Show that $\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$ is complete and compact with respect to given metric.
let the metric $d: \mathbb{T}^n \times \mathbb{T}^n \rightarrow \mathbb{R}$ be given by $d(x+\mathbb{Z}^n, y+\mathbb{Z}^n)$ = inf$\{ \Vert{v-w}\Vert; v \in x+\mathbb{Z}^n, w \in y+\mathbb{Z}^n\}$.
Here is what I've trying to do:
Since $\mathbb{T}^n$ is defined as a coset of $\mathbb{R}^n$ by relation such that $x\sim y$ if and only if $x-y$ is an integer, I guess the element of $\mathbb{T}^n$ can be uniquely written such as $x+\mathbb{Z}^n=\bar x=(x_1,x_2, ..., x_n)$, where $0 \leq x_i \lt 1$ for each $i$ .
and the distance between two points $\bar x, \bar y \in \mathbb{T}^n$ should be $\Vert (x_1-y_1,x_2-y_2,...,x_n-y_n)\Vert$ where $0 \leq x_i,\, y_i \lt 1$ for each $i$.
I think that it is possible to consider each $i$-th component of cauchy sequence $(x_k)$ on $\mathbb{T}^n$ as a cauchy sequence $({x_i}_k)$ on $[0,1]$.
Since $[0,1]$ is complete, each $i$-th component should converge to some point, say $z_i$, and then we can show the cauchy sequence on $\mathbb{T}^n$ converge to the point $z=(z_1, z_2, ..., z_n)$.
Also I know that in the metric space, compactness is equal to completeness and totally boundedness, so if I can show this space is totally bounded, then the compactness will follow.
I'd like to see whether the my above argument is right or not, and know more simple argument to this problem.
Thank you.
Consider the projection map $\pi:\Bbb R^n\to\Bbb R^n/\Bbb Z^n$. It satisfies $\|\pi(x)-\pi(y)\|\le\|x-y\|$, basically by definition, and so is Lipschitz and therefore continuous. Then $\Bbb T^n=\pi([0,1]^n)$ is the continuous image of a compact set and so is compact. Compact metric spaces are complete.