Let $X$ be any set and $\tau_c$ the cofinite topology in $X$ and $T$ be the set of Hausdorff topologies on $X$. Prove that $\tau_c=\inf T$ where the relation is inclusion in the set of all topologies on $X$, i.e. prove that:
- $\tau_c\subseteq \tau$ for any $\tau\in T$.
- If $\tau'$ is any other topology in $X$ satisfying 1. then $\tau'\subseteq \tau_c$.
It's easy to prove 1: For any $U\in T$, $U^c$ is finite (or $X$) and thus closed in any Hausdorff topology in $X$.
The problem is 2. Let $\tau'$ be any topology satisfying 1., one has to prove that if $U\in \tau'$ then $U^c$ is finite. One may proceed by contradiction and prove that if $U^c$ is infinite then there is a Hausdorff topology in $X$ in which $U$ is not open. So it comes down to:
Let $A$ be an infinite subset of $X$ different from $X$. Prove there is a Hausdorff topology in $X$ in which $A$ is not closed.
I think that's true but I couldn't construct $X$.