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 ?