Quotient Space Hausdorff Why is $\mathbb{R}/\mathbb{Z}$ a Hausdorff space, if we assume that $\mathbb{R}$ has the canonical topology, and $\mathbb{Z}$ the induced subspacetopology?
More generally, which requirements for a topologal space $S$ ans it's subspace $U$ allow to conclude that $S/U$ is hausdorff?
 A: As others have said, this quotient is hausdorff iff the relation is closed.
Note that the quotient $\mathbb{R}/\mathbb{Z}$ is in this case (the case where it defines the circle group $S^1$) a group theoretical quotient. The relation is thus
$$\{(a,b)\in \mathbb{R}^2 \mid a-b \in \mathbb{Z}\} = f^{-1}(\mathbb{Z})$$
where
$$f:\mathbb{R}^2 \rightarrow \mathbb{R}: (a,b) \mapsto a-b$$
Now, because $\mathbb{Z}$ is closed in $\mathbb{R}$, and $f$ is continuous (as $(\mathbb{R},+)$ is a topological group), $f^{-1}(\mathbb{Z})$ must be closed as well.
In general we have: if $X$ is a topological group that is hausdorff, and $Y$ is a closed normal subgroup, then $X/Y$ is hausdorff
A: Notice that $\mathbb R/\mathbb Z \cong (S^1,\tau)$, where $\tau$ is the subspace topology. This is the quickest way I can think of to see the result. On the other hand, one can use the fact that if $\sim$ is a closed subset of $\mathbb R^2$, then the quotient is hausdorff.
A: Fact: Let $X$ be a $T_3$ space (regular and $T_1$) and $A$ a subspace of $ X$, then $X{/}A$ in the quotient topology is Hausdorff iff $A$ is closed.
This assumes the usual quotient of modding out by a subspace: identifying that set to a point, so the equivalence classes of $X{/}A$ are $\{A\} \cup \{\{x\}: x \notin A\}$
