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Accumulation point proof

  1. Nov 10, 2011 #1
    I need to show that if every neighborhood of some [itex]x\in A[/itex] for some [itex]A\subseteq\mathbb{R}[/itex] contains infinitely many points of [itex]A[/itex], then [itex]x[/itex] is an accumulation point of [itex]A[/itex].

    So far, I have:

    Let [itex]A\subseteq\mathbb{R}[/itex]. I want to show that if every neighborhood of [itex]x\in A[/itex] has infintely many points of A, there exists a [itex]y\in\mathbb{R}[/itex] such that [itex]y\in((x-\epsilon,x+\epsilon)\bigcap A[/itex]\{x}).

    Am I on the right track?
  2. jcsd
  3. Nov 10, 2011 #2


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    Science Advisor

    Looks good.
  4. Nov 10, 2011 #3
    I'm having some trouble finding that [itex]y[/itex].

    Let A⊆R. Define the neighborhood of x∈A as (x−ϵ,x+ϵ) [itex]\forall\epsilon>0[/itex]. Since x∈A, (x−ϵ,x+ϵ)⋂A is not empty. Let y∈(x−ϵ,x+ϵ)⋂A...how do I show that [itex]y\not=x[/itex]?

  5. Nov 10, 2011 #4
    You can't in general. It could very well be that you chose y=x. Your choice doesn't disallow this. However, you know that (x−ϵ,x+ϵ)⋂A is infinite. So it doesn't only contain x, does it??
  6. Nov 10, 2011 #5
    Right, but don't I have to prove that this intersection is infinite? I'm trying to show that x is an accumulation point, but (x−ϵ,x+ϵ)⋂A being infinite presupposes x being an accumulation point, right?
  7. Nov 10, 2011 #6


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    I might be missing something, but you're trying to prove the definition of an accumulation point (which you can't do). The assumption is that every neighborhood of some x in A contains infinitely many points of A. Therefore, it is an accumulation point by definition.
  8. Nov 10, 2011 #7
    Isn't that given?? The first line of the post is

    So you are given that the neigbourhood (x−ϵ,x+ϵ) contains infinitely many points of A.

    Or am I totally misunderstanding your question?
  9. Nov 10, 2011 #8
    No, you're right. the proof seemed too simple so I thought I was missing something. Makes sense now.
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