Zorn’s Lemma and Basis

[@blackpenredpen]

Your thumbnail is on point!!!!

[@drpeyam]

Thank you!!!!

[@Anteater23]

Excellent video. Instead of spending an hour trying to read and understand the proof, you explain it super good in under 30 minutes. Thanks.

[@Leidl.Michael]

Zorns Lemma is such a powerful tool to prove things like this because you only have to prove something for totally ordered subsets of a partially ordered set, which makes things way easier. For an oral exam I had to learn the proof of the equivalence of the axiom of choice and Zorns Lemma, which didn't made me zornig at all. :D

[@MewPurPur]

This vid put me to sleep at first. Now I finally got it! Super helpful and thank you.

[@MsDanni123456789]

I have been thinking about these things for months, finally I understand these concepts because of this video, you are so positive and make it more fun, well done!

[@highermathematics-bx4mi]

I am finding this theorem from 24 hours but I couldn't find such a video now alhamdulillah now I find your lecture and enjoy it 🥳🌹🥰 very amazing

[@alvinlepik5265]

It also turns out that every non-zero vector space having a basis is equivalent to the axiom of choice. The non-zero bit is important, because the zero vector doesn't constitute a linearly independent subset.

[@nudelsuppe2090]

Usually the span of an empty set is considered to be zero so you basically have an "empty basis" for that

[@alvinlepik5265]

@@nudelsuppe2090 good point. It's consistent, too. A finite linear combination over the empty set yields 0 vector.

[@cruzazul2609]

Omg, I don't "dominate" the English and I learned more with you than with my professor who explained in my mother language SPANISH.

[@ces-1200]

Same :')

[@broccoloodle]

This guy's presentation is so charming :)

[@schwarzeseis4031]

Makes tons of sense. Gonna forget it again. Shall return.

[@dgrandlapinblanc]

I've listened and watched three times this video. Thanks.

[@jonasdaverio9369]

Zorn's lemma... That's reminding me some nightmares from when I was self-studying foundations of mathematics...

[@lagduck2209]

When he said it's equivalent to axiom of choice I've got strong flashbacks to when I was self-studying foundations of mathematics and lots of other topics bingeing wikipedia for hours, days, weeks and months; don't think I actually understood even 5% of what I studied, but it was fantastic trip through the beauty of maths, all of this interconnected world. And I can just barely follow this video nowadays, but I can so it's great. Also thanks for great interesting content and original ideas Dr

[@jonasdaverio9369]

I've always thought Zermelo's theorem doesn't make any sense (I mean, why would there be a well-order in R?) , axiom of choice isn't perticularly counterintuitive, and Zorn's lemma doesn't mean anything in term of intuition.

[@russellsharpe288]

@@jonasdaverio9369: The wikipedia article on the Axiom of Choice contains the following quotation: "The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" - Jerry Bona"

[@sandorszabo2470]

I'm very missed this theorem, and now I could find this video. Soon start the next semester :-)

[@dhunt6618]

Thanks for another great video. Sorry, but I can't resist mentioning the 2 related problems that Zorn couldn't prove - Zorn's dilemma... (sorry again, yes, it's an old pun)

[@drpeyam]

Hahahahaha

[@highermathematics-bx4mi]

🥳🥳🥳 very interesting manner of teaching

[@ravitejassu]

Very beautifully explained

[@dthhfjjjj]

I love this video, you explain it very clearly

[@christianharriviktorreibol5521]

Is an union of sets in a chain not automatically in a chain and so automatically a member of the family of sets?

[@gdsfish3214]

Our professor showed us in week two of first semester how zorn's lemma implies the cantor bernstein theorem or something like that... I wasn't able to keep up, but I still remember the headaches after the lecture... And I've been too scared to look at the notes I made back then.

[@HasXXXInCrocs]

So essentially for any infinite dimensional vector space, a combination of ideals of V is the basis of V? Is that what a chain is? Note that any principal ideal is independent because of the absorption property.

[@goatmatata2798]

Hello Dr peyam, I have a humble request please create a video where u prove gauss's multiplication theorem for the gamma function, I have tried it a million times but just can't get it right, please 🙏 if you do it do it in a not so complicated way, thank you in advance, from 🇰🇪 kenya

[@mrnarason]

Great explanation

[@Handelsbilanzdefizit]

Does it make sense, to define a derivative like this? lim h-->0 (f(x+h)/f(x))^(1/h) = f*(x) As far as I figured out, you can easily calculate attenuation or "elementary" probabilities with that. It seems, the Inverse is some kind of Product Integral (infinity multiplying) If the chance I go to the party is 50%, and my friend visits the party 50%, then we meet us there 25%. But multiplying such probabilities infinity times, then you get some bellcurve e^-x² (for many cases)

[@drpeyam]

Hehe, that’ll be part of another video to be posted sometime :)

[@weinihao3632]

Hi Dr. Peyam, thank you very much for the nice video. I found it a bit confusing that you described your sets with the term ring, since that is an algebraic structure by itself. In the beginning you shortlymention that the proof could also cover infinite dimensional vectorspaces, but in the end it is required that each member of the family is finite. Do you plan to add a follow up video to proof it for infinitely many dimensions as well?

[@drpeyam]

But I proved it for infinite dimensions

[@weinihao3632]

@@drpeyam Oh, sorry, you are absolutely right! I missed your statement at 21:43. Thank you for the clarification!

