For compact $A$, $\inf\{\varrho(y,x) : y \in A\}=\varrho(a,x)$

Question

I need help with prooving that if non empty $A$ $\subset(X,\varrho)$ is compact, then:

$(\forall x \in X) (\exists a \in A) \inf\{\varrho(y,x) : y \in A\}=\varrho(a,x) $

I found this solution:

$\varrho_{A}(x) = \inf\{\varrho(y,x) : y \in A\}=\lim_{n} \varrho(y_{n},x)$ for some sequence $y_{n} \in A$.

We pick subsequence $y_{n_{k}}\rightarrow a$ and then:

$\varrho_{A}(x) = \lim_{n}\varrho(y_{n},x)=\lim_{n} \varrho(y_{n_{k}},x)=\varrho(a,x)$

I don't get this solution. Is there any other way to prove it? Also, wouldn't it be enough if $A$ was closed, not compact? The way of proving that $\inf\{\varrho(a,b): a\in A, b\in B\}=\varrho(A,B)$ for compact $A$ and $B$ would be similar?

Thanks in advance!

Answer 1

It seems the following.

$\varrho_{A}(x) = \lim_{n}\varrho(y_{n},x)=\lim_{n} \varrho(y_{n_{k}},x)=\varrho(a,x) $

This solution seems to be OK.

Is there any other way to prove it?

Yes. If a point $x\in X$ is fixed then $\varrho(x, a)$ is a continous function on a compact set $A$, therefore it attains its infimum at some point $a\in A$.

Also, wouldn't it be enough if $A$ was closed, not compact?

No. Let $(X,\varrho)$ be a $\Bbb Q$ endowed with the standard metric, $A=\{a\in\Bbb Q: a>\sqrt{2}\}$, and $x=0$. Then $A$ is a closed subset of the space $X$, but the distance from the point $x$ to a set $A$ is not realized by a point $a\in A$.

The way of prooving that $\inf\{\varrho(a,b): a\in A, b\in B\}=\varrho(A,B)$ for compact $A$ and $B$ would be similar?

Yes.