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Let $(X, d) $ be metric space and $A\subset X$ and $x_0\in X$

Define $d(x_0, A) =\inf\{d(x_0, a) :a\in A\}$

What condition(s) on $X$ and $A$ can ensure the existence of an element $a\in A$ such that $d(x_0, a)=d(x_0, A) $ ?

Can we take care of uniqueness as well?

Motivation:

Suppose $X$ is a Hilbert space and $A\subset X$ non empty closed convex set then for all $x_0\in X$ $\exists! a\in A$ such that $d(x_0,a) =d(x_0, A) $

$X$ is reflexive and strictly convex iff $ A\subset X$ non empty closed convex set in $X$ is a Chebyshev set.

Particular cases:

Suppose $(X,d)$ be any metric spaces $A\subset X$ non empty compact set then $\forall x_0\in X, \exists a\in A$ such that $d(x_0, a) =d(x_0, A) $

Sketch:

$d_{x_0}:X\to\Bbb{R}$ is continuous and $A\subset X$ compact implies $\inf\{d(x_0, a) :a\in A\}$ is attained in A.

Going deep into the proof:

$d(x_0, A)=\inf\{d(x_0, a) :a\in A\}$ implies $\exists (a_n) \in A$ such that $d(x_0, a_n) \to d(x_0, A) $

$\lim _{n\to\infty}d(x_0, a_n)= d(x_0, A) $

Implies $ d(x_0,\lim_{n\to\infty} a_n)= d(x_0, A) $

We only need to make sure that $(a_n)\subset A$ converges to $a\in A$ for some $a\in A$

$$\text{OR}$$

$(a_n) $ has a convergent subsequence in $A$

Further treatment:

$(X, d) $ complete and $A\subset X$ non empty closed totally bounded set.

But no improvement at all! (Complete and totally boundedness $\iff$ compact.)

Mostafa Ayaz
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Sourav Ghosh
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    A slightly weaker condition is $A\cap B$ compact for any bounded set $B$. For example, $\mathbb N$ is not compact but its intersection with any bounded set is. – geetha290krm May 14 '23 at 08:02
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    I don't think you can find a general useful condition. For example say $X=\mathbb{R}^n$. Now fix a point $v_0$ of norm $1$. Then for any $A\subseteq\mathbb{R}^n$ outside of the unit disk we will have $d(0,A\cup{v_0})=d(0,v_0)$. So, this condition is extremely weak. It would be more interesting if it would hold for all $x_0$. – freakish May 14 '23 at 08:21
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    If $A$ is closed then we can guarantee the existence of $a\in A$ satisfying the condition, no? What I am missing? – Daniel Huff May 14 '23 at 18:28
  • @DanielHuff what about $X = [0,1) \cup {2}$ with the induced metric from $\mathbb{R}$ and $A = [0,1)$? Then $d(2, A) = 1$ but there is no $a$ such that $d(2, a) = 1$. – ronno May 15 '23 at 09:21
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    Along the same lines as @geetha290krm's comment, if $x_0$ varies then a necessary condition is for $A$ to be closed, and a sufficient condition is to have $A \cap \overline{B}$ compact for closed balls $\overline{B}$. So if $X$ is such that its closed balls are compact ($\implies$ $X$ is complete, but not equivalent) then it is necessary and sufficient that $A$ be closed. – ronno May 15 '23 at 09:31
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    The question is too general to be answerable, in this generality, the only reasonable answer is "sometimes yes, sometimes not." See also my answer here in the context of closed linear subspaces of Banach spaces. – Moishe Kohan May 15 '23 at 12:54

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