# proving something about the infimum distance between 2 set

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 $\inf\{d(s,t):s\in S, t\in T\}$.

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You can use \{\} for braces in LaTeX. –  Asaf Karagila Oct 15 '12 at 8:03
–  Martin Sleziak Oct 18 '12 at 7:22

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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Let $m = \inf \{ d(s,t) \mid s \in S, t \in T \}$. Then for every $\varepsilon > 0$ there is $(s_n,t_n) \in S \times T$ such that $|d(s_n,t_n)-m|<\varepsilon$. Which means $\lim_{n \to \infty} d(s_n, t_n) = m$. Can we now do $\lim_{n \to \infty} d(s_n, t_n) = d(s_n,\lim_{n \to \infty} t_n)$? Or how do we get something of the form $d(s_n, T)$? How does one get from the product thing to something only depending on $S$? Thanks. –  Matt N. Oct 15 '12 at 7:09
@MattN.: It can be tricky to work with limits here, so you are better off working with infima instead. First, we should note a missing definition from the above: $d(s,T)=\inf\{d(s,t):t\in T\}$. Now it is clear that $d(s,t)\ge d(S,T)$ always, and therefore $d(s,T)\ge d(S,T)$ as well. So $\inf\{d(s,T):s\in S\}\ge d(S,T)$. We want equality, so we need to show the opposite inequality. To do that, pick any $M>\inf\{d(s,T):s\in S\}$. By definition, there is some $s\in S$ with $d(s,T)<M$. Again from defintions, there is a $t\in T$ with $d(s,t)<M$. And so $d(S,T)<M$. –  Harald Hanche-Olsen Oct 15 '12 at 9:10
(There are still details to be filled in, of course, but I hope the previous comment helps.) –  Harald Hanche-Olsen Oct 15 '12 at 9:11
Thank you very much. It took me very long to understand, I'm sorry for the late reply. Now I think I understand. Would you mind if I posted a second (full) answer? If you do, I might still post it but shortly after delete it so that only +10k users can see it. But I think I'd benefit from writing it up. –  Matt N. Oct 15 '12 at 17:59
@MattN.: You're welcome. By all means, write up the full answer. I don't quite see the point of deleting it afterwards, though. Oh, yeah, the homework angle. Well, if you do, then come back and undelete it after a few days. I'm sure you as the author of the answer will be able to do so. –  Harald Hanche-Olsen Oct 15 '12 at 19:27
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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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You write "Let $s$ denote the limit of $s_n$." Why does $s_n$ converge? And if it does, why should $d(s,T) \leq d(S,T)$? You seem to switch quantifiers in the last sentence: you start with "for each $t \in T$ there is a sequence" and you end up with "there is $s \in S$ such that for all $t$..." (paraphrased). Such an $s$ can't exist in general. You can salvage your approach: try working with two sequences $s_n$, $t_n$ such that $d(s_n,t_n) \to d(S,T)$. –  commenter Oct 18 '12 at 6:50
@commenter Exactly my question too. Why should $d(s,T) \leq d(S,T)$? I discussed the other direction on chat and what appears to be a switching of quantifiers is not. Thanks for the comment, I'll rewrite this later. Maybe. –  Matt N. Oct 18 '12 at 6:56
Oh, right, it's not switching quantifiers, it's only bad notation: using $s$ for two different things. Confusing. –  commenter Oct 18 '12 at 6:59
@commenter Slave driver : ) –  Matt N. Oct 23 '12 at 9:24
The definitions and the choice of $(s_k,t_k)$ show that $d(S,T) \leq d(s_k,t_k) \leq d(S,T) +\frac{1}{k}$. Choose integers $k(n) \geq n$ such that $d(s,s_{k(n)}) \leq \frac{1}{n}$ (convergence of some subsequence). Since $k(n) \geq n$ you have $$d(S,T) \leq d(s,t_{k(n)}) \leq d(s,s_{k(n)}) + d(s_{k(n)},t_{k(n)})\leq \frac{1}{n} + d(S,T) + \frac{1}{n} \xrightarrow{n\to\infty} d(S,T),$$ so $d(s,t_{k(n)}) \to d(S,T)$ by the squeezing lemma. Since $$d(S,T) \leq d(s,T) \leq \lim_{n\to\infty} d(s,t_{k(n)}) = d(S,T)$$ this gives $d(s,T) = d(S,T)$. –  commenter Oct 23 '12 at 10:05