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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?

Jens Renders
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user267839
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    I am wondering where you have encountered the notation $\mathbb{R}/\mathbb{Z}$, because it is usually not an incidence of the topological notation $S/U$. Instead it is an incidence of the group theory notation $G/N$ where $G$ is a (topological) group and $N$ is a (closed) normal subgroup. The topological quotient and the group theoretic quotient are different kinds of quotients. – Lee Mosher Jul 13 '17 at 16:59
  • This was postulated as the definition of the circle group. So it's clear if we identify it with ${z \in \mathbb{C} | 1 = |z| }$ that it's hausdorff but the background of my question was how to compute this from the properties of $\mathbb{R}$ and $\mathbb{Z}$. – user267839 Jul 13 '17 at 17:08

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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

Jens Renders
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  • I think he means identifing $\mathbb{Z}$ to a point, so a very different quotient. – Henno Brandsma Jul 13 '17 at 21:56
  • @HennoBrandsma Why do you think that? $\mathbb{R}/\mathbb{Z}$ is not the notation for that. He also says in the comments that he is talking about the circle group (which is the quotient I'm working with) – Jens Renders Jul 13 '17 at 22:23
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    The notation $\mathbb{R}/\mathbb{Z}$ is ambiguous: it could be a group quotient (in which case you're quite right), or a purely topological quotient, and then this is an interesting quotient space (it's not first countable but it is sequential). I didn't read the comments till later, and the question alone should be enough (or else edit it based on the comments). The final sentence reeked of topology, not groups. – Henno Brandsma Jul 13 '17 at 22:28
  • @HennoBrandsma I have never seen the notation being used for the space you mention (I do know the space, it is a classical counterexample). Every source I have read uses $\mathbb{R}/\mathbb{Z}$ for the topological group quotient (ie the circle group $S^1$). Can you point me to a source that uses the notation like you interpreted it? – Jens Renders Jul 13 '17 at 22:40
  • Engelking General Topology IIRC. – Henno Brandsma Jul 13 '17 at 22:41
  • @HennoBrandsma Thanks, I'll be more carefull with the notation in the future. It is difficult to edit the question as it seems that the OP himself wasn't aware of the fact that he was talking about topological groups (and respective quotient). I will point it out in my answer tough. – Jens Renders Jul 13 '17 at 22:45
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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.

Andres Mejia
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  • But how to see it with the latter criterion? This means to show that ${(a, b) | a -b \in \mathbb{Z}}$ is closed... – user267839 Jul 13 '17 at 17:12
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    Show that the complement is open. Suppose that $(x,y) \in \mathrm{complement}$. Then $x-y \notin \mathbb Z$, so take the floor and ceiling of it, pick the minimum distance and build a suitably small box around $(x,y)$. – Andres Mejia Jul 13 '17 at 17:16
  • @KarlPeter see my answer for a more general way – Jens Renders Jul 13 '17 at 17:34
  • Yes, I like Jens' method a lot more, it is very elegant, although maybe slightly harder to see initially, +1 – Andres Mejia Jul 13 '17 at 17:35
  • I think he means identifing $\mathbb{Z}$ to a point, so a very different quotient. – Henno Brandsma Jul 13 '17 at 21:56
  • @HennoBrandsma why do you think this? In this case the result is immediate, since the integers are closed in the reals. – Andres Mejia Jul 13 '17 at 22:06
  • He also wants to mod out by a subspace $U$, plus he does not mention groups in his question either. So I assume he meant the standard topology way. In the group case, also the closedness of the subgroup will be necessary and sufficient as @JensRenders already observed. – Henno Brandsma Jul 13 '17 at 22:09
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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 $\{A\} \cup \{\{x\}: x \notin A\}$

Henno Brandsma
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