TITLE: Metrization of the cofinite topology
QUESTION [5 upvotes]: Can you help me please with this question?
Let $X$ be a non-empty set with the cofinite topology.
Is $\left ( X,\tau_{\operatorname{cofinite}} \right ) $ a metrizable space?
Thanks a lot!

REPLY [15 votes]: If $X$ is finite, the cofinite topology is the discrete one, which is metrizable, for example using the distance $d$ defined by $d(x,y):=\begin{cases}0&\mbox{ if }x=y,\\
1&\mbox{ otherwise}
.\end{cases}$ 
If $X$ is infinite, it's not a Hausdorff space. Indeed, let $x,y\in X$, and assume that $U$ and $V$ are two disjoint open subsets of $X$ containing respectively $x$ and $y$, then $U=X\setminus F_1$ and $V=X\setminus F_2$ where $F_1$ and $F_2$ are finite. Hence $\emptyset =X\cap F_1^c\cap F_2^c=X\setminus (F_1\cup F_2)^c$, which is a contradiction since $X$ is infinite. But each singleton is closed in such a space.