Is there an infinite, connected space $(X,\tau)$ such that the only Hausdorff topology $\tau\_2$ with the property that $\tau\subseteq \tau\_2$ is the discrete topology?



Refining topologies to make them Hausdorff
**Remark.** Every non-discrete $\text T\_1$ topology has a non-discrete Hausdorff refinement.


**Proof.** Let $(X,\tau)$ be a non-discrete $\text T\_1$ space. Choose $p\in X$ such that $\{p\}\notin\tau,$ and let $\tau\_2=\{A\cup B:A\in\tau,\ B\subseteq X\setminus\{p\}\}.$ Then $\tau\_2$ is a Hausdorff topology on $X$ and $\tau\subseteq\tau\_2$ and $\{p\}\notin\tau\_2.$


**Example.** On any infinite set $X$ there is a connected $\text T\_1$ topology $\tau$ such that no Hausdorff topology refining $\tau$ has more than one non-isolated point.


Namely, let $\tau=\mathcal U\cup\{\emptyset\}$ where $\mathcal U$ is a free ultrafilter on $X.$


**Example.** On any nonempty set $X$ there is a connected $\text T\_0$ topology $\tau$ such that the only $\text T\_1$ topology refining $\tau$ is the discrete topology.


Namely, choose $p\in X$ and let $\tau=\{A:p\in A\subseteq X\}\cup\{\emptyset\}.$


