Im looking for a counter example for the following problem:
Let $ S_{1}\neq\emptyset $ be a compact set and $S_{2}\neq\emptyset $ a closed set. (Those are subsets of $ \mathbb{R}^n $ for some $ n $).
So there exists $ s_{1}\in S_{1},\thinspace\thinspace\thinspace s_{2}\in S_{2} $ such that for the distance between the sets:
$ d\left(S_{1},S_{2}\right):=\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $
we have $ \inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right)=d\left(s_{1},s_{2}\right) $
Originally this was a prove/disprove question, but Im pretty sure that this is false, so Im looking for a counter example. Thanks in advance.
After reading the comments:
This is my attempt to prove the statement (rather then disprove it):
Let $ d\left(x_{n}^{1},x_{n}^{2}\right)$ be a sequence such that $ \lim_{n\to\infty}d\left(x_{n}^{1},x_{n}^{2}\right)=\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $.
Since $S_1 $ is compact, there exists a convergent subsequence of $ x_{n}^{1} $, say $ x_{n_{k}}^{1} $.Denote its limit by $ x_1 $. One can verify that
$ d\left(x^{1},x_{n_{k}}^{2}\right)\underset{n\to\infty}{\longrightarrow}\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $
I tried to prove that $ x_{n_{k}}^{2} $ also have a convergent sequence and that would end the proof. But I cant see how.
I think it's true now with the nonempty condition
Note that a subset of $\mathbb{R}^n$ is compact $\iff$ it is closed and bounded $\iff$ it is sequentially compact
The first important step is to realise that we can make it so that we only care about a bounded part of $S_2$
Choose any $x_1 \in S_1$ and $x_2 \in S_2$ to get an upper bound $D ≔ d(x_1, x_2)$ on $d(S_1, S_2)$
Since $S_1$ is bounded, it is contained in some large closed ball $\overline{B}(0, M)$
Hence the only points that we care about from $S_2$ are going to be within a distance $D$ of this ball
That is to say, we can w.l.o.g. consider ${S_2}' ≔ S_2 \cap \overline{B}(0, M+D)$ instead of $S_2$, which is now (sequentially) compact
From here we can take a sequence of pairs $a_n \in S_1$ and $b_n \in S_2$ with $d(a_n, b_n) \rightarrow d(S_1, S_2)$, and appeal to sequential compactness of our sets and continuity of $d$ to find the desired pair of points