The set of convergent subsequences

[zbr]

Hello all,

(*) I have a question about convergent subsequences. Specifically I am looking for an example of a sequence that is unbounded but who has convergent subsequences in the interval [0,1].

A similar question would be to have an unbounded sequence, but who has a convergent subsequence to a specific number, let's say 0.

For this I would take the sequence:
a_n = -1,0,1,-2,0,2,-3,0,3,...,-n,0,n

Can I do a similar thing for (*)? i.e Can I take the following sequence:

a_n = -1, [0,1], 1, -2, [0,1], 2,...,-n, [0,1], n

My intuition tell me no since there will be infinite terms in each of the intervals [0,1].

Thanks in advance for any and all help!

 

[Dick]

Hello all,

(*) I have a question about convergent subsequences. Specifically I am looking for an example of a sequence that is unbounded but who has convergent subsequences in the interval [0,1].

A similar question would be to have an unbounded sequence, but who has a convergent subsequence to a specific number, let's say 0.

For this I would take the sequence:
a_n = -1,0,1,-2,0,2,-3,0,3,...,-n,0,n

Can I do a similar thing for (*)? i.e Can I take the following sequence:

a_n = -1, [0,1], 1, -2, [0,1], 2,...,-n, [0,1], n

My intuition tell me no since there will be infinite terms in each of the intervals [0,1].

Thanks in advance for any and all help!

You can arrange all of the rational number in [0,1] into a sequence, right? Now throw some extra numbers into make it unbounded, just like you did for the zero sequence.

 

[zbr]

Ah, yes of coarse, set sequence 1, 1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, ... would satisfy the sequence with convergent subsequences within the interval [0,1], then just throwing in divergent terms would satisfy the second requirement.

Thanks so much for your help Dick!