0
$\begingroup$

Let $X$ be a metric space, $A$ a compact subset of $X$ and $B$ a closed subset such that $A\bigcap B=\emptyset$. Prove that $d(A,B)=d(x_{0},y_{0})$ for some $x_{0}\in A$ and $y_{0}\in B$. Please help me

$\endgroup$
  • 9
    $\begingroup$ Stimulus: What have you tried so far? Reaction: adds Please help me to the question. O well... $\endgroup$ – Did Jan 12 '13 at 18:04
  • $\begingroup$ I'm guessing $B \subset X$. If so, $B$ is also compact as the closed subsets of compact sets are again compact. $\endgroup$ – emka Jan 13 '13 at 3:11
  • $\begingroup$ possible duplicate of real analysis question about compactness $\endgroup$ – Ittay Weiss Jan 13 '13 at 7:04
  • $\begingroup$ @Ittay: Very close, but not a duplicate, I think. $A$ and $B$ are assumed to be compact in the other question. $\endgroup$ – Martin Jan 13 '13 at 8:18
  • $\begingroup$ very true Martin. My apologies. $\endgroup$ – Ittay Weiss Jan 13 '13 at 8:22
2
$\begingroup$

The statement in the question is wrong.

Example 1: Let $X = \{0\} \cup (1,\infty)$ with the metric induced by $\mathbb{R}$. Let $A = \{0\}$ and $B = (1,\infty)$. Note that $A$ is compact and open, and $B$ is closed. Then $d(A,B) = 1$ but there is no $y_0 \in B$ such that $d(0,y_0) = 1$.

Example 2: It was suggested that completeness of $X$ helps. Consider the closed subspace $X = \{0\} \cup \{(1+2^{-n})e_n \mid n \in \mathbb{N}\} $ of $\ell^2(\mathbb{N})$ where $e_n$ is the $n$-th standard basis vector $e_n = (0,\dots,0,1,0,\dots)$. Set $A = \{0\}$ and $B = \{(1+2^{-n})e_n \mid n \in \mathbb{N}\}$. Then $d(A,B) = 1$ and $A$ is compact, but there is no $y_0 \in B$ such that $d(0,y_0) = 1$.

The statement of the question is true if closed balls in $X$ are compact, or if $B$ is assumed to be compact. To see this, take $(a_n) \subset A$ and $(b_n) \subset B$ such that $d(a_n,b_n) \to d(A,B)$. Pass to suitable convergent subsequences of $a_n$ (by compactness of $A$) and $b_n$ (either by compactness of $B$ or by noting that this is a bounded sequence, hence it is contained in some ball). Show that the limit points lie in $A$ and $B$, respectively, and that they have distance $d(A,B)$.

$\endgroup$
2
$\begingroup$

First show that if $d(x, B) = inf\{d(x, y) : y \in B\} = s$ then for some $y_0 \in B$, $d(x, y_0) = s$ (This will use the fact that $B$ is complete so you also need to assume that $X$ is complete). Next show that $x \rightarrow d(x, B)$ defines a continuous function on $X$. Finally use the fact continuous real valued functions attain their extrema on compact spaces.

Edit: As Martin points out below, $d(x, B)$ may not be attained in general. So let me just add that the above argument works only when either $B$ is also totally bounded or closed balls in $X$ are compact.

$\endgroup$
  • 1
    $\begingroup$ Completeness of $B$ is not enough. There is no reason for a sequence $(y_n) \subset B$ such that $d(x,y_n) \to d(x,B)$ to be a Cauchy sequence. $\endgroup$ – Martin Jan 13 '13 at 8:14
  • $\begingroup$ You are absolutely right. I edited my answer accordingly. $\endgroup$ – Anonymous Jan 13 '13 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.