# Bounded sequence that diverges, convergent subsequence

1. Oct 7, 2010

### miren324

1. The problem statement, all variables and given/known data
Let (sn) be a sequence in R that is bounded but diverges. Show that (sn) has (at least) two convergent subsequences, the limits of which are different.

2. Relevant equations

3. The attempt at a solution
I know that a convergent subsequence exists by Bolzano-Weierstrass. I'm having trouble showing that multiple convergent subsequences exist, and how to show that they would have different limits. I know that (sn) will probably look something like (-1)^n (so it will be a sequence that neither increases or decreases, but rather most likely "bounces around"), and if I had an actual sequence to prove this on I could probably do it, but to prove it in general is giving me some difficulty.

The only thought I had was that maybe defining two subsequences of (sn) that have only one element, i.e. (an) = {s1} and (bn) = {s2}. I'm not sure this would work, however, since I am not sure that an infinite constant sequence, such as (an), would be a subsequence of (sn), seeing as how, (s1) would occur much fewer times in (sn) than it would in (an) (for example: (sn) = sin(n), (an) = sin(1), then (an) = {sin1, sin1, sin1, sin1, ...} and (sn) = {sin1, sin2, sin3, sin4, ...}).

Any help would be greatly appreciated.

2. Oct 8, 2010

### JThompson

The approach to this proof depends on what theorems you are allowed to use. For instance, if you let S be the set of subsequential limits of $$s_{n}$$ then there is a theorem which states

$$\sup S = \lim\sup s_{n}\mbox{ and }\inf S = \lim\inf s_{n}$$

and all you need to do is prove $$\lim\sup s_{n}\neq\lim\inf s_{n}$$

which will be easy since you know that $$s_{n}$$ isn't convergent.

Last edited: Oct 8, 2010
3. Oct 8, 2010

### ╔(σ_σ)╝

None of my proof are correct. Only one sequence converges the other I made up did not. But we need both to converge.

4. Oct 8, 2010

### ╔(σ_σ)╝

The correct solution is to show that there are two sequences that converge to the infima and suprema and infima is not equal to the suprema.

Another correct approach is to show that there is a subsequence of Sn that does not converge to the same limit as your bolzano weierstrass sequence and then show that that subsequence has a convergence subsequence that converges also to a different limit than your original BW sequence.

5. Oct 8, 2010

### JThompson

You can do an infinite constant subsequence, just not the way illustrate in your sin example. You can't take (an)=(s1) as a subsequence simply because of the definition of a subsequence; however, you can defined
$$a_{n}=sin(2\pi n)$$ and then your sequence is the constant 0. This you are allowed to do this because sin(n) takes on the value of 0 infinitely many times.

Last edited: Oct 8, 2010