# Closed sets

#### poorna

Hey, I was just trying out some questions from Rudin, when I had this doubt as to whether my answer is right, can you please tell me ifI am right?

Let E' be the set of all limit points of a set E. Prove that E' is closed.

Let x be a limit point of E'. So every nbd of x contains infinitely many points of E'. But in every nbd of every point of E' are infinitely many points of E, since every point of E' is a limit point of E.
Thereby every nbd of x contains infinitely many points of E. So x is a limit point of E (ie) x is in E'.
Hence E' is closed. Am i right?

#### Drexel28

MHF Hall of Honor
Hey, I was just trying out some questions from Rudin, when I had this doubt as to whether my answer is right, can you please tell me ifI am right?

Let E' be the set of all limit points of a set E. Prove that E' is closed.

Let x be a limit point of E'. So every nbd of x contains infinitely many points of E'. But in every nbd of every point of E' are infinitely many points of E, since every point of E' is a limit point of E.
Thereby every nbd of x contains infinitely many points of E. So x is a limit point of E (ie) x is in E'.
Hence E' is closed. Am i right?
This is true in any $$\displaystyle T_1$$ space, but since it's Rudin it must be that you are speaking of a metric space.

What you said is pretty much it, but I think you need to say it slightly better. Every neighborhood $$\displaystyle U$$ contains a point $$\displaystyle y$$ of $$\displaystyle E'$$ different from itself but since $$\displaystyle U$$ is also a neighborhood of $$\displaystyle y$$ it follows (since this is a metric space) that $$\displaystyle U$$ contains infinitely many points of $$\displaystyle E$$, in particular it must contain a point of $$\displaystyle E$$ distinct from $$\displaystyle x$$ and thus $$\displaystyle x$$ is a limit point of $$\displaystyle E$$ so $$\displaystyle x\in E'$$

#### nimon

I think it is best just to show that the complement of $$\displaystyle E^{\prime}$$ is open. So let the surrounding topological space (or metric space) be denoted by $$\displaystyle X$$ and assume that $$\displaystyle E^{\prime}\subset X$$ is a proper subset. Then let $$\displaystyle x\in X\backslash E^{\prime}$$, so that $$\displaystyle x$$ is not a limit point of $$\displaystyle E$$. Then by definition there must be a neighbourhood $$\displaystyle U$$ of $$\displaystyle x$$ such that $$\displaystyle U\cap E^{\prime}= \emptyset$$, i.e. $$\displaystyle x\in U \subseteq X\backslash E^{\prime}$$. Since this holds for any such $$\displaystyle x\in X\backslash E^{\prime}$$, this means that $$\displaystyle X\backslash E^{\prime}$$ is open, so that $$\displaystyle E^{\prime}$$ is closed.

Last edited:

#### Drexel28

MHF Hall of Honor
Then by definition there must be a neighbourhood $$\displaystyle U$$ of $$\displaystyle x$$ such that $$\displaystyle U\cap E^{\prime}= \emptyset$$, i.e. $$\displaystyle x\in U \subseteq X\backslash E^{\prime}$$.
That isn't true at all. It is true that there is a neighborhood such that $$\displaystyle U\cap E\subseteq\{x\}$$

• nimon