Can a bounded subsequence have infinitely many convergent subsequences?

[Szichedelic]

I'm not sure if I am confusing myself or not, but a friend and I were trying to figure this out. Basically, I know that if a sequence is bounded, we are guaranteed at least one convergent subsequences. However, is it possible for a bounded sequence to have infinitely many of such subsequences?

 

[mathman]

Yes: 1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, ...

1/n is a convergent subsequence limit for every n.

 

[jgens]

A bounded sequence can actually have uncountably many ($2^{\aleph_0}$) convergent subsequences. In particular, for each i in N, let {x_i,n} be an enumeration of the rational numbers in [0,1]. Then consider the sequence:

x_1,1, x_1,2, x_2,1, x_1,3, x_2,2, x_3,1, ...

That is, enumerate all the x_i,n such that i+n = 2, then all elements such that i+n = 3, then all elements such that i+n = 4, and so on. Since the rationals are dense in R, this gives us a convergent subsequence for every real number in [0,1].