I've been asked to prove that a metric space $(X,d)$ is complete, where $X=\Bbb R^n$ and the metric $d$ is defined by $d(x,y)=\max_{j}|x_j-y_j|$ where $y=(y_1,...,y_n)$.
I tried to complete this by using two theorems
Theorem 1: The Euclidean space $\mathbb{R}^D$ is complete
Theorem 2: Let $(X,d)$ be a metric space. Let $E$ be a subset of $X$. If $X$ is complete and if $E$ is closed, then $E$ is complete
The only approach I could think of is to show that $X\subset \mathbb{R}^D$, then to show that for an arbitrary limit point $c=(c_1,...,c_n)$ of $X$, we have that $c \in X$. Would this approach work?
Edit: I am assuming that this approach is incorrect. I have also added the detail that the set X is the set of all real-valued n-tuples. My (very likely incorrect) attempt at the proof I shall also add here:
$(\mathbb{R}^D,e)$ where $e$ is the Euclidean metric is complete by Theorem 1
$X=\{x=(x_1,...,x_n)|x_i\in \mathbb{R}\}$
Let $D=n$. Then $\mathbb{R}^n=\{z=(z_1,...,z_n)|z_i\in \mathbb{R}\} \supset X$
Now, let $c=(c_1,...,c_n)\in \mathbb{R}^n$ be a limit point of $X$. Then there exists a sequence $(x_k)_{k\in \mathbb{N}}$ in $X$ for which $d(x_k,c)\rightarrow 0$. Note that $c=(c_1,...,c_n)\in X.$ Since $c$ was chosen arbitrarily and is a limit point of $X$, we have that $X$ contains all its limit points and is therefore closed.
So, by theorem 2, since $X\subset \mathbb{R}^n$, $\mathbb{R}^n$ is complete, and $X$ is closed, we have that $X$ is complete.
So by Th.1 $\mathbb{R^n}$ with $||.||_2$ is complete. As your your metric is induced by the norm $||x||_\infty=\max_{k=1,...,n}|x_k|$ and as the norms are equivalent (You see this by $||.||_2 \leq n ||.||_\infty $ and $||.||_\infty\leq ||.||_2 $): we can conclude due to this equivalence that your space $X$ must be also complete.