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This has been asked heavily in the past, for instance here and here.

Let $x\in X$ and $\operatorname{dist}(x,A)=\inf\lbrace \rho(x,a): a\in A\rbrace$ for any nonempty subset $A$ of X.

It is a fact that if $A$ is compact, then there exists an $a\in A$ such that $\operatorname{dist}(x,A)=\rho(x,a)$.

If $X=\mathbb{R}^n$ and $A$ is closed, then the conclusion of $b$ holds.

I don't see quite well why I need $\mathbb{R}^n$ besides the obvious. If $A$ is also bounded, then $A$ is compact and there is nothing to prove. If $A$ is closed but unbounded, how can I prove the statement? I'm confused because closed subsets can be quite different. For instance I can choose $\mathbb{N}$, but also $\cup [0,a], a\geq0$. I think both are closed and unbounded, but one has a nonempty set of limit points while the other not.

user2820579
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1 Answers1

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Let's say $A$ is closed and $\text{dist}(x,A)=d\ge0$. Now, for each $d+\frac{1}{n}$ there has to be some element $y_n$ in $A$ such that $\rho(x,y_n)<d+\frac{1}{n}$.

Consider the sequence $(y_n)_{n\in\Bbb N}$. We have that $\|y_n\|=\rho(y_n,0)\le\rho(y_n,x)+\rho(x,0)=\rho(x,y_n)+\|x\|<d+\frac{1}{n}+\|x\|\le d+1+\|x\|$, so the sequence is bounded. By Bolzano-Weierstrass theorem (which is valid in $\mathbb{R^n}$) it has a convergent subsequence, let's say $(y_{n_k})$ with limit $y$. Then $\rho(y_{n_k},x)<d+\frac{1}{n_k}\xrightarrow{n_k\to\infty}d$, which means $\rho(y,x)=d$, since $A$ is closed and $(y_{n_k})$ is in $A$, so $y\in A$.

user2820579
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Darsen
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  • But where's the explicit use of $\mathbb{R}^n$? – user2820579 Nov 03 '20 at 23:25
  • In the Bolzano-Weierstrass theorem. – Darsen Nov 03 '20 at 23:26
  • By hypothesis. If $d=\operatorname{dist}(x,A)=\inf\lbrace \rho(x,a): a\in A\rbrace$, then for every $n$ there must be some element in $A$ closer to $x$ than $d+\frac{1}{n}$, since otherwise there would be some $n_0$ such that every element $a$ would give us $\rho(x,a)\ge d+\frac{1}{n_0}$, which isn't possible, since taking the infimum we wouldn't get $d$ in the first place. – Darsen Nov 03 '20 at 23:36
  • Yeah, sorry I saw because it was for the infimum. – user2820579 Nov 03 '20 at 23:37