# Closed sets and accumulation points

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 ?

• Does it really matter whether you are proving this in complex analysis or any topological space? What is the definition of a closed set? Feb 3, 2016 at 5:58
• A set $S$ is said to be closed subset of $\mathbb{C} if it contain all of its boundary points . This definition is from Brown &Churchil Feb 4, 2016 at 16:46 ## 1 Answer Your idea works. In detail: Suppose$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.