For [[knowledge-mathematics-analysis-functional-analysis-metric-spaces|metric space]] \((X,d)\) a subset \(A\subseteq X\) is closed iff for every convergent sequence \(\{a_n\}_{n\in\mathbb{N}}\subseteq A\) one has \(\lim_{n\rightarrow\infty}a_{n}\in A\).
\begin{proof} We construct a proof by contraposition. Assume that \(A\subseteq X\) is open which implies that \(A^c\) is closed. Therefore, there is a limit point \(x\in A^c\) such that \(B_\epsilon(x)\cap A\neq \emptyset\) for all \(\epsilon>0\). \end{proof}