Quotient space of unit sphere is Hausdorff

Question

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

The quotient space of a topological space is Hausdorff if its graph is closed. That is, if $R=\{(x,y)\in S^1\times S^1|x \sim y \}$ is closed, then $S^1/\sim$ is Hausdorff. I am struggling to show that $R$ is closed.

$R=\{(x,y)\in S^1\times S^1|x \sim y \}=\{(e^{2\pi it_1},e^{2\pi it_2})\in S^1\times S^1|t_1-t_2=k,k \in\mathbb{Z}\}$. Easiest way I think, is to show that $S^1\times S^1-R$ is open. If $(x,y)\in S^1\times S^1-R$ where $x=e^{2\pi it_1},y=e^{2\pi it_2}$ $\implies$ $t_1-t_2=c,c\notin \mathbb{Z}$. But I do not know how to proceed from here. If this approach is too complicated, do suggest a different method. Thank you in advance. i changed the r to a t

Answer 1

Define $f:\mathbb S^1\times\mathbb S^1\to\mathbb S^1$ by $f(x,y)=xy^{-1}$, where if $y^{-1}:=e^{-ia}$ if $y=e^{ia}$. Since $f$ is continuous, the preimage $f^{-1}(\{1\})$ is closed. But this is precisely $R$.

Answer 2

This equivalence relation depends only on the parity of $k$, this is the equality relation if $k$ is even and if $k$ is odd it glue antipodal points. So we can only consider $k=1$, i.e $z \sim -z$.

We will proof that $S^1/\sim \cong S^1$. Indeed, consider the map $f : S^1 \to S^1, z \mapsto z^2$. This is constant on the orbits, injective on the quotient space. This induce a continuous bijection $S^1/\sim \to S^1$, but since $S^1/\sim$ is compact and $S^1$ is Haudorff, we have an homeomorphism. In particular the quotient is Hausdorff.

Answer 3

Your question is misstated .if t_1 - t_2 = k then e^2(pi)it_1 =e^2(pi)it_2 so the equivalence relation is the identity on S^1 and S^1 as a subspace of R^2 is clearly Hausdorff -It's a metric space . Your question is probably about the isomorphism between S^1 and R/Z induced by the map t--> e^2(pi)it which has kernel Z,