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Can anyone please explain to me how one could apply the nearest theorem to deduce if there is a unique point, nearest to a set.

E.g., Show that if $F$ is a non-empty closed set in ${\bf R}^p$ and if $x$ is not an element of $F$, there is a unique point of $F$ that is nearest to $x$.

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Did you forget a convexity assumption on $F$? – t.b. May 7 '12 at 8:58
The existence of at least one nearest point follows from $F\cap \overline{B_r(x)}$ being compact and nonempty for some $r$. As noted above, in general this might not be unique though. – Alex Becker May 7 '12 at 9:01
well, my books don't say anything about convextity, but i am assuming it is a convex set.In any case, does that alter the solution to the abve? – Patrova May 7 '12 at 9:11
Without convexity assumptions, there is no uniqueness. Consider $F$ the complement of an open ball and $x$ the center of this ball. – anon May 7 '12 at 9:27
You write that you want to apply "the nearest theorem." I assume you want to apply "the nearest point theorem." If that's right, can you edit your question accordingly, and also state the nearest point theorem, so we know what we are trying to apply? – Gerry Myerson May 7 '12 at 12:39

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