# Closed set

• Sep 1st 2007, 03:10 PM
putnam120
Closed set
I am having trouble solving this problem from Principles of Mathematical Analysis by Rudin.

*Let $\displaystyle E'$ be the set of all limit points of the set $\displaystyle E$. Prove that $\displaystyle E'$ is closed.

What I am trying to do is prove that the complement of the set is open thus making the set itself closed. So far I have

$\displaystyle A=(E')^c$. if $\displaystyle x\in A$ then $\displaystyle x$ is not a limit point of $\displaystyle E$ so there exist a neighborhood $\displaystyle N_r(x)$ such that $\displaystyle N_r(x)\subset E^c$ for some $\displaystyle r>0$. Therefore, $\displaystyle \forall x\in A$ we have that $\displaystyle x$ is a limit point of $\displaystyle E^c$.

Any help or pointers on what I should do next would be greatly appreciated. :)
• Sep 1st 2007, 05:22 PM
ThePerfectHacker
There is a similar question here.
• Sep 1st 2007, 10:19 PM
CaptainBlack
Quote:

Originally Posted by putnam120
I am having trouble solving this problem from Principles of Mathematical Analysis by Rudin.

*Let $\displaystyle E'$ be the set of all limit points of the set $\displaystyle E$. Prove that $\displaystyle E'$ is closed.

What I am trying to do is prove that the complement of the set is open thus making the set itself closed. So far I have

$\displaystyle A=(E')^c$. if $\displaystyle x\in A$ then $\displaystyle x$ is not a limit point of $\displaystyle E$ so there exist a neighborhood $\displaystyle N_r(x)$ such that $\displaystyle N_r(x)\subset E^c$ for some $\displaystyle r>0$. Therefore, $\displaystyle \forall x\in A$ we have that $\displaystyle x$ is a limit point of $\displaystyle E^c$.

Any help or pointers on what I should do next would be greatly appreciated. :)

You have gone slightly to far. What you have shown is that for all $\displaystyle x \in A$ there exists a neighbourhood of $\displaystyle x$ which contains no points of $\displaystyle E'$, so is a subset of $\displaystyle A$, which proves that $\displaystyle A$ is open.

RonL