Mathematical Analysis Fundamentals

Proof

For necessity, assume the contrary: let $A$ be a closed set containing the convergent sequence $\{p_n\}$ with $p={\lim }_{n\to \infty }{p}_n\notin A$. Then $p\in A^c$. Since $A^c$ is open, there is a neighborhood of $p$ that is a subset of $A^c$. By Theorem 4.22, this neighborhood contains ...