Prove that set of all limit points is closed?

Hey guys,

If A is a subset of R, let L be the set of all limit points. We want to show that L is closed. Here's how I was thinking about proving it:

To show that L is closed, we can show that the complement of L (call it c(L)) is open. c(L) is open if we can put an epsilon-neighborhood around each of the isolated points of L and always have it included in L. I'm not exactly sure where to go from here.

Any hints would be much appreciated.

Thanks,

Mariogs

Re: Prove that set of all limit points is closed?

Quote:

Originally Posted by

**mariogs379** If A is a subset of R, let L be the set of all limit points. We want to show that L is closed. Here's how I was thinking about proving it: To show that L is closed

Surely you mean that $\displaystyle L$ is the set of all limit points of $\displaystyle A~?$

That is usually called the derived set of $\displaystyle A$ and is noted by $\displaystyle A'$

Now if $\displaystyle t\notin L$ then $\displaystyle t$ is not a limit point of $\displaystyle A$.

So there is open set $\displaystyle Q_t$ such that $\displaystyle O_t\cap (A\setminus\{t\})=\emptyset$.

Can finish?

Re: Prove that set of all limit points is closed?

Yeah, that's what I meant. So Qt must be open and, because it is the complement of L, L must be closed. Yeah?