Let \(X\) be a first countable space and \(A\subseteq X\). Then \(x\in \bar{A}\) if and only if \(x\) is the limit of a convergent sequence lying in \(A\).
Proof
Let \(x_n\to x\) with \(x_n\in A\) for each \(n\). Then according to (0x677afd2a) is in \(\bar{A}\).
On the contrary let \(x\in \bar{A}\). Since \(X\) is first countable there is a nested neigbourhood basis . Again using (0x677afd2a) we can choose for every basis element \(U_n\) a point \(x_n\in A\). Then \(x_n\to x\) and \(x_n\in A\) for all \(n\).