In complex analysis how to prove that if $S$ is closed in $\mathbb{C}$ then it contain all of its accumulation points.
If $S$ is closed then $S$ contain all its boundary points.(If $z_{0} $ is a boundary point then every neighbourhood of it will contain both the points in $S$ and $S'$.) Suppose $z_{0}$ is an acccumulation point of $S$ then every neighbourhood of it will contain at least one point of $S$ other than $z_{0}$. Then $z_{0}$ will become a boundary point hence must be in $S$.
But how to prove the converse ?
If we assume that our set satisfies the condition that it contains every accumulation point of it and let $z_{0}$ be a boundary point of $S$ which is not in $S$ we can argue that it is an accumulation point and arrive at a contradiction ?
Best Answer
Your idea works. In detail:
Suppose $S$ contains all its accumulation points, and we want to show $S$ is closed.
So let $z$ be any boundary point of $S$. Suppose for a contradiction that $z \notin S$.
Then let $B$ be a ball around $z$ (or any open set containing $z$, if you want to be more general). Then $B$ intersects $S$ in some $s$ and this $s \neq z$, as we assumed $z \notin S$ and we know $s \in S$. As $B$ was arbitrary, $z$ is an accumulation point of $S$.
By assumption on $S$, $z \in S$, but we have a contradiction. So in fact $z \in S$ must hold. As $z$ was any boundary point, $S$ contains all of them so is closed (in your definition of closed).
This shows the reverse implication.