1. ## Complete metric spaces.

If completeness of a metric space is defined as,

A metric space $(M, d)$ is complete if every Cauchy sequence ${x_n}$ in M converges.

I am stuck trying to verify the following metric space is complete.
$\mathbb{R}^k$ with the metric defined as $d(x,y)=\max _{1\leqslant i\leqslant k}\left|x_i-y_i\right|$

Cleary this is a metric space, but how can I see if its complete?

I thought a verification of completeness might be similar to testing whether the euclidean space is compete (that is with aid of the Bolzano Weierstrass theorem). But that theorem doesn't hold for general metric spaces.

2. Originally Posted by aukie
If completeness of a metric space is defined as,

A metric space $(M, d)$ is complete if every Cauchy sequence ${x_n}$ in M converges.

I am stuck trying to verify the following metric space is complete.
$\mathbb{R}^k$ with the metric defined as $d(x,y)=\max _{1\leqslant i\leqslant k}\left|x_i-y_i\right|$

Cleary this is a metric space, but how can I see if its complete?

I thought a verification of completeness might be similar to testing whether the euclidean space is compete (that is with aid of the Bolzano Weierstrass theorem). But that theorem doesn't hold for general metric spaces.
Let $\{x_n\}\subset\mathbb{R}^k$ be a Cauchy sequence. For each $x_j\in\{x_n\}$, define $x_j^i$, $1\leq i\leq k$, to be the $i$th component of $x_j$.

Prove that for each $i$ the sequences $\{x_n^i\}$ are all Cauchy under the standard metric over $\mathbb{R}$ and are therefore convergent to numbers $y_1,...,y_k\in\mathbb{R}$. Then show that the sequence $\{x_n\}$ converges to $y=(y_1,...,y_k)$.