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

1. Feb 16, 2010

### jrsweet

1. The problem statement, all variables and given/known data
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.

3. 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!

2. Feb 16, 2010

### Mandark

Suppose the subsequence we want is $$a_{n_1}, a_{n_2}, \ldots$$ with $$n_1 < n_2 < \cdots$$. If you could ensure that $$x-1/k < a_{n_k} < x+1/k$$ for all k you'd be done (why?). Can you think of how you can ensure such a sequence exists?

3. Feb 17, 2010

### jrsweet

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?