How do you prove that some point is the *only* accumulation point in a set?

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Homework Statement



I'm using the following definition of an accumulation point:

A point [itex]a \in \textbb{R}[/itex] is an accumulation point of a set [itex]A\subset \textbb{R}[/itex] if every [itex]\epsilon-[/itex]neighborhood of [itex]a[/itex] contains at least one element of [itex]A[/itex] distinct from [itex]a[/itex].

Now, given the set [itex]A=\{\frac{1}{n}:n\in \textbb{Z}^+\}[/itex], I'm trying to prove that the only accumulation point of [itex]A[/itex] is 0.

2. The attempt at a solution

I was able to prove that 0 is an accumulation point, but my question is about proving there can't be any others.

Intuitively, it makes sense to me because any element between 0 and 1 is either an element of [itex]A[/itex] or between two elements of [itex]A[/itex]. In either case, you can take [itex]\epsilon[/itex] small enough to not include any elements of [itex]A[/itex].

The problem I have is that I'm not sure how to formulate this in a way that would be considered rigorous.
 

Answers and Replies

  • #2
Dick
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Having the right picture of the problem is the first step. And you've got that. Given n, what's an epsilon the would be less than both (1/n-1/(n+1)) and (1/(n-1)-1/n)? If you can find one, that would fulfill your intuitive sense in a rigorous way. Yes?
 
  • #3
HallsofIvy
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Let [itex]\alpha[/itex] be any number other than 0. Show that [itex]\alpha[/itex] is not an accumulation point.

You probably should do two cases: (1) [itex]alpha> 0[/itex] and (2) [itex]\alpha[/itex]< 0. Sjhow that there are only finite number of n such that [itex]|1/n|> \alpha[/itex].
 
  • #4
arildno
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Well, how about letting z be an accumulation point, and then prove that it must equal zero?
That is, prove the uniqueness of zero as an accumulation point!
 

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