# Cocountable Topology is not Hausdorff.

This question has been answered already, this is an attempt to rephrase it.

Let $$(X,\tau)$$ be a cocountable topology with $$X$$ uncountable.

Show that $$(X, \tau)$$ is not Hausdorff.

Let $$r\not=s$$, and $$r\in U$$ and $$s \in V$$, where $$U, V$$ are open and not empty.

Assume $$U \cap V = \emptyset$$.

$$X\setminus (U \cap V) =$$

$$(X\setminus U)\cup (X\setminus V)$$.

RHS is a union of countable sets, hence countable.

LHS is $$X\setminus\emptyset=X$$ uncountable, a contradiction.

Perhaps nitpicking :

$$\emptyset$$ is open and an element of the topological space, but $$X\setminus\emptyset$$ is not countable.

The complement of any non-empty open set is countable. Since $$r \in U$$ and $$s\in V$$ your open sets $$U$$ and $$V$$ are non-empty. So your argument works fine.
• Thanks for pointing out explicitly that one considers $X$ \ $U$, for $U$ non-empty. Nov 28 '21 at 12:12