Is x an Accumulation Point of A in the Real Numbers?

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Discussion Overview

The discussion revolves around the concept of accumulation points in the context of real numbers. Participants are exploring the definition and proof related to whether a point \( x \) is an accumulation point of a set \( A \subseteq \mathbb{R} \) based on the presence of infinitely many points of \( A \) in every neighborhood of \( x \).

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that if every neighborhood of \( x \in A \) contains infinitely many points of \( A \), then \( x \) should be an accumulation point of \( A \).
  • Another participant expresses difficulty in identifying a point \( y \) within the neighborhood that is not equal to \( x \).
  • There is a discussion about the nature of the intersection \( (x-\epsilon, x+\epsilon) \cap A \) and whether it can be shown to be infinite without assuming \( x \) is an accumulation point.
  • Some participants clarify that the assumption of the intersection being infinite is part of the definition of an accumulation point, leading to confusion about the proof's validity.
  • One participant acknowledges that the proof seemed overly simple and expresses relief upon realizing the reasoning aligns with the definition.

Areas of Agreement / Disagreement

Participants generally agree on the definition of an accumulation point but express differing views on the proof process and the necessity of demonstrating the intersection's infinitude without assuming the conclusion.

Contextual Notes

There is uncertainty regarding the proof structure and whether the assumption about the neighborhood containing infinitely many points is sufficient to conclude that \( x \) is an accumulation point without further justification.

autre
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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 infinitely 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?
 
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Looks good.
 
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]?

Thanks!
 
autre said:
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]?

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??
 
However, you know that (x−ϵ,x+ϵ)⋂A is infinite.

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?
 
autre said:
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].

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.
 
autre said:
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?

Isn't that given?? The first line of the post is

if every neighborhood of some x∈A for some A⊆R contains infinitely many points of A

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

Or am I totally misunderstanding your question?
 
Or am I totally misunderstanding your question?

No, you're right. the proof seemed too simple so I thought I was missing something. Makes sense now.
 

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