Let $S,T$ be 2 set. Prove that there is a $x\in S$ s.t, $d(x,T)=d(S,T)$ if $S$ is a compact set. Here, $d(S,T)$ denoted the $\inf\{d(s,t):s\in S, t\in T\}$.
proving something about the infimum distance between 2 set
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0is there any way to attack this problem without use of functions, limits etc? Ie is there a way to prove this using concepts of open balls, closed balls, compactness? – 2016-09-22
2 Answers
Let $S,T$ be subsets of a metric space $(X,d)$. Let $S$ be compact. Let $d(x,T) := \inf_{t \in T} d(x,t)$ and $d(S,T) := \inf_{(s,t) \in S \times T} d(s,t)$.
Claim: There exists $s$ in $S$ such that $d(s,T) = d(S,T)$.
Proof: We have $d(S,T) \leq d(s,T)$ for all $s$ in $S$. Hence it is enough to show that there exists $s$ in $S$ such that $d(S,T) \geq d(s,T)$.
$d(S,T) = \inf_{(s,t) \in S \times T} d(s,t)$ means that for $k$ we can find $(s_k, t_k)$ such that $d(s_k , t_k ) \leq d(S,T) + \frac{1}{k}$. Then $ \inf_{t \in T} d(s_k, t) = d(s_k, T) \leq d(S,T) + \frac{1}{k} $ Since $S$ is compact it is sequentially compact hence $s_k$ has a convergent subsequence. Let its limit be denoted by $s$. Then $\lim_{k \to \infty} d(s_k, T) = d(s,T) \leq \lim_{k \to \infty} d(S,T) + \frac{1}{k} = d(S,T)$
Which proves the claim.
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0@commenter Thank you! That's what I was after. – 2012-10-23
It smells like homework, so I'll just give a hint: $d(x,T)$ is a continuous function of $x$ (if you don't know it, you need to prove it, of course).
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0I've been trying all day to write down something that doesn't use continuity of the metric. Yesterday I thought I had it but I didn't write down the idea and by now I'm convinced it wasn't right. I've read your argument using $M$ several times but still don't understand it. I think I have a (hopefully correct) proof of one direction (the easy one) which I posted as an answer. Unfortunately, I'm stuck for the other direction. – 2012-10-17