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I have two subsets of $\mathbb{R}^n$, given by $K$ and $F$, $K$ is compact and $F$ is closed. I'm trying to show that $\inf\{ d(x,y) : x \in K, y \in F \}$ is attained.

My ideas so far:

  • I know that a continuous function on a compact set attains it's bounds. So that $f_y: K \to \mathbb{R}$ given by $f_y(x) = d(x,y)$ attains it's bounds for each $y$ but this doesn't quite give the result.

  • I need to use the fact we are working in $\mathbb{R}^n$ somehow, perhaps by using the equivalence between compactness and sequential compactness?

  • The fact $\mathbb{R}^n$ is Hausdorff, means that $K$ is also closed. So both sets contain their limit points.

Unfortunately, I am struggling to put these ideas together, thank you for any help

Vincenzo Tibullo
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Wooster
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  • Look at the function $h(x) = \inf { d(x,y) : y \in F}$. – Daniel Fischer Jun 24 '14 at 13:10
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    Choose sequences $(f_n)$ in $F$ and $(k_n)$ in $K$ with $\lim d(f_n,k_n)$ equal to your $\inf$. Choose $k$ a limit point of $(k_n)$. Then $(d(k, f_n))$ converges to your $\inf$. This implies $(f_n)$ is bounded. So... – David Mitra Jun 24 '14 at 13:14
  • @DanielFischer, I did try to look at that function, but found it a bit tricky to show that it is continuous! – Wooster Jun 24 '14 at 13:14
  • @DavidMitra I'm just struggling to see why choosing $k$ a limit point means $d(k,f_n)$ converges to the inf? – Wooster Jun 24 '14 at 13:18
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    @Wooster Show more, that's easier: In any metric space, for a nonempty set $A$, define the function $f_A\colon x \mapsto \inf { d(x,a) : a \in A}$. Show that $\lvert f_A(x) - f_A(y)\rvert \leqslant d(x,y)$. (And note $f_A(x) = 0 \iff x \in \overline{A}$.) – Daniel Fischer Jun 24 '14 at 13:20
  • Sorry, I slipped. I should have said: if $k_{n_l}\rightarrow k$, then $(d((k, f_{n_l}))$ converges to the $\inf$. – David Mitra Jun 24 '14 at 13:27
  • @DavidMitra sorry to ask more questions, but I'm still not sure I follow. How do we know that $k_n$ has a convergent subsequence? – Wooster Jun 24 '14 at 13:35
  • Because $K$ is sequentially compact. – David Mitra Jun 24 '14 at 13:37
  • Of course sorry, I was thinking it was in $F$ for some reason. – Wooster Jun 24 '14 at 13:38
  • @DavidMitra Would you mind possibly elaborating on this argument a bit, I've managed to understand Daniel's proof but I am struggling with this one. How do we know that $d(k,f_{n_j})$ converges to the inf? – Wooster Jun 24 '14 at 14:04
  • Daniel's approach is better. But: Let $\alpha$ denote the $\inf$. We have $\alpha\le d(k, f_{n_l})\le d(f_{n_l}, k_{n_l}) + d( k_{n_l},k)$ for all $l$. The right-hand side of this inequality converges to $\alpha$. So, the result follows by the Squeeze Theorem. – David Mitra Jun 24 '14 at 14:29
  • Okay yes, and then because $F$ is closed, the limit $f$ is in $F$, and so the inf is attained – Wooster Jun 24 '14 at 14:36

1 Answers1

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Hint

The property of $\mathbb{R}^{n}$ you need to use is that a set is compact if and only if it is bounded and closed.

Consider the set $F'= \{y \in F \mid d(K,y) \leqslant d(K,F)+1\}$. You can easily show that $d(K,F)=d(K,F')$ and I'm sure you can take it from there.

Martin Sleziak
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Sergio
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