Dude I just wanted to say that I fell in love with this channel within a single video. I was looking for help on a homework for my intro to analysis class and stumbled across your video on cauchy sequences. I was almost in disbelief at how efficiently you explain these concepts. Keep rocking on man!
I was told that Bolzano-Weierstrass says that any infinite and bounded subset of R^n has at least one accumulation point. This video looks more like a corollary to me.
@VeryEvilPettingZoo We proved that every bounded sequence has a convergent subsequence by proving that every bounded sequence has a monotone subsequence. Bolzano-Weierstrass Theorem was another lecture and I've never realised that the two are equivalent.
I just wanna mention, a sequence is not a set! In a set the order or repeated appearance of the member is irrlevant, i.e {1,3,2,3,3,4}={1,2,3,4}, whereas in a sequence these factors indeed play a role. Thats why the notation (a_n) is more common than {a_n}, since formally, a K-valued sequence is a map IN --> K, hence an element of the IN-fold cartesian product of K, thus more akin to a vector with infinitely many entries, than a set.
5:37 I'm unsure if this is correct but here's my proof. Assuming n_k is bounded, Since it is strictly increasing, n_k has a limit. Therefore; L - e < n_k 0.5. That means you can find an epsilon small enough but greater than zero for which our expression is less than 1. This leads to a contradiction.
Every increasing sequence of natural numbers is unbounded: Let n_i be an increasing sequence of natural numbers (N = {1, 2, ...}), and assume for contradiction that there is a natural number M such that n_i < M for all i. Since the n_i are increasing we have n_i < n_i + 1 = n_{i+k} + 1 >= (k + n_i) + 1 = n_i + (k+1) so in general n_{k+d} >= d + n_k by induction. But then n_{M+1} >= M + n_1 > M, contradicting n_{M+1} < M.
I'm on my way to real analysis right after Calc 3, so I cannot thank u enough for having advanced topics like this ready for me and everyone else who'll need them.
@@taopaille-paille4992 Real analysis is an introduction to advanced calculus because it is the theory of calculus on real numbers. You cannot get more advanced than the theory itself I think. Further i'm not speaking on this specific video topic, he's creating a full real analysis playlist
If M/2^(k-1) is the length of the I_k th set, then shouldn't the length of the I_K th (k>K) set be bigger? This means that the step at the end ( 21:16 ) when he says that the length of the I_K th set is less than epsilon is not correct. length of I_k < length of I_K So even if (length of I_k ) is less than ε, this does not imply that the length of I_K is also less than epsilon.
Michael Pen, I looked at this video my first time and I quit it coz possibly my mind had not yet settled but just again have looked at it, I have clearly understood what is a subsequence and how to come up with one. This is so easy. Thanks alot Pen, it is a puzzle that has been bothering me alot. I have understood the idea. Thanks so much. May the Good LORD maker and creator of heavens and the Earth bless u and really bless u much.
I was roaming around to this channel to find some inspiration to prove that, in the case of metric space, the complement of a closed subset S of a metric space (M, d) is open, using Cauchy sequence. And 04:45 on this video gave me the idea. Great video! Thanks sir!
Also professor if u don't mind the question, are u able to do research papers when running a channel like this? or has the channel become a full priority now? thx! I'm just trying to gauge how time consuming it is to be a researcher
Use the fact that natural numbers are unbounded, and we are picking from the partition that has infinitely many terms. We may choose to pick a_nk where n_k is bigger than everything we have previously picked. This is possible because in each step we only have chosen finitely many terms and there are infinitely many terms to choose another term from.
Michael can you please help me on a question where we have a constant sequence and are asked about number of subsequences of it. Do we count the subsequences as distinct or is it only one?
@VeryEvilPettingZoo I hope there is a nice way to count the quantity of all reducible polynomials. But I cannot find it. I also don't like the solution you described.
Dude I just wanted to say that I fell in love with this channel within a single video. I was looking for help on a homework for my intro to analysis class and stumbled across your video on cauchy sequences. I was almost in disbelief at how efficiently you explain these concepts. Keep rocking on man!
Btw the fact that every bounded sequence has a convergent subsequence is called the Bolzano-Weierstrass Theorem
I was told that Bolzano-Weierstrass says that any infinite and bounded subset of R^n has at least one accumulation point. This video looks more like a corollary to me.
@VeryEvilPettingZoo We proved that every bounded sequence has a convergent subsequence by proving that every bounded sequence has a monotone subsequence. Bolzano-Weierstrass Theorem was another lecture and I've never realised that the two are equivalent.
Thank you for doing these videos. These are the best real analysis lectures I can find!
