apr 28 '15 at 11:30
  • $\begingroup$ Write "$\mathcal{T}^+$" for the coarsest topology finer than each element of $\mathcal{T}$; so $\mathcal{T}^+$ is generated by $\bigcup\mathcal{T}$ via unions. It's clear that there is no element in $\bigcup\mathcal{T}$ separating $x$ and $y$; since $\bigcup\mathcal{T}$ generates $\mathcal{T}^+$ via unions, this implies $x$ and $y$ remain unseparated in $\mathcal{T}$. Note that in general we can't pass from claims about $\bigcup\mathcal{T}$ to claims about $\mathcal{T}^+$. $\endgroup$ – Noah Schweber Apr 28 '15 at 12:08

Here is what I think is a more conceptual explanation of what is going on in Jeremy Rickard's answer. There are two different (and dual) ways to "generate" a topology: you can specify that some sets are open, and take all the open sets that they generate, or you can specify that some ultrafilters converge to some points, and take the notion of convergence that this generates. The first notion gives you the coarsest topology in which certain sets are open; the latter notion gives you the finest topology in which certain ultrafilters converge to certain points. These are dual in that the first is giving the universal topology on $X$ that makes certain maps out of $X$ continuous, while the latter is giving the universal topology that makes certain maps into $X$ continuous. For a question like this, where you are looking for a very fine topology with a certain property, it is thus more natural to use the latter approach.

So suppose $\tau$ is a non-Hausdorff topology on $X$, and let $U$ be an ultrafilter on $X$ which $\tau$-converges to two different points $x,y\in X$. Let $\tau'$ be the topology which is generated by saying that $U$ converges to both $x$ and $y$. Explicitly, a set is in $\tau'$ if either it does not contain $x$ or $y$ or it is in $U$. The only ultrafilter that $\tau'$-converges (other than principal ultrafilters converging to the corresponding point) is $U$, which converges to both $x$ and $y$. Thus the only way to make the topology finer is to make $U$ not converge to either $x$ or $y$, which will make it Hausdorff.

This argument shows that the two cases ($T_1$ and non-$T_1$) of Jeremy Rickard's answer are really the same; the only difference is that in the non-$T_1$ case you may have to take $U$ to be a principal ultrafilter.

| cite | improve this answer | |
  • 1
    $\begingroup$ Nice observation. I'd noticed that the $T_1$ case of my answer was pretty much choosing an ultrafilter, but hadn't formalized it quite as neatly as you did. $\endgroup$ – Jeremy Rickard Apr 28 '15 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.