Refining topologies to make them Hausdorff

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?

• what if $X=\mathbb{N}$ and $\tau$ be the collection of sets of the form $\{n,n+1,...\}$? – erz Sep 21 '17 at 5:58
• @erz Seems to me any $\text T_1$ topology on $\mathbb N$ will be a refinement of your topology, e.g., a topology which is homeomorphic to the usual topology of $\mathbb Q,$ which is Hausdorff but not discrete. – bof Sep 21 '17 at 6:03

1 Answer

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 $\tau=\{A:p\in A\subseteq X\}\cup\{\emptyset\}.$