TITLE: Proof $\mathbb{R}^n$ is a complete metric space.
QUESTION [7 upvotes]: $\mathbb{R}^n$ is a complete metric space.

Consider a Cauchy sequence $\{\mathbf{x}_k\}$ in $\mathbb{R}^n$, we want to show it converges to a point $\mathbf{x} \in \mathbb{R}^n$. That is to say, if $|\mathbf{x - x_k}| \to 0$ as $k \to \infty$.
Hence, we let $\epsilon \to 0$, and we get $\mathbf{|x_k - x_j|} < \epsilon$ by Cauchy sequence, and let $\mathbf{x = x_j}$ we showed the desired result.

Definition $\mathbb{R}^n$ is a complete metric space. Every Cauchy sequence in $\mathbb{R}^n$ converges to a point of $\mathbb{R}^n$.
Definition Cauchy sequence. Given $\epsilon > 0$, there is an integer $K$ such that $\mathbf{|x_k - x_j|} < \epsilon$ for all $k,j \geq K$.


I am not fond of my proof, because I am not certain if I can approach $\epsilon$ to be zero, nor if I can equate $\mathbf{x}$ to be $\mathbf{x_j}$ since $\mathbf{x_j}$ is changing while $\epsilon$ changes.
Edit Especially, I am baffled that why we need to do it in coordinates? I think they can be subtracted directly, as the definition of Cauchy sequence I added a short while ago?

REPLY [8 votes]: Every Cauchy sequence is bounded, hence contained in a compact cube, admits thus a convergent subsequence and therefore converges itself.