And with that, our Cauchy sequence lesson sequence comes to an end for now! Thanks for watching and give it a share if you want to help the channel grow, so I can make more real analysis lessons!
Nos every Cauchy Sequences are convergent, it just applies when you are taking the whole R^n like your metric space. Because you can take a subspace of R^n where your sequence be Cauchy but not convergent.
I’m probably missing something here, but wouldn’t log n be a counter example of that? Like isn’t it Cauchy, yet non convergent? Or so I thought at least.
Thanks for watching, that's a great example because the terms of log(n) get arbitrarily close to their neighbors. That is, a_n - a_{n-1} gets arbitrarily small, but a_n - a_m will not for any n and m greater than any N. However far we go in log(n), we can always find numbers arbitrarily far away. Then, as we would expect, it is not convergent.
Looks circular to me 😭😭 Proving cauchy sequence are convergent using Bolzano Weierstrass theorem. But proving bolzano Weierstrass theorem requires monotone convergence theorem. And the proof i saw on Wikipedia relies on cauchy sequence being convergent
It's actually axiomatic I think. Monotone convergence theorem relies on the axiom that R is a complete field. That is, that every non empty subset of R has a "supreme" (don't know how they call it in English). From that you can proof this theorem. However, as my teacher said in one of my classes, this theorem and the supreme axiom are equivalent. So much so that we call this theorem the R completeness theorem. Take this with a grain of salt, tho, I'm just a first year student at Uni
And with that, our Cauchy sequence lesson sequence comes to an end for now! Thanks for watching and give it a share if you want to help the channel grow, so I can make more real analysis lessons!
Not every cauchy sequences are convergent. Only in complete set cauchy sequences are convergent.
So the question is: offer a counter example to the statement :)
Nos every Cauchy Sequences are convergent, it just applies when you are taking the whole R^n like your metric space. Because you can take a subspace of R^n where your sequence be Cauchy but not convergent.
4:49 big N tooth term
N🦷
GREAT VIDEOOOOoOoOOooOOo
I’m probably missing something here, but wouldn’t log n be a counter example of that? Like isn’t it Cauchy, yet non convergent? Or so I thought at least.
Oh oops, log n isn’t Cauchy. Well forget I said anything.
Thanks for watching, that's a great example because the terms of log(n) get arbitrarily close to their neighbors. That is, a_n - a_{n-1} gets arbitrarily small, but a_n - a_m will not for any n and m greater than any N. However far we go in log(n), we can always find numbers arbitrarily far away. Then, as we would expect, it is not convergent.
Sir.. i can't understand, please once again give the cleared explanation. Why the sequence log n is not a cauchy sequence.
Sir... please explain... why the sequence log n is not a cauchy 😢
@@17matboy if its cauchy there exist N ,such that for all n>N log m-logn
woah that was a bit fast, just a bit
Yeah, it's because there are a lot of information at once. I'm about to watch the video to the third time so the concept can stick.
Looks circular to me 😭😭 Proving cauchy sequence are convergent using Bolzano Weierstrass theorem. But proving bolzano Weierstrass theorem requires monotone convergence theorem. And the proof i saw on Wikipedia relies on cauchy sequence being convergent
Real analysis is ending me bro
monotonne convergence theorem doesn't require at all cauchy sequences... it's usually covered way before cauchy.
The truth is that Ramanujan saw it in his dreams and told this to Cauchy
It's actually axiomatic I think. Monotone convergence theorem relies on the axiom that R is a complete field. That is, that every non empty subset of R has a "supreme" (don't know how they call it in English). From that you can proof this theorem. However, as my teacher said in one of my classes, this theorem and the supreme axiom are equivalent. So much so that we call this theorem the R completeness theorem. Take this with a grain of salt, tho, I'm just a first year student at Uni