# If a set contains all accumulation points then it is closed

It is a question from my complex analysis courses;

If a set contains all accumulation points then it is closed

Our accumulation point definition is “if a point is an accumulation point of set $$S$$, every deleted neighborhood of it contains at least one point of $$S$$

Our closed set definition is “if a set is closed then it contains all boundary points”

I cannot prove it without contradiction. I need direct proof. I am confused in does accumulation point mean boundary point? I need your helps. Thanks in advance

Let $$z$$ be a boundary point. By definition it means any neighborhood of $$z$$ contains points in $$S$$ and points outside of $$S$$. So for each $$n\in\mathbb{N}$$ you can take $$z_n\in S$$ such that $$|z-z_n|<\frac{1}{n}$$. Now look at the sequence $$z_n$$. If $$z$$ is an element of the sequence then from the way we built the sequence it follows that $$z\in S$$. If $$z$$ is not an element of the sequence then it follows that for each $$n\in\mathbb{N}$$ there is a point $$z_n\ne z$$ such that $$z_n\in S$$ and $$|z-z_n|<\frac{1}{n}$$. But that implies $$z$$ is an accumulation point of $$S$$ and hence $$z\in S$$. So in any case you get $$z\in S$$.

If you don't want to use sequences here is another way: if there exists $$\epsilon>0$$ such that for all $$w\in\mathbb{C}$$ that satisfy $$0<|w-z|<\epsilon$$ we have $$w\notin S$$ then it follows that $$z\in S$$ because a neighborhood of $$z$$ still must contain a point of $$S$$. Otherwise, for all $$\epsilon>0$$ there exists $$w\in S$$ such that $$0<|w-z|<\epsilon$$. But from here it follows that $$z$$ is an accumulation point and hence $$z\in S$$. So anyway we get $$z\in S$$.

• Thanks for answer. I cannot use sequences now. We have not defined sequences for complex numbers yet. Could you please rewrite it? Commented Oct 14, 2018 at 19:55
• Ok, I added another way to solve it.
– Mark
Commented Oct 14, 2018 at 20:03
• thanks a lot sir Commented Oct 15, 2018 at 8:19

It's easier to show that the complement of $$S$$ is open: if $$z\in S^c$$ then $$z$$ is not an accumulation point of $$S$$ (because $$S$$ has all its accumulation points, by hypothesis). But then there must be an neighborhood $$U$$ of $$z$$ such that $$U\cap S=\emptyset\Rightarrow U\subseteq S^c$$. That is to say, $$S^c$$ is open.

• The question is can OP use that. He gave a specific definition of a closed set.
– Mark
Commented Oct 14, 2018 at 20:26