[@SierszulskiM]

As a Pole I'd just like to mention it would be nice to call the lemma "Kuratowski-Zorn lemma" - Kuratowski came up with it eevn earlier than Zorn, if I'm not mistaken.

[@kiprejoel-kgjplatform8463]

I have some exercises...could you help me Dr. ?

[@highermathematics-bx4mi]

Very amazing 🥰🥰🥰🌹🌹🌹Assalam-o-alaikum may Allah protect you from hardships ameeeeeeeeen

[@edwardhuff4727]

"Contains" is used as a synonym of "includes." It isn't. A "contains" B when B ∈A. A "includes" B when B ⊂ A. A partially ordered family of sets can be "ordered by inclusion" but "ordered by containsion" is different if it even means anything. A von Neumann ordinal set "includes everything it contains": if B ∈ A then B ⊂ A, or P(A) ⊂ A. where P(A) = 2^A = set of all subsets of A = power set of A. You can't have a set that "contains everything it includes" (Russell's paradox, the set isn't well founded, A ∈ A). The sets of Zorn's lemma are posets, partially ordered, and "ordered by inclusion".

[@drpeyam]

Ok, but I’ve always learned contained and includes as the same thing. I’d explicitly use B is an element of A

[@edwardhuff4727]

@@drpeyam I found it confusing too, and when I couldn't figure which is which, looking in Halmos Naive Set Theory it's hard to find the definition of "includes." The distinction might be less vital when elements are atoms like { 1, 2 } rather than { {{}}, {{},{{}}} }.

[@alexandersanchez9138]

Also, Dr. Peyam said “circles” instead of “disks.” Indeed, concentric circles don’t contain one another. Alas, it doesn’t really matter because this is an informal (not to be confused with ‘not rigorous’) proof, and we get the gist.

[@hungnguyen-fn8gm]

how did you do it can you share with me , thank you

[@tarunpurohit6522]

Lv u docter

[@justjacqueline2004]

I am so ******* dim!Whilst I can follow it I am damned if I could solve a problem requiring this work.

[@drpeyam]

I agree, me neither! I always get stuck on those problems

[@ezras7997]

Is there a zenos schilling heel paradox type of thing

[@ezras7997]

Don’t say there is something that ridiculous lol

[@ellobo9999]

Thank you!

[@hyunwoopark9241]

Thats quite a awesome thumbnail for a mth video

[@Anteater23]

Do you do Galois Theory?

[@drpeyam]

Sadly not at all

[@highermathematics-bx4mi]

My ny gusy wala face dakh kar video on ki thi because mojy bi gusa a raha tha is per😡 Now I am happy🤣🤣🤣🤣😇😇😇

[@NitinVerma-nh9vp]

Sir plz prove Sylow's 2nd theorem using Zorn's lemma

[@jonasdaverio9369]

By the way, I hate (some of the) classical proofs. They really make no sense! Per example, you can show you will never find any basis on some vector spaces, but in the same time you know there has to be one. You know for sure it exists, but you know for sure you'll never find it. What is the point of it all?! Constructivism (aka intuitionism) is far superior and more intuitive (who would have thought?).

[@drpeyam]

Totally agree, I prefer constructivism, but sometimes there’s nothing we can do

[@jonasdaverio9369]

Then talk about it in your videos!

[@drpeyam]

lol, why? It’s just an opinion, not a mathematical fact, haha. Also I don’t say I don’t like it, sometimes it’s necessary

[@jonasdaverio9369]

There are whole foundations of mathematics based on constructivism. homotopytypetheory.org/book/ It is not a matter of taste. No model is true, some are useful. You can totally imagine mathematics without law of excluded middle. Saying it is necessary is like saying "Sometimes it is necessary to assume A because we want to show A."

[@jonasdaverio9369]

@@drpeyam Since you know French and you seem interested, you should check out a series of video made by Science4all, it is quite old but it thing you should like it: m.kzbin.info/aero/PLtzmb84AoqRRgqV5DfE_ykuGQK-vCJ_0t From the 15th video, it begins to talk almost only about constructivism and why constructivism is better than classical logic.

[@foreachepsilon]

Isn't it V not(subset) span(beta)?

[@drpeyam]

But beta is a subset of V, so span(beta) is a subspace of V

[@foreachepsilon]

You wrote span(beta) not(subset) V which would imply there is a v in span(beta) not in V but looks like you said there is a v in V not in span(beta), right?

[@drpeyam]

It’s the latter, there is v in V not in span(beta)

[@foreachepsilon]

Dr Peyam so wouldn’t that be V not(subset) span(beta) as opposed to span(beta) not(sunset) V then?

[@drpeyam]

Ooooh, I get the confusion now. I didn’t say span(beta) is not a subset, the notation means span(beta) is a strict subset of V, so span(beta) is a subset of V but not equal to V, so there’s v in V not in span(beta)

[@newtonnewtonnewton1587]

Nice video thanks D peyam السلام عليكم رمضان كريم

[@waterfirecards5128]

First!

[@drpeyam]

Megha Jogithaya It’s been unlisted for a month, that’s why

[@jonasdaverio9369]

@@drpeyam Noooo, you revealed our secret. That's really mean to us! (harmonic mean, not the arithmetic one)

[@Anteater23]

You look like Kaka footballer.

[@beback_]

Why do you look Iranian but sound Indian