can set $\mathcal{A}$ be written as union of countable set which are rare sets $\mathcal{A}$=the set of all fnite sequences in $l_1$
$l_1$:  the space of sequences of $x_n$ s.t. : $\sum^\infty_1 |x_n|<\infty$
$A_n$ is rare set if the interior of the closure is empty,${Int}\bar A_n=\emptyset$,
is it possible that: $\mathcal{A}=\cup^\infty_{n=1} A_n$ :countable union and $A_n$ is rare

Try $A_n$ = the set of all sequences of length $n$. It is clear that the union of $A_n$ is $\mathcal A$. It can also be shown that $A_n$ is closed and has an empty interior.
