Making a topological space hausdorff 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?
 A: 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.
