The problem is the following:
Let $(X,d)$ be a compact metric space, and $A$, $B \subseteq X$ disjoint sets such that
$d(A,B):=inf\{d(a,b):a \in A, b \in B\} = 0$
Prove that $\partial A \cap \partial B \neq \emptyset$
This was my reasoning: if $d(A,B)=0$ then $\exists a_n \subseteq A, b_n \subseteq B$ such that $D_n := \lim_{n\rightarrow\infty} d(a_n,b_n)=0$
As $X$ is compact, and $(a_n) \subseteq A \subseteq X$, then $\exists a_{n_k} \rightarrow a \in X$ (Is this $a$ particularly in $\overline{A}$ ?)
As $D_{n_k}$ is a subsequence of $D_n$, it tends to 0, but as $d(a,b_{n_k}) \leq d(a,a_{n_k})+d(a_{n_k},b_{n_k}) \rightarrow 0$, then $b_{n_k} \rightarrow a$
(which, if my suspicion is right, is in $\overline{A}$)
Then $a \in \overline{A} \cap \overline{B}$, and therefore $\overline{A} \cap \overline{B} \neq \emptyset$
I have two doubts:
- Is it ok to imply that $a\in \overline{A}$ or $a \in \overline{B}$?(same argument for both)
- Can I show that $\overline{A} \cap \overline{B} \neq \emptyset \Rightarrow \partial A \cap \partial B \neq \emptyset$ ?