# Proving that a quotient space is Hausdorff

This question is influenced by Quotient space of unit sphere is Hausdorff. I have slightly modified this to make it somewhat difficult.

Let $S^1=\{e^{2\pi it}|r\in\mathbb{R}\}$ be the unit sphere. Define $\sim$ on $S^1$ where two points are identified if $t_1-t_2=\sqrt{2}k$, for some $k\in\mathbb{Z}$. It must be shown that $S^1/\sim$ is Hausdorff.

Define $f:S^1\times S^1\rightarrow S^1$ by $f(x,y)=xy^{-1}$. Clearly $f$ is continuous. Let $R=\{(x,y)\in S^1\times S^1|x \sim y \}=\{(e^{2\pi it_1},e^{2\pi it_2})\in S^1\times S^1|t_1-t_2=\sqrt{2}k,k \in\mathbb{Z}\}$.

Also, let $A=\{ e^{2\sqrt{2}k\pi i}|k\in \mathbb{Z} \}$. Then $f^{-1}(A)=R$. $A$ being a countable set implies that $A$ is closed, therefore $R$ is closed since $f$ is continuous. Thus $S/\sim$ is Hausdorff. Is this proof correct? Thank you.

• The equivalence relation has each equivalence class being dense in $S^1$. So, my first guess would be that the quotient topology is actually the trivial topology, which wouldn't be Hausdorff. – Daniel Schepler Jun 16 '17 at 18:19
• Why would countable imply closed? – Hagen von Eitzen Jun 16 '17 at 18:23
• Is $\Bbb{Q}$ closed in $\Bbb{R}? – Neal Jun 16 '17 at 18:37 • And in fact, assuming the equivalence classes are dense, here's the proof the quotient topology is the trivial topology: if$A \subseteq S^1 / \sim$is closed and nonempty, then$\pi^{-1}(A)$is closed and contains an equivalence class, so it is all of$S^1$. That implies$A$is all of$S^1 / \sim$. – Daniel Schepler Jun 16 '17 at 18:53 • @DonaldEdwards It would be a corollary of the fact that$\{ m + n\sqrt{2} : m, n \in \mathbb{Z} \}$is dense in$\mathbb{R}$. I would imagine that fact has probably been proved before somewhere on this site. – Daniel Schepler Jun 16 '17 at 19:48 ## 2 Answers First,$S^1=\{e^{2\pi it}:t\in\mathbb{R}\}$is a unit circle rather than a unit sphere. Second, for$t_1,t_2\in\mathbb{R}$we have$e^{2\pi it_1}=e^{2\pi it_2}$if and only if$t_2-t_1\in\mathbb{Z}$, hence we can also write$S^1=\{e^{2\pi it}:t\in[0,1)\}$. Third, in order to correctly define a binary relation$\sim$on$S^1$by identifying points$e^{2\pi it_1}$and$e^{2\pi it_2}$whenever$t_1,t_2$satisfy a property$P(t_1,t_2)$, one has to ensure that the property$P$has the same logical value for all pairs$(t_1,t_2)\in\mathbb{R}^2$that yield the same pair of points$(e^{2\pi it_1},e^{2\pi it_2})$on the circle. Namely, that for any$t_1,t_2,t'_1,t'_2\in\mathbb{R}$, if$t'_1-t_1\in\mathbb{Z}$and$t'_2-t_2\in\mathbb{Z}$, then$P(t_1,t_2)\equiv P(t'_1,t'_2)$. Your property$P(t_1,t_2)\equiv(\exists k\in\mathbb{Z})(t_1-t_2=\sqrt{2}k)$is not such. Let$t_1=1+\sqrt{2}$,$t'_1=\sqrt{2}$,$t_2=t'_2=1$. Then$t'_1-t_1$and$t'_2-t_2$are integers,$P(t_1,t_2)$holds true since$t_1-t_2=\sqrt{2}$, but$P(t'_1,t'_2)$is false since$t'_1-t'_2=\sqrt{2}-1$and there is no$k\in\mathbb{Z}$satisfying$\sqrt{2}k=\sqrt{2}-1$, as this would mean that$k=1-1/\sqrt{2}$. So your relation is not correctly defined and it has no sense to ask whether$S^1/\mathord{\sim}$is Hausdorff. • So what you are saying is that this relation is not transitive>. – user81883 Jun 18 '17 at 23:19 •$t'_1-t'_1$,$t'_2-t'_2$being integers is irrelevant – user81883 Jun 18 '17 at 23:35 • @DonaldEdwards No, I say that the relation is not defined properly. Every point of$S^1$can be described in infinitely many ways, and your definition of the relation depends on which description you take into account. By other words,$S_1$already is a quotient of$\mathbb{R}$. Property$P$defines an equivalence on$\mathbb{R}$, but that is not congruent (or in accordance) to the equivalence used to define$S^1$. – Peter Elias Jun 19 '17 at 2:39 • @DonaldEdwards And what did you mean by saying that zeros beeing integers is irrelevant? – Peter Elias Jun 19 '17 at 2:42 The only way for points to be equivalent is for k=o so your equivalence relation is the identity (again ) . Try t_1-t_2 =k/2 which would identify antipodal points. Then I think your argument will work . detail ? If t_1-t_2=2^(1/2)k then e^2(PI)i$\sqrt2$k =1 . Then$\sqrt2\$k is an integer so the square root of 2 is rational ( impossible) unless k=0 So k=0 and the equivalence relation is the identity

• t_1-t_2=2^(1/2)k then e^2(PI)i2–√2k =1 does not make sense – Heisenberg Aug 1 '17 at 5:21