The set of limit points is closed

  • Thread starter R.P.F.
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Homework Statement


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


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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 [tex]\frac{\epsilon}{2}[/tex] neighborhood of l and x[tex]\neq[/tex]l
x is a limit point of A => exists a in A such that a is in the [tex]\frac{\epsilon}{2}[/tex] neighborhood of x and a[tex]\neq[/tex]x
a is in the [tex]\epsilon[/tex] neighborhood of l. The problem is, how do i show that a[tex]\neq[/tex]l?

Thanks!
 

Answers and Replies

  • #2
tiny-tim
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Hi R.P.F.! :smile:

(have an epsilon: ε :wink:)

Hint: try it with ε = 1/n, for all values of n. :wink:
 
  • #3
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Alternative suggestion: what about proving that the complement of L is open? Wouldn't it be easier?
 
  • #4
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Hi R.P.F.! :smile:

(have an epsilon: ε :wink:)

Hint: try it with ε = 1/n, for all values of n. :wink:
Hi. But aren't we supposed to prove this for every epsilon?
 
  • #5
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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?
 
  • #6
tiny-tim
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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? :smile:
 
  • #7
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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.
 
  • #8
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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!
 
  • #9
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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? :smile:
So you mean if we do this for every n then we can use the Archimedian property to show it works for every epsilon?
 
  • #10
tiny-tim
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wot's the Archimedian property? :redface:
 
  • #11
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wot's the Archimedian property? :redface:
Sorry it should be Archimedean property: For every [tex]r \in R[/tex], there exists [tex]n \in N[/tex] s.t. r>1/n.
 
  • #12
tiny-tim
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ahh!

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

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

carry on from there :smile:
 

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