I just wanna mention, a sequence is not a set! In a set the order or repeated appearance of the member is irrlevant, i.e {1,3,2,3,3,4}={1,2,3,4}, whereas in a sequence these factors indeed play a role. Thats why the notation (a_n) is more common than {a_n}, since formally, a K-valued sequence is a map IN --> K, hence an element of the IN-fold cartesian product of K, thus more akin to a vector with infinitely many entries, than a set.
Yes! Notation gang 👊
Good Place to Start 0:14
Good Place to Stop 22:14
Needless to say, the whole video is a Good Thing to Watch
So now it becomes a race? Lmao, I think we go a little too far that with meme 😂
@@goodplacetostop2973 lol
TBH I just wanted to surprise you a little
5:37 I'm unsure if this is correct but here's my proof. Assuming n_k is bounded, Since it is strictly increasing, n_k has a limit. Therefore;
L - e < n_k 0.5. That means you can find an epsilon small enough but greater than zero for which our expression is less than 1. This leads to a contradiction.
Every increasing sequence of natural numbers is unbounded:
Let n_i be an increasing sequence of natural numbers (N = {1, 2, ...}), and assume for contradiction that there is a natural number M such that n_i < M for all i.
Since the n_i are increasing we have
n_i < n_i + 1 = n_{i+k} + 1 >= (k + n_i) + 1 = n_i + (k+1)
so in general n_{k+d} >= d + n_k by induction.
But then n_{M+1} >= M + n_1 > M, contradicting n_{M+1} < M.
I'm on my way to real analysis right after Calc 3, so I cannot thank u enough for having advanced topics like this ready for me and everyone else who'll need them.
it is not advanced though.
@@taopaille-paille4992 Real analysis is an introduction to advanced calculus because it is the theory of calculus on real numbers. You cannot get more advanced than the theory itself I think. Further i'm not speaking on this specific video topic, he's creating a full real analysis playlist
If M/2^(k-1) is the length of the I_k th set, then shouldn't the length of the I_K th (k>K) set be bigger? This means that the step at the end ( 21:16 ) when he says that the length of the I_K th set is less than epsilon is not correct.
length of I_k < length of I_K
So even if (length of I_k ) is less than ε, this does not imply that the length of I_K is also less than epsilon.
if k>K then lenght(I_k) < lenght(I_K)
These lectures are incredible, the explanations are so clear. Thanks for these videos.
Michael Pen, I looked at this video my first time and I quit it coz possibly my mind had not yet settled but just again have looked at it, I have clearly understood what is a subsequence and how to come up with one. This is so easy. Thanks alot Pen, it is a puzzle that has been bothering me alot. I have understood the idea. Thanks so much. May the Good LORD maker and creator of heavens and the Earth bless u and really bless u much.
I was roaming around to this channel to find some inspiration to prove that, in the case of metric space, the complement of a closed subset S of a metric space (M, d) is open, using Cauchy sequence. And 04:45 on this video gave me the idea. Great video! Thanks sir!
22:13
in the proof of B-W theorem ,you must take n1
I was taugh a more complicated proof of the last result. 😅 This video was very useful. 😊
9:24 a lot of letters to keep track of, lower and upper case
The notation is quite confusing, but fortunately now I got it, thanks!
Also professor if u don't mind the question, are u able to do research papers when running a channel like this? or has the channel become a full priority now? thx! I'm just trying to gauge how time consuming it is to be a researcher
I think a strictly increasing sequence can be bounded?
I don't really know any set theory. But, question: Was that the Axiom of Choice in the second proof?
How do you make sure that the n_1,n_2,... you picked is strictly increasing?
Use the fact that natural numbers are unbounded, and we are picking from the partition that has infinitely many terms. We may choose to pick a_nk where n_k is bigger than everything we have previously picked. This is possible because in each step we only have chosen finitely many terms and there are infinitely many terms to choose another term from.
@@jimallysonnevado3973 thanks!
Michael can you please help me on a question where we have a constant sequence and are asked about number of subsequences of it. Do we count the subsequences as distinct or is it only one?
What if the sequence is alternating sign? I don't think the convergence holds for that, or am I wrong? 😅
You can take a subsequence that's not alternating. For example, if you have (-1)^n you can take the subsequence (-1)^{n_k} where n_k is even.
What is the quantity of all irreducible polynomials of degree n with coefficients from Z/pZ, where p is prime?
@VeryEvilPettingZoo I hope there is a nice way to count the quantity of all reducible polynomials. But I cannot find it. I also don't like the solution you described.
I prefer the classic proof of the Bolzano Weierstrass theorem
For a sequence to be bounded I thought you have less than or equal to sign instead of less than and you made a mistake and you used the open interval
You did a very good job. Years after college I’m beginning to understand analysis, thank you!