# How to show that $\mathbb{R}^N$ is a complete metric space

How may I show that for $N \geq 2$, $\mathbb{R}^N$ is a complete metric space given that the metric defined is $d = ||x-y||_2$?

A complete metric space is a metric space where every cauchy sequence converges in the space.

I think I would start off saying that given our metric $d = ||x-y||_2$, and some sequence $x_n$, then the sequence is cauchy if $||x_n - x_m||_2 < \epsilon$ for some $n,m > M$ where $M$ is a positive integer. Since I must show that the sequences converge in the metric space, I will have to of course use the definition of a converging sequence. That is, the converging sequences definition is: $\forall_{\epsilon > 0} \exists_{N \in \mathbb{N}}, \; \forall_{n > N} \left ( (x_n - L) < \epsilon \right)$.

Though, i'm not quite sure how to tie my information together.

I imagine that you know that $\mathbb{R}$ is complete. Right?
Then the idea is to look at the coordinates (there are $N$) of your sequence.
Suppose that $\{x_n\}$ is Cauchy. Then, each sequence of components $\{x_{n,i}\}$ ($i=1,\ldots,N$) is Cauchy and, by the completeness of $\mathbb{R}$, convergent to some $L_i$. Let $L=(L_1,\ldots,L_N)$. Then, $x_n\to L$.