1. ## subsequence help.

Let {sn} be a sequence and let E be the set containing members x such that x is a limit of a subsequence of sn. Let S*=sup(E).
Prove S* is a member of E.

I do not understand the first part of the proof (from a textbook). It goes if S* = infinity then E is unbounded (fine so far). Then {sn} is unbounded, hence there is a subsequence -> infinity.

This raised a few questions.

If {sn} is unbounded, does sn -> + or - infinity?
The proof also seems to imply {sn} -> infinity iff xn -> infinity, where xn is a subsequence. Correct?

Thanks

2. ## Re: subsequence help.

I guess $\displaystyle \{s_n\}$ is a sequence of real numbers. Well, if $\displaystyle \{s_n\}$ is unbounded, then the sequence is either unbounded below or unbounded above. In the first case, we can find a subsequence which converges to $\displaystyle -\infty$, and in the second case we can find a subsequence which converges to $\displaystyle +\infty$. Here, since $\displaystyle \sup E=+\infty$ in the first part, then $\displaystyle \{s_n\}$ is not bounded above.
If we have $\displaystyle \lim_n x_n=+\infty$ for a subsequence $\displaystyle \{x_n\}$ of $\displaystyle \{s_n\}$, it doesn't mean that $\displaystyle \lim_n s_n=+\infty$. For example, take $\displaystyle s_n=(-1)^nn$ (the limit doesn't exist).

3. ## Re: subsequence help.

So you are saying {sn} is unbounded iff it has a subsequence which is unbounded?

If so, since sn is unbounded it tends to infinity, hence it has a subsequence tending to infinity. Correct?

4. ## Re: subsequence help.

Originally Posted by Duke
Let {sn} be a sequence and let E be the set containing members x such that x is a limit of a subsequence of sn. Let S*=sup(E). Prove S* is a member of E.
The proof (from a textbook). It goes if S* = infinity then E is unbounded (fine so far). Then {sn} is unbounded, hence there is a subsequence -> infinity.
In these cases it is best to say what textbook you are using.
Not knowing where that proof is going makes it really hard to comment.

Here is the way I know the proof works.
Suppose that $\displaystyle S^* = \sup (E)$ and that $\displaystyle S^* < \infty \;\& \,S^* \notin E$.

Construct a sequence $\displaystyle x_1<x_2<\cdots<S^*$ so that each $\displaystyle x_n\in E~\&~\left| {x_n - S^*} \right| < \frac{1}{n}$
A bounded montone sequence has a limit: $\displaystyle (x_n)\to y$.
It is easy to see $\displaystyle y\in E.~\&~y<S^*$. But that is a contradiction.
Thus $\displaystyle S^* = \infty \;\text{ or } \,S^* \in E$

5. ## Re: subsequence help.

I'm using Rudin's analysis. I was just struggling to see how he goes from (1) if S* = infinity then E is unbounded to (2) Then {sn} is unbounded, and then finally (3) hence there is a subsequence -> infinity. Can you tell me the results he is using to get from (1) to (2) and then (2) to (3) ? Thanks for your alternative proof as well.