It [turns out](https://mathoverflow.net/questions/36085/minimal-hausdorff) 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 $T\_2$?



Minimal Hausdorffness reversed
Yes.


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.


