1. ## Closed sets

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?

2. Originally Posted by 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?
This is true in any $T_1$ 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 $U$ contains a point $y$ of $E'$ different from itself but since $U$ is also a neighborhood of $y$ it follows (since this is a metric space) that $U$ contains infinitely many points of $E$, in particular it must contain a point of $E$ distinct from $x$ and thus $x$ is a limit point of $E$ so $x\in E'$

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

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