Theorem. A set is closed if and if it contains all its acomulation point.
proof.
$\Leftarrow$ Assume that the set $ E $ is not closed, that is. there is a sequence converget $x_n \rightarrow c $ in $E$, but $c \notin E$ , then $x_n \neq c $, hence $c$ is a point of acomulation.
that is to say $E$ does not contain all its points of accumulation.
Hint: Given $\mathfrak C$, let $I=\mathfrak C$ and $A_i=i$ for all $i\in\mathfrak C$ ...