showing $T^n = \mathbb{R}^n / \mathbb{Z}^n$, $n$ dimensional torus is complete I want to prove $n$ dimensional torus $T^n = \mathbb{R}^n / \mathbb{Z}^n$ is complete?  [I know torus is compact since torus is a compact metric space, so from "Every compact metric space is complete" $T^n$ is complete.]
Can one prove this without using the compactness of $T^n$?

First I know this $n$ dimensional torus is a metric space, i.e.,
\begin{align}
  d(x+\Bbb Z^n, y+\Bbb Z^n) = \inf\{ \|v-w\| : v \in x+\Bbb Z^n \textrm{ and } w\in y + \Bbb Z^n\} 
\end{align}
and metric space is called complete if every Cauchy sequence has a limit.
Also I know $\mathbb{R}^n$ is a complete space. But I am not sure how its quotient space is complete.
 A: Let $(x_m) \subset \mathbb{R}^n/\mathbb{Z}^n$ be a Cauchy sequence. Let $(y_m) \subset \mathbb{R}^n$ be a lift of $(x_m)$. Note that we have some freedom in defining each $y_m$. Indeed,
$$y'_m = y_m + a_m$$
where $a_m \in \mathbb{Z}^n$ for $m \in \mathbb{N}$ defines another lift $(y'_m)$ of $(x_m)$. I claim that one may choose $(y_m)$ such that it is Cauchy in $\mathbb{R}^n$.
Indeed, since $(x_m)$ is Cauchy, then, taking $\epsilon = 1/4$, there is an $M \in \mathbb{N}$ such that
$$ l, m \geq M \implies d(x_l, x_m) < \frac{1}{4}$$.
In particular,
$$ \forall m \geq M, d(x_m, x_M) < \frac{1}{4}. $$
Let $y_M \in \mathbb{R}^n$ be any lift of $x_M$. Then for $m > M$, let $y_m$ be the unique lift of $x_m$ such that $d(y_m, y_M) < 1/4$. We have some freedom in defining the beginning of the sequence $(y_m)$, namely for $m < M$, but of course, this does not affect convergence, nor being Cauchy.
I claim that $(y_m)$, thus constructed, is Cauchy. Indeed, the tail of $(y_m)$, for $m \geq M$, is isometric to the tail of $(x_m)$, for $m \geq M$, so that $(y_m)$ is also Cauchy, albeit in $\mathbb{R}^n$, rather than the $n$-torus $\mathbb{R}^n/\mathbb{Z}^n$. But $\mathbb{R}^n$ is complete, so $(y_m)$ converges to, say, $y \in \mathbb{R}^n$.
By continuity of the projection map
$$ p: \mathbb{R}^n \to \mathbb{R}^n/\mathbb{Z}^n $$
since $(y_m)$ converges to $y$, it follows that $(p(y_m) = x_m)$ converges to $p(y)$.
