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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.

Anne Bauval
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    So $X=Y^n$, but what is $Y$? Mind that it is false that a metric space can be isometrically embedded in $\mathbb{R}^d$. – Andrea Mori Apr 06 '23 at 07:57
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    To rephrase Andrea's comment: "$X$ is the set of all $n$-tuples" does not tell us enough about what $X$ is. Where do the elements of the $n$-tuples come from? Are $x_1,\dots,x_n$ real numbers? Rational numbers? Functions? Unicorns?... – 5xum Apr 06 '23 at 08:53
  • They are real-valued. I've updated the question to show this. –  Apr 06 '23 at 09:15
  • So $X = \mathbb R^n$. – Paul Frost Apr 06 '23 at 09:32
  • Yes, I noticed this when answering. It may be the case that I shouldn't be allowed to use what I've listed as theorem 1 when answering this question but I am nonetheless given it in my lecture notes. I was under the impression that since the two metric spaces $(\mathbb{R}^n,e)$ and $(X,d)$ aren't equal that I would be required to prove that $(X,d)$ is closed regardless of equality between the underlying sets –  Apr 06 '23 at 09:54
  • Does this answer your question? Cartesian Product of Two Complete Metric Spaces is Complete – Anne Bauval May 01 '23 at 13:07

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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.

Maksim
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  • The argument here should be: these two norms are bi-Lipschitz, so they define the same Cauchy sequences. – Didier May 01 '23 at 16:17
  • Well, if norms are equivalent then that implies bi-Lipschitz. – Maksim May 04 '23 at 14:45
  • Sure, but that's the point for proving completeness. – Didier May 05 '23 at 06:04
  • Do not think so. I'd rather say that the norm equivalence implies the same topology, which leads to the result. – Maksim May 05 '23 at 14:51
  • Well, no, and that's my entire point. Completeness is not a topological property, but a metric one. $(\Bbb R,|\cdot|)$ and $(\Bbb R, \frac{|\cdot|}{1+|\cdot|})$ are homeomorphic metric spaces. The first is complete, the second is not. – Didier May 06 '23 at 18:34