Let \(X\) be a first countable space and \(A\subseteq X\). Then \(A\) is open if and only if every sequence converging to a point of \(A\) is eventually in \(A\).
Proof
Let \(A\) be open. Then \(A=\Int A\) and every sequence converging to a point of \(A\) is eventually in \(A\) (0x677c3160) .
Otherwise, let \(x\in A\). According to (0x677c3160) we obtain \(x\in \Int A\). This implies \(A=\Int A\) which proves the claim.