# Cofinite topology on infinite sets can't be induced from a metric space.

I'm a very beginner in topology, and I have a question.

The book I'm studying says that cofinite topologies on any infinite sets can't be induced from any metric spaces, but how do I show this?

I haven't learned anything about hausdorf, compactness, separability, or connectedness yet.

I feel that I need to use the fact that I can't make any open ball in (X,d) whose complement is finite. In here X is a set and d is a metric given.

• Hausdorffness is absolutely the way to go. You really can't appeal to that? – Randall Aug 29 '18 at 3:00
• @Randall I haven't learned anything about Hausdorffness yet, the book requires to prove it without that notion. – Linus Aug 29 '18 at 3:02

Assume for contradiction that there is an infinite set $X$ with metric $d$ such that the induced topology on $X$ is the cofinite topology. Let $x,y$ be distinct points of $X$. Then let $\varepsilon=d(x,y)>0$. Let $U=B_{\varepsilon/2}(x)$, $V=B_{\varepsilon/2}(y)$. By the triangle inequality, $U\cap V=\varnothing$. However, $U$ and $V$ are nonempty open sets, hence $U$ and $V$ are both cofinite. Hence $(U\cap V)^C = U^C\cup V^C$ is also finite, so $U\cap V$ is cofinite. In particular, it is nonempty. Contradiction.
In the cofinite topology on the infinite set $X$, any two non-empty open sets have a non-empty intersection. This should be reasonably clear: if $U$ and $V$ are non-empty and open and $U \cap V$ is empty, then $$X = X -(U \cap V) = (X-U) \cup (X-V).$$ But now the infinite set $X$ is a union of two finite sets, a contradiction.