Can limsup be attained?

[@mathwithjanine]

I AM READY!! 😄

[@phukinho]

At 20:20. If there are finitely many terms equal to t, they have to come before SN0 right? Otherwise the last one of them would be another dominant term coming after SN0. Therefore we don't really need N1. Am I wrong?

[@thomasborgsmidt9801]

I consider this video as one of the most important video's I've seen. I will have to view it several times to let it sink in. To me the important thing is that the convergence of of one function to another should lend itself to a more sensible description of the world - of f.i. monetary policy. The is the general economic problem, that banks perform short term arbitrage which in itself is stabilising - but only in the short term! There can be earned money from that; but not many! In the long term the return on capital (hence interest) is determined by increased in productivity and inflation/deflation. This means that in the long term these short term deviations should converge to a function that is monotenous. Generally speaking I assume it will be a logarithmic function - perhaps ln(ln(t))? The important thing is to separate the simultaneous function dependent on different parameters - but interacting. The other area (that will probably involve complex numbers??) is music! Now analisys of music generally involves a lot of selfimportant gibberish that uses several additional functions - also unknown functions to explain. Generally known to get further away from what is truth.

[@wtt274]

Thank you very much for this great video . Without any doubt , your lecture is impeccable !

[@dramwertz4833]

At 29:30 it should be Snk+1 > Snk right? (and not

[@drpeyam]

It’s a typo, I was checking if people were paying attention

[@dramwertz4833]

@@drpeyam ;)

[@Pyeface12]

Nice one Dr P

[@elephantdinosaur2284]

"Are you ready for the challenge!? I can't hear you ... I really can't because it's youtube but let's assume you're ready" XD Thumbs up from me within the first 30 secs. The jokes crack me up :) Hey Peyam, Nice proof and another nice use of considering dominant terms. I think you could make this a corollary to one of your previous videos. Maybe something like this: (1) Choose a subsequence {s'_n} of {s_n} that converges to t=limsup s_n; this is possible since we can always find a term s_m that's within an epsilon from t. (2) Take a second subsequence {s''_n} of {s'_n} that is monotonic; I think you proved this more generally in an earlier video by considering dominant terms. (3) Then since every subsequence of a convergent sequence has the same limit, the second subsequence {s''_n} has the desired properties :)

[@drpeyam]

Thank you! And that’s the point, (1) is not obvious at all, it’s proving that part that takes forever

[@elephantdinosaur2284]

@@drpeyam That's so true! I tried working on it and there's gaps all over the place!! It takes ages filling in the holes. How about this to justify (1)? Let epsilon>0, set S = sup{s_n:n>N} where N is so large that |S - t|infty} sup{s_n:n>N}. It's enough to find some m>N such that |s_m - S| < epsilon/2, cause then from the triangle inequality (a.k.a the best inequality in the world) we have |s_m - t| < |s_m - S| + |S - t| < epsilon/2 + epsilon/2=epsilon. Claim: |s_m - S|N. Proof: Suppose the opposite, then since s_m < S = sup{s_n:n>N}, we must have s_m < S - epsilon/2 for all m > N. Takes sups over m>N, gives S

[@drpeyam]

So I think the proof is correct, although I didn’t check every detail, but you’re missing an important point: I agree that there is always an sm that is epsilon/2 apart, but that doesn’t show that there is a subsequence. Your choice of sm needs to be ordered in some way, that’s why you need an inductive construction. In any case, there’s a reason this proof is 30 mins long, there are no shortcuts here unfortunately

[@drpeyam]

Also what makes the proof nontrivial is that the subsequence is monotonic, that’s what makes it 10x harder. I agree that if you remove monotonic, your proof is easier, but it’s not really as trivial as saying that I have shown this before

[@elephantdinosaur2284]

@@drpeyam Yeah I agree with you 100%. This is definitely non-trivial!!! Wrestling with the weeds, gives me a greater appreciation of the video. I've thought about this one for almost 5 hours and seen this video 4 times already and you did a great job with it. I'm going to try and commit the main steps of your proof to memory. At this stage, I'm surprised it wasn't a 50 min video and you did it on ONLY about a dozen slides!! LEGENDDD! Thank you for taking the time in answering my questions. You're an awesome prof and I wish I had been in your class. Thanks again Peyam :)

[@blackpenredpen]

Is there a theorem that you can't prove? I mean... wtf (want to find a theorem that you cannot prove)

[@LiangLao2]

why the axiom of choice is independent of ZF? is there easy way to show it?

[@mikeredd2608]

love you dr peyam

[@joeeeee8738]

Hey Dr Peyam, is there a way of showing the difference between a sumatory of f(x) and integral of the same f(x) ?

[@madhavpr]

This is beautiful !! Just a quick technical point- I guess the sequence terms {s_nk} have been constructed with the epsilon property of the supremum [ if t = sup(STUFF), then t - epsilon is not an upper bound of STUFF]. If we pick a natural number {n_k} in the open interval (t - 1/k , t + 1/k), how do we ensure that the sequence of natural numbers produced this way is strictly increasing in order to form the required subsequence? That is, how do we guarantee that n1 < n2 < .... < n_k < n_k+1 < ... ? I guess that this involves a sneaky use of the well ordering property of natural numbers, but I'm not sure.

[@drpeyam]

You don’t need the well ordering principle, since there are only finitely many values from n1 up to nk. It’s called an inductive construction and there’s a video on that too in the playlist

[@madhavpr]

@@drpeyam okay, that makes sense.. But (in particular), how can we justify the existence of the natural number n_2 which is greater than n_1 without induction ? Once this is justified, obviously the subsequence can be constructed easily

[@drpeyam]

Archimedean principle

[@godthinkun]

Good work

[@dgrandlapinblanc]

Really not evident. Thank you very much.

[@sumitgupta6905]

In these cases if the limsup of a sequence is finite (t here) how can we formalise the fact that its subsequence {snk} of peaks that is decreasing is bounded below? Since we need {snk} to be bounded below in addition to be decreasing to be convergent.

[@drpeyam]

Here sn and tn are bounded, I think

[@mikecohen5887]

the subsequence {snk} of peaks is a subsequence of the sequence which converges to limsup, hence also converges.

[@luna9200]

You said you were gonna do some PDE vids! We expecting those soon?

[@drpeyam]

I’ve already made some (convolution, dominated convergence theorem)

[@NotoriousSRG]

I was born ready.

[@drpeyam]

Hahahaha