# The set of limit points is closed

## Homework Statement

L is the set of limit point of A in the real space, prove that L is closed.

## The Attempt at a Solution

L may or may not have limit points. If L does not have limit points, then it's obviously closed.

If L has limit points, the let l be a limit points of L. => exists x in L such that x is in the $$\frac{\epsilon}{2}$$ neighborhood of l and x$$\neq$$l
x is a limit point of A => exists a in A such that a is in the $$\frac{\epsilon}{2}$$ neighborhood of x and a$$\neq$$x
a is in the $$\epsilon$$ neighborhood of l. The problem is, how do i show that a$$\neq$$l?

Thanks!

tiny-tim
Homework Helper
Hi R.P.F.! (have an epsilon: ε )

Hint: try it with ε = 1/n, for all values of n. Alternative suggestion: what about proving that the complement of L is open? Wouldn't it be easier?

Hi R.P.F.! (have an epsilon: ε )

Hint: try it with ε = 1/n, for all values of n. Hi. But aren't we supposed to prove this for every epsilon?

Alternative suggestion: what about proving that the complement of L is open? Wouldn't it be easier?
I don't see how this could me easier..hints?

tiny-tim
Homework Helper
Hi. But aren't we supposed to prove this for every epsilon?
I'm thinking of a slightly different method to the one you're thinking of (which I think doesn't work).

What can you get if you do it for every n? I don't see how this could me easier..hints?
Let M be the complement of L. You want to prove that M is open. A point is in M if there is a neighborhood U of x that does not contain any point of A except perhaps x itself. Can we show that none of the points in U can be a limit point of A? Or, if you wish, take epsilon/2 of this neighborhood. It's seems to me that it will work.

Let M be the complement of L. You want to prove that M is open. A point is in M if there is a neighborhood U of x that does not contain any point of A except perhaps x itself. Can we show that none of the points in U can be a limit point of A? Or, if you wish, take epsilon/2 of this neighborhood. It's seems to me that it will work.
Yes it does work. I just wrote up the proof. Thank you!

I'm thinking of a slightly different method to the one you're thinking of (which I think doesn't work).

What can you get if you do it for every n? So you mean if we do this for every n then we can use the Archimedian property to show it works for every epsilon?

tiny-tim
Homework Helper
wot's the Archimedian property? wot's the Archimedian property? Sorry it should be Archimedean property: For every $$r \in R$$, there exists $$n \in N$$ s.t. r>1/n.

tiny-tim
carry on from there 