Closed sets via sequences (proof I)

Proposition (Sequential characterization of closed sets, proof I). Let $(X,d)$ be a metric space and let $A\subset X$. Then $A$ is closed if and only if whenever $(a_n)$ is a sequence in $A$ and $a_n\to a$, we have $a\in A$.

Proof sketch (using closure).

• If $A$ is closed, then $\overline{A}=A$. If $a_n\in A$ and $a_n\to a$, then by closure via sequences we have $a\in\overline{A}=A$.
• Conversely, assume the “contains limits” property. Take any $x\in\overline{A}$. By the same closure-via-sequences proposition, there exists $a_n\in A$ with $a_n\to x$. By hypothesis $x\in A$. Hence $\overline{A}\subset A$, so $\overline{A}=A$ and $A$ is closed.