Is the distance between those compact sets equal to $0$?

Question

Let $(S,d)$ be a metric space such that there exists a sequence of compact sets of $S$ such that $K_1 \subset K_2 \subset ... $

Let $M=\bigcup_{i=1}^{\infty} K_i$. Let $D$ be a countable dense set of $M$ ($M$ is separable).

Define $\mathcal{A} = \{ B(d,q) : q \in \mathbb Q^+_0 , d \in D\}$, where $B(d,q)$ is the open ball of center $d$ and radius $q$. This set is countable.

Define $\mathcal C = \{\emptyset\} \cup \{\bigcup_{i=1}^{n} (\bar A_i \cap K_i) : A_i \in \mathcal{A}, n \ge 1\}$

$\mathcal C$ is countable and its elements are compact sets. It is also included in the Borel $\sigma$-algebra of $S$.

Is it true that if we take two non-empty compact sets $C_1,C_2 \in \mathcal C$ such that $C_1 \cap C_2 = \emptyset$, we could have $d(C_1,C_2) = 0 $ ?

Answer 1

This is impossible for any two compact sets, regardless of whether they are coming out of your question or otherwise.

The function $d:C_1\times C_2\to\mathbb R$ is a continuous function on the compact space $C_1\times C_2$, so must reach a minimum. Thus, $d(C_ 1,C_2)=d(c_1,c_ 2)$ for some $c_1\in C_1, c_2\in C_2$. If $d(C_1,C_2)=0$, then $d(c_1,c_2)=0$, so $c_1=c_2\in C_1\cap C_2$, which is impossible if $C_1\cap C_2=\emptyset$.

Answer 2

No. Suppose we did, in which case for each $n\in\mathbb{N}$ there is $x_n\in C_1$ and $y_n\in C_2$ such that $d(x_n,y_n)<\frac{1}{n}$. By compactnes let $x\in C_1$ be the limit of a converging subsequence $(x_{n_m})_m$ of $(x_n)_n$. By compactness let $y\in C_2$ be the limit of a convergent subsequence of $(y_{n_m})_m$. Since $d:S\times S \rightarrow \mathbb{R}$ is a continuous function it is easy to derive that $d(x, y)=0$.