# Subsequential Limits

• Oct 8th 2012, 10:47 AM
renolovexoxo
Subsequential Limits
Construct a sequence {sn} for which the subsequential limits are {-infinity, -2,1}

Construct a sequence {sn} for which the set of subsequential limitsof the sequence are countable.

Any help getting started would be great, I don't know where to begin.
• Oct 8th 2012, 11:15 AM
johnsomeone
Re: Subsequential Limits
Let me give you a few examples:
To make a sequence with subsequential limits 0 and 1, how about:
0, 1, 0, 1, 0, 1, 0, 1, ....
or
1/3, 2/3, 1/4, 3/4, 1/5, 4/5, ..., 1/n, (1-1/n), ...

To make a sequence with subsequential limits 0 and 1 and -1, how about:

$\displaystyle a_n = \sin\left(\frac{\pi n}{2}\right), n \ge 0$, which is 0, 1, 0, -1, 0, 1, 0, -1, ...

To make a sequence with subsequential limits e and infinity, how about:

$\displaystyle a_n = \left(1+\frac{1}{n}\right)^n$ for $\displaystyle n$ even and $\displaystyle a_n = n$ for n odd.

For the subsequential limit set being countable, I'd suggest trying to make it the positive integers - something easy.

Can you blend the following infinite set of sequences into a *single* sequence?
1, 1, 1, 1, ...
2, 2, 2, 2, ...
3, 3, 3, 3, ...
4, 4, 4, 4, ...
etc.
(Hint: do you recall the method that's usually used to prove that a countable union of countable sets is a countable set?)
• Oct 9th 2012, 02:04 PM
renolovexoxo
Re: Subsequential Limits
I'm afraid I don't really understand. The idea of subsequential limits makes sense, but for the first part I am having a lot of trouble seeing how to put in negative infinity. For part b, I'm afraid I still don't really understand.
• Oct 9th 2012, 02:21 PM
johnsomeone
Re: Subsequential Limits
Quote:

Originally Posted by renolovexoxo
I'm afraid I don't really understand. The idea of subsequential limits makes sense...

Since you're having trouble doing this, I'd have to disagree. What's going on hasn't really become clear to you yet.
1) Did you look at my examples, and think about why they work?
2) Can you do part a without the negative infinity? A sequence whose subsequential limits are -2 and 1?
3) How about a sequence whose subsequential limits are -10, -2, and 1?
4) Can you produce any simple ordinary sequence that goes to negative infinity? If so, then replace the terms of the sequence in #3 that are going to -10 with the terms of your sequence going to negative infinity, and you should have it.

Quote:

Originally Posted by renolovexoxo
For part b, I'm afraid I still don't really understand.

It's more important to worry about part a, since you can't run before you can walk.
If you figure out part a, and think you really understand it, then maybe try to tackle part b. Maybe reread my first post's ideas about part b then.
Until part a seems absurdly simple to you, there's no point in worrying about part b. Don't worry - it'll click eventually if you continue to think about and actively work on it.
• Oct 9th 2012, 02:31 PM
Plato
Re: Subsequential Limits
Quote:

Originally Posted by renolovexoxo
I am having a lot of trouble seeing how to put in negative infinity.

$\displaystyle s_n = \left\{ {\begin{array}{rl} { - n,} & {n \equiv _3 0} \\ { - 2 + \frac{1}{n}.} & {n \equiv _3 1} \\ {1 - \frac{1}{n},} & {n \equiv _3 2} \\\end{array}} \right.$ where $\displaystyle n \equiv _3 0$ means $\displaystyle \mod(n,3)=0$

So we have three sub-sequential limits: $\displaystyle -\infty,~-2,~\&~1$