Zorn's Lemma, The Well-Ordering Theorem, and Undefinability | Nathan Dalaklis 

Thanks! I’m glad you thought so!!

Yup, ordinals for types of orderings (counting numbers, omega family...) and cardinals for types of quantity (sizes of finite sets, aleph, beth families...)

Ian Grant I’ve never heard a specific name given for the theorem but the Gödel encoding line of thinking is one way to approach it. There is also a measure theoretic way to address the argument for this with Vitali sets.

Can you give some bibliography about how the well ordering is general undefinable?

You can think of ω that way. It is a limit ordinal which intuitively means that a limiting process is required to define it, in the case of ω it’s motivated by exactly what you have said here there is no upper bounding natural number of the naturals so ω is defined as the ordinal that corresponds to a number that is larger than all the other naturals.

@@CHALKND Okay...trying to go ahead from what I think I understood (I've switched from lignuistics to math): oméga is something that exists so we can assign an oméga-th position that delimits ALL the naturals? (Guess I need to sleep over this -- infinities are fine, but structured infinities are...interesting^^)

I concur, however, I wasn't particularly focused on the equivalence here. The well-ordering theorem (or Zermelo's Theorem) is also equivalent to the AoC too which I think is totally bonkers!

You have the book about this equivalence? This lecture is so cool, but it is so dense. It's hard for me because I saw the first time now. I'd like to understand the equivalence between zorns lemma and axiom of choice

I actually don't have a particular textbook that primarily influenced this one. 😕 This video was primarily inspired by a talk a friend of mine gave. I thought that it was so cool that I created this video so I could return to the mathematical idea in the future! 😀

1) The ordinals form a collection called a 'proper class'. Intuitively, this means that there are too many of them to fit them into a set. This is an odd idea but a familiar one when looking at mathematical objects defined by certain properties. For example, the collection of Groups and the collection of Dynamical Systems both form proper classes. To say that there are the same of them as there are ordinals can be thought of as an equivalence between the chain in the Hasse diagram and the ordinals when thought of in a categorical sense. 2) The (0,1) example fails to fit the scenario of the lemma, and thus does not contradict the arguement given here. That is, for the lemma to apply, every chain in our partially ordered set P has an upper bound in P. However, (0,1) does not have an upper bound in (0,1). To be a bit more direct about the situation of the proof: What is assumed: partial ordering on P and all chains have upper bounds in P, but no maximal element (the last thing being what is assumed in hopes of contradiction) What is initially deduced: (starting from the lack of a maximal element) In the Hasse Diagram there must be a chain that does not end, that is we can follow this chain to make larger and larger elements. Hopefully this helps!

Ah I see, so the upper bound must be in the chain itself? Is it necessary then to require that *all* chains in the poset P have an upper bound for Zorn's lemma to apply? I'm probably wrong but if a *single* chain in P has an upper bound x, is this not then a maximal element in P? (Since every element in the chain is less than or equal to x, and every other element in P is not comparable to x, hence x is maximal?)

In the case where your set is a chain, the notion of maximal element and maximum coincide. So if the upper bound is in the set it would be maximal. In terms of requiring all chains the devil is in the details. A chain is just a totally ordered subset. So as a counter example to your claim take (0,1). Any closed or right closed subset ( like (0,.5] or [.25,.35]) is a chain in P with an upper bound in P that has a maximal element. However these elements that are maximal in the subsets are not maximal in P.

@@CHALKND Okay. Thanks for clarifying.

You are correct! Within the context of the video I think I was just trying to get at an intuitive notion of ω. That idea being if you were to put the size of the Naturals into an ordered list the corresponding position would be ω which comes after all the natural number.

CHALK gotcha :) I love your videos by the way, please make moreeee haha

@@JacobWillsonPhoenix Thanks!! I'm so glad you do! (it has been a while since I posted, but I am working on some newer ones :D )

Good thing theres a remade version of it on my channel where this doesn’t happen

Thanks!

Glad you thought so Yusuf!

I am still making videos! Since youtube, at my current size, is just a hobby I'm only really able to put out videos every 1-2 weeks. If everything goes as planned there will be a new video next week!

Gotta love google(/ or other search engines) 😅

Well in this context it is not a pure choice function as in the definition of choice, however, when we are picking from upper bounds of the chain to make the chain longer and longer that is where choice is used. In particular, to make the chain a copy of all the ordinals (and thus have a subset of a set which is not a set) we need to make uncountably many selections.

🧐 ...not sure how to respond to this 🤔

98

🤔 Who did 90 8?

These are all great points to emphasize yes.

Awesome!

I'm planning on doing a video on AoC soon, but I already have plans for my next video; maybe after that! Thanks for the suggestion!

many m's

I’ve actually never heard a professor refer to it in such a way.

This statement is true

Sir. This is a video about Zorn's Lemma

@@CHALKND thats assessment year stuff com on, this guy's a noob!

Not sure who “this guy” is or what “assessment year” is but again, sir, this is a video about Zorn’s Lemma.

@@CHALKND understandable , ok sir have a good day