1. ## limit superior/inferior proof

*x_n indicates x sub n
*>= indicates greater than or equal to

Let (x_n) be a bounded sequence and for every n in natural numbers, let s_n:= sup {x_k : k>=n} and t_n:= inf {x_k : k>=n}

Prove that (s_n) and (t_n) are monotone and convergent. Also prove that if lim(s_n) = lim(t_n), then (x_n) is convergent.

Can someone get me started in the right direction? Thank you!

2. You want to show that $s_n$ is decreasing and $t_n$ is increasing.

These should be fairly obvious since $\{ x_k|k\geq n+1\}\subseteq \{ x_k|k\geq n\}$.

Note that "convergent" allows for the possibility of converging to $\infty$ or $-\infty$.

I assume that we're working in a complete space (maybe the reals or complexes?) If so, then you only need show that the sequences are Cauchy sequences.

3. where do you get that the {x_k : k>= n +1} is a subset of {x_k : k>= n} ?? And we haven't gotten to Cauchy sequences yet

4. Originally Posted by mremwo
where do you get that the {x_k : k>= n +1} is a subset of {x_k : k>= n} ??
That is simply saying $\{x_6,x_7,x_8,\cdots\}\subset\{x_5,x_6,x_7,\cdots\ }$ for example.

Do you understand that if $A\subset B$ and $U=\sup(B)$ then $\sup(A)\le U~?$

5. Originally Posted by Plato
That is simply saying $\{x_5,x_6,x_7,\cdots\}\subset\{x_6,x_7,x_8,\cdots\ }$ for example.
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You got your sets backwards. It should be $\{x_6,x_7,x_8,\cdots\}\subset\{x_5,x_6,x_7,\cdots\ }$ .

6. Please see my edit. Thanks

7. I assume that we're working with real numbers here (you haven't clarified this).

Do you already have the theorem that an increasing sequence that is bounded above converges? (and similarly for decreasing and bounded below)

8. Yes, I'm working with real numbers, sorry.
I do have that theorem, but how would I go about proving that each of the sequences converge?
I did the proving they are monotone part (thanks), now I'm stuck at proving they converge.

9. Well once you prove monotone you're essentially done. For example, if you show that the sequence is increasing, then there are 2 cases.

Case 1 - the sequence is bounded above. Then the sequence converges to a finite number.
Case 2 - the sequence is not bounded above. Then the sequence converges to infinity.

Edit: I just noticed that you are given that original sequence is bounded. Thus the sup and inf of the sequence exist. So the other 2 sequence are bounded as well. So Case 1 only applies.

10. Thank you very much, I think I'm starting to get it.

One question: I am a little bit confused about the relationship the x_n sequence has with s_n and t_n. I mean I know that s_n is the sup the set that was given and t_n the inf, but I don't get what the k's have to do with it. Also, it then asks me to show if lim(s_n)=lim(t_n) then x_n is convergent. Why is that?

EDIT: As in, how does (x_n) being bounded tell you that the sup and inf of the other sequences are bounded as well?

11. Originally Posted by mremwo
EDIT: As in, how does (x_n) being bounded tell you that the sup and inf of the other sequences are bounded as well?
If the sequence $x_n$ is bounded then $\left( {\exists B > 0} \right)\left( {\forall n} \right)\left[ { - B \leqslant x_n \leqslant B} \right]$.
That insures the sequence has a supremum and an infimum.
So it is true of any of sub-sequence of the sequence.

12. Okay, I got that now. Now how would lim(s_n)=lim(t_n) show that (x_n) is convergent?