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If $K$ is a nonempty compact subset of $M$, and $A$ is any nonempty subset of $M$, I have to prove that there is a point $y$, an element of $K$ such that $\textrm{dist}(x, A) \leq \textrm{dist}(y, A)$ (where $\textrm{dist}(q, A) = \inf\{d(q, z) : z \in A\}$) for every $x \in K$, using the compactness in terms of open coverings.

I was trying to prove by contradiction, and it basically comes down the proving that the set of all distances from a point in $K$ to the set $A$ does not attain its supremum.

Any thoughts on how to proceed?

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No: if you want to proceed by contradiction you have to assume, not prove that the set of all distances from a point in $K$ to the set $A$ does not attain its supremum. (Well, it sort of depends on how you set up the proof, but in all likelyhood, this is what you want) –  Mariano Suárez-Alvarez Sep 30 '12 at 6:13

2 Answers 2

Let $s=\sup\{\operatorname{dist}(x,A)\mid x\in K\}$. For $t<s$ the set $U_t:=\{x\in K\mid \operatorname{dist}(x,A)<t\}$ is open (why?). If there is no $y\in K$ such that $\operatorname{dist}(y,A)=s$, then $K=\bigcup_{t<s}U_t$, hence there is a finite subcover $K=U_{t_1}\cup\ldots \cup U_{t_n}$ and we conclude $\operatorname{dist}(x,A)<\max\{t_1,\ldots,t_n\}$ for all $x\in K$, hence $s\le\max\{t_1,\ldots,t_n\}<s$, contradiction.

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You should be proving that the supremum is attained, not that it is not attained. Hint: if $\text{dist}(x,A) < \text{dist}(y,A)$, then $\text{dist}(z,A) < \text{dist}(y,A)$ for all $z$ in an open set containing $x$.

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