On the cardinality of an intersection Let us consider a divergent sequence $a_n:\Bbb N\to\Bbb N$ and a decreasing sequence $\{A_{n}\}_n$ of subsets of $\Bbb R^d$ such that $A_n$ is the disjoint union of $a_n$ balls $B(c_j^{n},r_n)$ of fixed radius $r_n>0$, such that
$$
\{c_j^{(n)}\}_{j=1}^{a_n}\subset\{c_j^{(n+1)}\}_{j=1}^{a_{n+1}}\\
r_n\to0
$$
and call
$$
A:=\bigcap_{n=1}^{+\infty}A_n
$$
the limit set.
It seems easy to deduce that
$$
A=\bigcup_{n\ge1}\{c_j^{(n)}\}_{j=1}^{a_n}\;\;,
$$
but my question is: is in this case $A$ is necessarily countable for any diverging $a_n$?
EDIT As Gae S pointed out
$$
A=\bigcup_{n\ge1}\{c_j^{(n)}\}_{j=1}^{a_n}\;\;,
$$
is not true.
But is it true that
$$
A=\operatorname{cl}\left(\bigcup_{n\ge1}\{c_j^{(n)}\}_{j=1}^{a_n}\right)\;\;?
$$
 A: Consider the following Cantor-like construction in $\Bbb R^1$:

*

*$A_0=(0,1)$;


*$A_{n+1}=\left(\frac13A_n\right)\cup\left(\frac12-\frac123^{-(n+1)},\frac12+\frac123^{-(n+1)}\right)\cup\left(\frac23+\frac13A_n\right)$
Notice that $A_n\supsetneqq A_{n+1}$ and that $\bigcap_nA_n$ contains the irrational points of Cantor's set, which are uncountably many.
This sequence $A_n$ fits your specifics (except, obviously, $\bigcap_n A_n$ being the set of the centres of the sequence of balls) by having $B\left(c_1^{(n)},r_n\right),\cdots, B\left(c^{(n)}_{a_n},r_n\right)$ be the connected components of $A_n$. For the record, here $a_{n}=2^{n+1}-1$. It is quite clear that each $A_n$ is union of disjoint intervals of length exactly $3^{-n}$ and therefore $r_n=\frac12 3^{-n}$. The reason why $\left\{c^{(n)}_j\right\}_{j=1}^{a_n}\subseteq \left\{c^{(n+1)}_j\right\}_{j=1}^{a_{n+1}}$ is that if we call $f(x)=\frac x3$, $g(x)=\frac23+\frac x3$, then the $c^{(n)}_j$-s are exactly the points of the form $(h_1\circ h_2\circ\cdots\circ h_m)(1/2)$ for some $0\le m\le n$ and $h_1,\cdots,h_m\in\{f,g\}$ (for $m=0$, we adopt the notation $h_1\circ\cdots\circ h_m:=id$). Therefore, all the $c^{(n)}_j$-s are evidently among the $c_j^{(n+1)}$-s.
Remark: I don't see an obstruction to doing this for an appropriate Cantor dust in $\Bbb R^d$ with $d\ge2$.
After edit: In this example, certainly $A:=\bigcap_nA_n\ne\operatorname{cl}\left\{c_j^{(n)}\,:\, n\in\Bbb N\land 1\le j\le a_n\right\}$, because $A$ is not closed: $\operatorname{cl}(A)\setminus A$ contains infinitely many rational numbers; for instance, the extremal points of the connected components of the complement of Cantor's set.
