Let $X$ a non-Hausdorff topological space and let $\sim$ denote an equivalence relation on $X$. Is it possible that $X/\sim$ be Hausdorff? Let $X$ a non-Hausdorff topological space and let $\sim$ denote an equivalence relation on $X$. Is it possible that $X/\sim$ be Hausdorff? Show an example.
I would like to know also if there is a sufficient condition for $X/\sim$ to be an non-Hausdorff space.
 A: For your first question, yes, this is possible. Take, for example, the topological space
$$(X, \tau) = (\{1, 2, 3\}, \{\emptyset, \{1, 2\}, \{3\}, \{1, 2, 3\}\})$$
with equivalence relation
$$\sim \; = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1) \}.$$
Then, if $\tau'$ is the quotient topology on $X / \sim$, then
$$(X / \sim, \tau') = (\{\{1, 2\}, \{3\}\}, \{\emptyset, \{\{1, 2\}\}, \{\{3\}\}, \{\{1, 2\}, \{3\}\}\}),$$
i.e. the discrete topology, which is definitely Hausdorff. However, $(X, \tau)$ is not Hausdorff, as $1$ and $2$ cannot be separated with open sets.
A: Let $X=\mathbb{N} \cup \{p,q\}$ ($p,q$ distinct points not in the set $\mathbb{N}$) with the topology that $O$ is open iff $O \cap \{p,q\}=\emptyset$ or $\mathbb{N} \setminus O$ is finite. The points $p$ and $q$ cannot be separated by open sets so $X$ is not Hausdorff. $X$ is compact and $T_1$, though.
But the equivalence relation $\sim$ on $X$ that has one non-singleton class $\{p,q\}$ has the property that $X/{\sim}$ is homeomorphic to a convergent sequence and thus compact and Hausdorff (also metrisable). 
