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When $K$ is a nonempty subset of a metric space $M$, $\forall x \in M$ let $P_K(x)=\{y \in K: d(x,y)=d_K(x)=\inf_{k \in K}d(x,k)\}$ (metric projection)

$\bullet$ The set $K$ is $proximinal$ in $M$ if $\forall x \in M$ the set $P_K(x) \neq \emptyset$.

$\bullet$ And The set $K$ is $\varepsilon-proximinal$ in $M$ if $P_K(x)\neq \emptyset$ for each point $x$ of $M$ with $d_K(x)>\varepsilon$.

It's clear every $proximinal$ set is $\varepsilon-proximinal$ set. I'm looking for a $\boldsymbol{\varepsilon}\textbf{$-proximinal$ that is not $proximinal$}$ set ?!

A.Ebadi
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  • It is recommended that you provide at least an outline about your own attempt and ideas at solving the problem. – Bunny Oct 07 '17 at 20:02
  • This is an interesting problem. Could you find an example? What makes you think that such an example exists? – Giuseppe Negro Jun 20 '18 at 16:05
  • @GiuseppeNegro Thank you for your attention. I thought it would be possible to find such an example among sets that lack some of their cluster points (unclosed). – A.Ebadi Jun 21 '18 at 21:45
  • @A.Ebadi: indeed, if you allow for nonclosed sets then there are easy examples; see my answer. – Giuseppe Negro Jun 21 '18 at 22:46

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The set $K=[-1, 0)\cup(0, 1]\subset \mathbb R$ is $\epsilon$-proximinal for every $\epsilon>0$ but it is not proximinal, because $P_K(0)=\varnothing$.

It would be harder and more interesting to find examples of a closed set that is $\epsilon$-proximinal for every $\epsilon>0$ but not proximinal. I don't know if such an example exists.

EDIT. Such an example does not exist for trivial reasons.

Claim. If a closed set $K$ is $\epsilon$-proximinal for all $\epsilon>0$, then it is proximinal.

Proof. If $x\notin K$, then there exists $\epsilon>0$ such that $\mathrm{dist}(x, K)=\epsilon$ because $K$ is closed. Therefore $P_K(x)\ne \varnothing$. $\Box$

Giuseppe Negro
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  • Exactly, might also be interesting to see Closed set that is not proximinal set. – A.Ebadi Jun 22 '18 at 06:43
  • Do you mean that since K is closed then every point in its complement has a nearest point in the K? Or I got it wrong? Is the $\epsilon-proximinality$ assumption used? – A.Ebadi Jun 29 '18 at 10:40
  • Yes, I use the assumption of $\epsilon$-proximinality. If $x\notin K$, then $x$ is in the complement of $K$, which is an open set by definition. Therefore there exists a ball of radius $\epsilon>0$ that contains $x$ and that does not intersect $K$. Since $K$ is $\epsilon$-proximinal, the metric projection $P_K(x)$ is not empty. And since $x$ was arbitrary, we have proven that $P_K(x)\ne \varnothing$ for all $x$. NOTE: This needs that $K$ is $\epsilon$-proximinal for all $\epsilon >0$. – Giuseppe Negro Jun 29 '18 at 11:00
  • (A side note. If you see an answer that you like, you can upvote it, by clicking on the up arrow on the left. If you are satisfied with an answer, you should accept it by clicking on the green tick sign. I am not saying that you should upvote and/or accept my answer, I am just informing in case you did not notice. Thank you) – Giuseppe Negro Jun 29 '18 at 11:03
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    Anyway: the interesting problem would now be to find a closed set $K$ in a metric space that is $\epsilon$-proximinal for a fixed value of $\epsilon$, but is not proximinal. (One can assume $\epsilon=1$). Such example must be infinite-dimensional as closed sets in finite-dimensional normed spaces are all proximinal. – Giuseppe Negro Jun 29 '18 at 11:19
  • Given the point you mentioned, yes, you are absolutely right . Thank you very much. – A.Ebadi Jun 29 '18 at 11:29
  • If we define the $\epsilon-proximinality$ as follows, we can also ask the same question:\K is $\epsilon−proximinal$ in M if $P_K(x)≠∅$ for each point x of M with $d_K(x)<ε$. – A.Ebadi Jun 29 '18 at 11:40