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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.


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$\min(f_x(K))$ exists if $K$ is closed. You haven't really used compactness yet. How could you take advantage of that assumption? – Andrés E. Caicedo May 4 '12 at 19:59
@AndresCaicedo well $f_x(K)$ is compact for every $x$, however, that still does not say anything about the union since it is infinite right? – sxd May 4 '12 at 20:02
Right. How about a different approach: If the infimum is 0, there is a sequence $x_n$ of points of $K$ and a sequence $y_n$ of points of $\Omega^c$ with $\rho(x_n,y_n)\to0$. – Andrés E. Caicedo May 4 '12 at 20:04
@AndresCaicedo The only two theorems that I know about sequences in compact spaces are that they have a subsequence which converges and that compact spaces are complete. I don't see how that is going to help me here? – sxd May 4 '12 at 20:20
Ok. The sequence of $x_n$ has a subsequence that converges. The limit of this subsequence, call it $x$, must be in $K$ as well. To simplify, let's just say that the whole sequence of $x_n$ converges. Now, can you ensure that a subsequence of the $y_n$ converges, and that the limit $y$ is in $\Omega^c$? What would $\rho(x,y)$ be? – Andrés E. Caicedo May 4 '12 at 20:29
up vote 1 down vote accepted

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?

share|cite|improve this answer
How do you know that $\inf\{d(x,y),y\in F\}$ is not $0$? – sxd May 4 '12 at 20:37
I've added the argument. – Davide Giraudo May 4 '12 at 20:39
Oh, is that because the infimum of that set is equal to the minimum? – sxd May 4 '12 at 20:42

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