The set of limit points is closed

[R.P.F.]

Homework Statement

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

Homework Equations

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\neql
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\neqx
a is in the \epsilon neighborhood of l. The problem is, how do i show that a\neql?

Thanks!

 

[tiny-tim]

Hi R.P.F.!

(have an epsilon: ε )

Hint: try it with ε = 1/n, for all values of n.

 

[arkajad]

Alternative suggestion: what about proving that the complement of L is open? Wouldn't it be easier?

 

[R.P.F.]

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?

 

[R.P.F.]

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]

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?

 

[arkajad]

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.

 

[R.P.F.]

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!

 

[R.P.F.]

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]

wot's the Archimedian property?

 

[R.P.F.]

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]

ahh!

in that cases, no, nothing to do with the Archimedean property

hint: for each ε = 1/n, you should get a point x_n …

carry on from there ​