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If $(X,d)$ is a complete metric space and $A$ is closed in $(X,d)$ then show that for each $x \in X$ there exists an element $a_0 \in A$ such that $d(x,A)=d(x,a_0).$

I tried this problem several times but always got stuck at some point or another. For instance, I initially thought about showing the existence of a Cauchy sequence $(a_n)$ in A such that $d(x,a_n) \to \operatorname{inf}\{d(x,a):a \in A\}=d(x,A)$ and then by completeness of $A$ and by continuity of the function $f(a)=d(x,a)$, to prove the existence of an element $a_0$ in $A$ such that $a_n \to a_0$ which would imply $d(x,a_n)=f(a_n) \to f(a_0)=d(x,a_0)=d(x,A)$. But the problem was that I could not construct such a Cauchy sequence in $A$. All my other attempts have similarly gone in vain.

If anybody has any answer, please reply. Thank you.

Sayantan
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    This is not true in general; se e.g. here :http://math.stackexchange.com/questions/481843/a-counter-example-of-best-approximation/481861#481861 However, this is true if $X$ is a finite-dimensional normed space, or more generally if every bounded sequence in $X$ has a convergent subsequence: in this cas your proof works. This is also true if $X$ is a Hilbert space and $A$ is convex. – Etienne Sep 29 '13 at 15:49

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Etienne is right, and here is another counterexample:

  • let $X$ be the sequence space $\ell^p$ (the choice of $p\in [1,\infty]$ does not matter)
  • $A=\{(1+1/n)e_n:n\ge 1\}$ where $e_n$ are standard basis vectors
  • $x=0$

Then the distance from $x$ to $A$ is $1$, but it is not attained. The set $A$ is closed, because any convergent sequence in $A$ is eventually constant.

You need some sort of compactness to get the infimum to be attained. In a reflexive space, you can use weak* compactness (Banach-Alaoglu) and then Mazur's lemma; as Etienne said, the second step requires $A$ to be convex.

user98130
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