Suppose we have a metric space $(X,d)$. Let $S$ be a compact subset of $X$. Provide me with an example of $X$, and $S$ (closed and bounded in $X$) such that $$\min \{d(p_0,p): p \in S\}$$ does not exist, where $p_0 \in X$.
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Martin Sleziak
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h.tutar
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Is $p_0$ fixed or not? – Vim May 02 '16 at 06:47
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1The question is not very well formulated. Since this is a question for a an example, it should start with asking for that, not with "Suppose" and "Let" (which would be appropriate for a question about a proof). – Marc van Leeuwen May 02 '16 at 08:28
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@MarcvanLeeuwen This is interesting (to me, anyway). How would you word a question like this? – Thomas May 02 '16 at 08:49
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Just start with: "Can anybody give me an example of the following situation" and then describe the conditions that you would like to have satisfied. With "suppose" or "let" one usually introduces a situation that is given and one is not allowed to tamper with; here it is the opposite. – Marc van Leeuwen May 02 '16 at 09:14
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2Instead of telling people what to do, tell them what your problem is. – Carsten S May 02 '16 at 09:14
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This question seems to be related: For any point $ a $ of a compact subset $ S $ of a metric space, prove that there exists a nearest point $ c $ to $ a $. (It was in the list of related questions on the right.) – Martin Sleziak May 02 '16 at 13:01
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I'll assume $S$ not empty.
Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum.
If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with the metric induced by $\mathbb{R}$. Then $S=(1,2]$ is closed and bounded, but $$ \inf_{p\in S} d(0,p)=1 $$ and there's no $p\in S$ such that $d(0,p)=1$.
egreg
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This is slick! Actually I've been having a hard time looking for a counterexample in $\ell^2$ but apparently I've gone the troublesome way. – Vim May 02 '16 at 10:16
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The function $x\mapsto d(p_0,x)$ is continuous on $X$, hence attains its minimum on $S$ since $S$ is compact.
carmichael561
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1No. $S$ isn't necessarily compact. And the min may not be attained either. See egreg or Antitheos answers for simple counterexamples. – Vim May 02 '16 at 10:22
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what I saw was "bounded and closed". But the OP seemed to have confused the two. – Vim May 02 '16 at 15:22