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

  • Thread starter jrsweet
  • Start date
  • #1
32
0

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!
 

Answers and Replies

  • #2
42
0
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
32
0
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?
 

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

Replies
25
Views
1K
Replies
3
Views
11K
Replies
2
Views
3K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
2
Replies
32
Views
8K
  • Last Post
Replies
4
Views
922
Replies
10
Views
5K
  • Last Post
Replies
5
Views
844
Top