Rod

anyone want to talk real analysis

Does anyone want to talk real analysis?

Can anyone help me clarify my understanding of a real analysis question?

@what'sup..math.stackexchange.com/questions/481595/…

sorry its raining outside, n that just distracted me haha

@PeterTamaroff, I just need to know weather the proof is correct.

hello wrld

y do we study the lim sup of a set

that's all

s

sorry largest limit of all subsequence

I know that it is the largest of subsequence of a given sequence, assuming that this given sequence is bounded

is there anything else to it, excuse me if this is too broad

I have a general idea of what it means, the largest subsequential limit

most recently we've been talking about lim sup

I'm an undergrad and I'm taking a real analysis course right now and the lectures seem a but scattered, I just want to talk about some definitions

does anyone want to talk real analysis?

hello world

Last comment -1^n is also not convincing enough because we assumed that this sequence only contains positive real numbers.

ok thx a lot

and after this statement we can say the sequence is unbounded...and proceed like I was doing?

Now suppose we have a bounded sequence in R; so there exists a monotone subsequence, necessarily bounded. It follows from the monotone convergence theorem that this subsequence must converge. this would contradict that there are no convergent subsequences?

so if I told you a sequence doesn't converge, I would need to prove that it isn't bounded above?

hey...Why can't you just say x$_{n}$>M+$\epsilon$ this statement should still hold, and reverting back to the limit definition |x$_{n}-M|>$\epsilon$

But we are already assuming that it has no convergent subsequence, so this example doesn't apply.

If a sequence has no convergent subsequence,you shouldn't have to prove that its not bounded above.If it were bounded you could pick a subsequence which converged to that so called upper bound. And that statement with $x_{n}$>M+$\epsilon$ holds for all n$\geq\textbf{N}$

@MichaelHardy, I tried to correct my proof above. Is it getting there now, or am i still missing something?

@StefanH.is this M not the same as L. Sequences can tend to a limit without exactly being theat so x$_{k}\leq M$ or L in my case.