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I have to show the following: $(V,\rho)$ be a metric space, $K\subset V$ compact and $\Omega \subset V$ is open, then $d(K,\Omega^c) = \inf\{\rho(x,x') \mid x \in K \textrm{ and } x' \in \Omega^c\} > 0$.

I had the following in mind.

For every $x \in V$ $f_x: K \rightarrow \mathbb{R}: a \mapsto \rho(a,x)$ is continuous. Since $K$ is compact, $\min(f_x(K))$ exists. So I thought that it might be possible that $\{\rho(x,x') \mid x \in K \textrm{ and } x' \in \Omega^c\}$ contains its infimum and therefore $d(K,\Omega^c)$ cannot be $0$ since $K$ and $\Omega^c$ are disjoint.

Also, $\{\rho(x,x') \mid x \in K \textrm{ and } x' \in \Omega^c\} = \displaystyle\bigcup_{x\in \Omega^c} f_x(K)$

But since the minimum of a infinite union does not necessarily exists I have no clue how to proceed. So I wonder if I am looking into the right direction. If so, could anyone give me a hint? If not, could anyone point me into the right direction.

Please, DO NOT POST FULL ANSWERS!

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    @AndresCaicedo Well $\rho(x,y)$ would equal 0, which would give us a contradiction. Let me think about how I can find such a sequence $y_n$.2012-05-04

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Let $F=\Omega^c$. Then $F$ is closed. For each $x\in K$, $\inf\{d(x,y),y\in F\}$ is positive (since we assumed $K$ and $F$ disjoint, otherwise it's not true). Indeed, if $F$ is a closed set, $x\in F$ if and only if $\inf\{d(x,y),y\in F\}=0$.

The map $x\mapsto \inf\{d(x,y),y\in F\}$ is continuous. What about a positive map over a compact set?

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    Oh, is that because the infimum of that set is equal to the minimum?2012-05-04