Let \(X\) be a first countable space and \(A\subseteq X\). Then \(A\) is closed if and only if \(A\) contains every limit of ever converging sequence of points in \(A\).
Proof
Let \(A\) be closed and \((x_n)\) a converging sequence in \(A\) with limit \(x\in X\). Then due to (0x677c315f) \(x\in \bar{A}\) and since \(A=\bar{A}\) (0x677c31a5) , we have \(x\in A\).
On the other hand assume \(A\) is not closed. Then \(A\neq \bar{A}\). Choose \(x\in \bar{A}\setminus A\). Then according to (0x677c315f) there is convergent sequence lying in \(A\) with limit \(x\). So \(A\) does not contain all convergent sequences of points in \(A\).