Quotient space of unit sphere is Hausdorff 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
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 $R$.
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,