Making a topological space hausdorff

Question

Imagine you have a topological space $(M,T)$ which is not Hausdorff. Then, there is a set $S$ of pairs of points $x,y$ so that there are no disjoint neighborhoods of $x$ and $y$. It seems like the relation $x \sim y \leftrightarrow (x,y)\in S$ is an equivalence relation, which meens you can form the quotient space $M/\sim $, and it seems like this space is Hausdorff. Am I right with my conclusion and is there a name for this construction?

Answer 1

No, this fails for a couple reasons. First, $S$ is not an equivalence relation in general: it can fail to be transitive.

Second, even if you let $\sim$ be the transitive closure of $S$, $M/{\sim}$ may not be Hausdorff. For instance, let $M=\mathbb{Z}\cup\{\infty\}$, topologized as follows. A set $U\subseteq M$ is open iff:

• If $n\in\mathbb{Z}$ is odd and $n\in U$, then $n+1\in U$ and $n-1\in U$.
• If $\infty\in U$, then there exists $n\in\mathbb{Z}$ such that $m\in U$ for all $m\geq n$.

Then $(x,y)\in S$ iff either $x=y=\infty $ or $x,y\in\mathbb{Z}$ and either $|x-y|\leq 1$ or $x$ and $y$ are both odd and $|x-y|\leq 2$. The transitive closure of this relation has just two equivalence classes, $a=\{\infty\}$ and $b=\mathbb{Z}$. So the quotient $M/{\sim}=\{a,b\}$ has two points. Moreover, the only open set containing $a$ is the entire space, since any neighborhood of $\infty$ must contain points of $\mathbb{Z}$. Thus $M/{\sim}$ is not Hausdorff.

To get a Hausdorff space, you must instead iterate this process transfinitely, until you eventually reach a quotient which is Hausdorff. This quotient is the quotient by the smallest equivalence relation on $M$ with a Hausdorff quotient. This is sometimes called the "Hausdorffification" of $M$; see Construction of a Hausdorff space from a topological space for other constructions and more information on this quotient.