X is an accumulation point show there is subsequence that converges to x

In summary, the conversation discusses how to show that there is a subsequence of (an) that converges to an accumulation point x. The idea is to construct a subsequence where each term is within a specific interval around x, which ensures that the subsequence will converge to x. The conversation also mentions using the definition of an accumulation point repeatedly to narrow the interval and approach x.
  • #1
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Homework Statement


Suppose x is an accumulation point of {an: n is a member of integers}. Show there is a subsequence of (an) that converges to x.

The Attempt at a Solution


I'm a little stuck on this one. I know that since x is an accumulation point then every neighborhood around x, (x-e,x+e) contains infinitely many points of an. I guess I just don't know how to construct the subsequence.

Any help would be greatly appreciated!
 
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  • #2
Suppose the subsequence we want is [tex]a_{n_1}, a_{n_2}, \ldots[/tex] with [tex]n_1 < n_2 < \cdots[/tex]. If you could ensure that [tex]x-1/k < a_{n_k} < x+1/k[/tex] for all k you'd be done (why?). Can you think of how you can ensure such a sequence exists?
 
  • #3
Can we say the sequence is {x-1/k2} for integers>0 ? I'm still a little confused.

I know that the interval (x-1/k, x+1/k) will be contained in a neighborhood of x, and thus has infinitely many points. I'm not sure why we are using 1/k though. I guess we are in essence using the definition of an accumulation point repeatedly and narrowing the interval and making it converge to x?
 

1. What does it mean for a point to be an accumulation point?

An accumulation point is a point in a sequence where infinitely many terms of the sequence are close to it. This means that any neighborhood around the point contains infinitely many terms of the sequence.

2. How is an accumulation point different from a limit point?

An accumulation point and a limit point are essentially the same concept. The only difference is that an accumulation point may or may not be a term in the sequence, while a limit point must be a term in the sequence.

3. Why does a subsequence need to converge to the accumulation point?

A subsequence that converges to the accumulation point is a subset of the original sequence that becomes arbitrarily close to the accumulation point as the sequence progresses. This shows that the accumulation point is a significant point in the sequence and is therefore important to consider.

4. Can there be more than one subsequence that converges to the same accumulation point?

Yes, there can be multiple subsequences that converge to the same accumulation point. This is because a sequence can have multiple points that are considered accumulation points, and each of these points can have its own subsequence that converges to it.

5. How is the existence of a subsequence that converges to an accumulation point proven?

To prove the existence of a subsequence that converges to an accumulation point, we use the definition of an accumulation point and the Bolzano-Weierstrass theorem. This theorem states that every bounded sequence has a convergent subsequence. Therefore, since the accumulation point is a bounded point in the sequence, there must be a subsequence that converges to it.

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