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

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SUMMARY

The discussion centers on proving that if every neighborhood of a point x in a subset A of the real numbers ℝ contains infinitely many points of A, then x is an accumulation point of A. Participants clarify that the definition of an accumulation point inherently includes the condition that neighborhoods contain infinitely many points of A. The confusion arises from the need to demonstrate that the intersection of the neighborhood (x−ϵ, x+ϵ) with A is infinite, which is already given by the problem's premise.

PREREQUISITES
  • Understanding of accumulation points in topology
  • Familiarity with neighborhoods in the context of real numbers
  • Basic knowledge of set theory and intersections
  • Proficiency in mathematical proofs and definitions
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  • Study the formal definition of accumulation points in topology
  • Explore examples of neighborhoods in ℝ and their properties
  • Learn about infinite sets and their implications in real analysis
  • Review common proof techniques used in mathematical analysis
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Mathematics students, particularly those studying real analysis or topology, educators teaching these concepts, and anyone interested in understanding the properties of accumulation points in mathematical sets.

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I need to show that if every neighborhood of some x\in A for some A\subseteq\mathbb{R} contains infinitely many points of A, then x is an accumulation point of A.

So far, I have:

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

Am I on the right track?
 
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Looks good.
 
I'm having some trouble finding that y.

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

Thanks!
 
autre said:
I'm having some trouble finding that y.

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

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 x\in A for some A\subseteq\mathbb{R} contains infinitely many points of A, then x is an accumulation point of A.

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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