I am trying to find the answer to the following question:
Let $X$ be a Hausdorff space and $x \neq y \in X$. Does there exist $U$ and $V$ open such that $x \in U$ and $y \in V$ such that $\overline{U} \cap \overline{V} = \emptyset$.
I know without the closure it is true using the Hausdorff-ness. But how does it follow with the closure as well?
Thanks in advance!
You cannot: the relatively prime integer topology on $\mathbb{Z}$ is a topology that is Hausdorff but not $T_{2\frac{1}{2}}$, which is another name for the separation of points by closed neighbourhoods. More such examples can be found using a query in $\pi$-base, which is based on the book "Counterexamples in Topology".