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:

  1. $\tau_c\subseteq \tau$ for any $\tau\in T$.
  2. 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 $X$.


Suppose that $\tau'$ is a topology on $X$ such that $\tau'\nsubseteq\tau_c$, and let $U\in\tau'\setminus\tau_c$. Let $A=X\setminus U$; then $A$ is infinite, and as you say, we want to find a $\tau\in T$ such that $A$ is not closed in $\tau$.

We actually know a little more about $A$: $A\ne X$, for if $A=X$, then $U=\varnothing\in\tau_c$. Thus, we can find a point $p\in U$. We’ll be done if we can find a Hausdorff topology $\tau$ on $X$ such that $p\in\operatorname{cl}_\tau A$. The simplest way to do this is to make every point of $X$ isolated except the point $p$, and make $\tau$ look like the cofinite topology at the point $p$:

$$\tau=\{V\subseteq X:p\notin V\text{ or }X\setminus V\text{ is finite}\}\;.$$

I’ll leave it to you to check that $\tau$ really is Hausdorff, and that $p\in\operatorname{cl}_\tau A$.


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