Maybe someone can help me out, I thought I had an idea but I don't think it is right.
I am trying to prove the above. My attempt:
I am given that the lim sup Xn is finite, which I think means it exists, which I will call L. So I have a sequence [Xn] = [x1, x2, x3,... xn, xn+1,....] and another sequence [Xm] = [xn, xn+1,...]. This is where I am stuck, I don't know where the next step should be. I have an idea that if I can show [Xm] is monotonically decreasing then the sup of [Xm] is Xn and then taking the limit of Xn gives me L, which would complete the proof, I think. But, I don't know how to show that the sequence is monotonically decreasing, or if it even is... it was just an idea. Any help/hints would be appreciated!
Thanks for replying Plato, and I know it seems weird to prove a def, I guess I mean to just show that it is true. I've defined lim sup to be the supremum of the set of subsequential limit points of the sequence, {Xn}... The problem lets me assume that the set is finite, so I've been calling the sup of the set, L.
So the set of subsequential limits = [..., L] I have to show somehow that the lim sup Xm (m>=n) is also L.
I read somewhere that if I have a sequence [An] then I can define a new sequence [Bn] = sup Ak, k>=m = sup [An, An+1, An+2,...] I am trying to incorporate this and get something out of it because then the sequence Bn is montone decreasing
Okay the last post is getting cluttered, I am trying to work on a new solution. I have:
Given a sequence Sn = {S1, S2, S3, ... Sn, Sn+1,...}
Let E be the set of subsequential limits of Sn. Assume E is finite then E= {....L}
This means that the Lim sup Sn = L.
Now for the right half of the equation. Set Tn = sup Sk , k>=n. Then by substitution we have Lim Tn on the right.
So as of now it is:
L = Lim Tn (n goes to infinity) ... I just need to show this somehow.
L = Lim sup (Sn, Sn+1,...)
I know that Tn is a decreasing and bounded sequence so it does converge, but I just don't know how to show it is equal to L or the limit superior of Sn.
Okay, I thought it was the last term, but if it's maximum I will just reword it... All I meant to do was denote the maximum as L.
also that last line in quoted text should be Sn, not Xn.
Can I say as n is going to infinity, n must be getting closer and closer to k... so at some point I will have that sup Sn -> Sup Sk which would be lim (Tn) => lim sup (Sk) => lim sup (Sn) => L. ... On second thought this might not make much sense.
I just need to find a way to show that the decreasing and bounded Tn satisfies:
Lim Tn as n-> infinity = L
Where Tn is sup Sk for k>=n
Just thinking... since Tn is decreasing and bounded then the Lim of Tn must be = inf Tn.
Can I say that the existence of the limit superior means that Sn is bounded above, L. If I define Tn as sup Sk k>=n then:
T1 = sup {S1, S2, S3, S4,...}
T2 = sup {S2, S3, S4, ...}
T3 = sup {S3, S4,...}
This shows that each one is a subset of the one before and thus are decreasing. So: T1>T2>T3>... Tk-1>Tk>... Each T is bounded below and above (by the original Sn) So the Limit of Tn is then the same as the upper bound, L. Does this make sense?
I have been asked to prove the same thing... I too do not understand WHY we have been asked to prove a definition but there it is! I suppose it's sufficient to show that since E is bounded above (by s*=lim sup(xn)), if any subsequence were to have a limit (sup xm) greater than the upper bound of E, that would contradict the definition of E and its boundedness? I'm not coming up with any better ideas!