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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.

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    $\begingroup$ 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. $\endgroup$ – Daniel Schepler Jun 16 '17 at 18:19
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    $\begingroup$ Why would countable imply closed? $\endgroup$ – Hagen von Eitzen Jun 16 '17 at 18:23
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    $\begingroup$ Is $\Bbb{Q}$ closed in $\Bbb{R}? $\endgroup$ – Neal Jun 16 '17 at 18:37
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    $\begingroup$ 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 $S^1 / \sim$. $\endgroup$ – Daniel Schepler Jun 16 '17 at 18:53
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    $\begingroup$ @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. $\endgroup$ – Daniel Schepler Jun 16 '17 at 19:48
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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.

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  • $\begingroup$ So what you are saying is that this relation is not transitive>. $\endgroup$ – user81883 Jun 18 '17 at 23:19
  • $\begingroup$ $t'_1-t'_1$, $t'_2-t'_2$ being integers is irrelevant $\endgroup$ – user81883 Jun 18 '17 at 23:35
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    $\begingroup$ @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 $S^1$. $\endgroup$ – Peter Elias Jun 19 '17 at 2:39
  • $\begingroup$ @DonaldEdwards And what did you mean by saying that zeros beeing integers is irrelevant? $\endgroup$ – Peter Elias Jun 19 '17 at 2:42
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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

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

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