I want to prove that $T^n=\mathbb{R}^n/\mathbb{Z}^n$ is Hausdorff.
We define in $\mathbb{R}^n$ the equivalence relation $\sim$ by $x\sim y$ if and only if $y=z+x$ where $z\in\mathbb{Z}^n$. Endowing the space $\mathbb{R}^n/_\sim$ with the quotient topology, the projection map $p:\mathbb{R}^n\to\mathbb{R}^n/_\sim$ is open. In fact, if $U\subset\mathbb{R}^n$ is open, then $p^{-1}(p(U))=\bigcup_{z\in\mathbb{Z}^n} U+z$, which is an union of open subsets ($\therefore$ open). Now given that $\mathbb{R}^n$ is Hausdorff and $p$ is open, to prove that $T^n$ is Hausdorff, we can prove that the graph of $\sim$ defined by $R_\sim=\{(x,y)\in\mathbb{R}^n\times\mathbb{R}^n: x\sim y\}$ is closed in $\mathbb{R}^n\times\mathbb{R}^n$.
I would like to know if my proof is correct:
Let $f:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ defined by $f(x,y)=f(x_1,\ldots,x_n,y_1,\ldots,y_n)=\sum_{i=1}^n \sin(\pi(y_i-x_i))^2$. Then, clearly $f$ is continuous, if $y=x+z$ with $z\in\mathbb{Z}^n$ this implies that $y_i=x_i+z_i$ for all $1\leq i\leq n$ and thus $f(x,x+z)=\sum_{i=1}^n \sin(\pi z_i)^2=0$. Conversely, if $f(x,y)=0$ then $\sin(\pi(y_i-x_i))=0$ for all $i$. Therefore $y_i-x_i\in\mathbb{Z}$ for all $i$ and thus $x\sim y$. In conclusion $f^{-1}(0)=R_\sim$ which is closed.
Thanks!
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Yes, your proof is correct.