I was reading this theorem in my analysis text and came across something I could not quite follow:

Theorem: Let S be a subset of the metric space E. Then S is closed if and only if, whenever $\displaystyle p_1, p_2, ...$ is a sequence of points of S that is convergent in E, we have $\displaystyle \lim_{n\to\infty} p_n \in S $.

I understand the "only if" part. Now for the "if" part the text states:

"Suppose S is not closed. Then the complement is not open, and there exists a point p in the complement of S such that any open ball of center p contains points of S. *Hence for each positive integer n we can choose$\displaystyle p_n \in S$such that$\displaystyle d(p, p_n)<\frac{1}{n}$.Then$\displaystyle \lim_{n\to\infty} p_n = p $, with each$\displaystyle p_n \in S$and$\displaystyle p \notin S$.*This shows that if the hypothesis on convergent sequences holds, then S must be closed, completing the proof. "

^^ I've bolded the text that I don't understand. I've thought about this and I can't figure out where the 1/n comes from as well as the conclusion $\displaystyle \lim_{n\to\infty} p_n = p $. Thanks for the help!