Ordinals, Cardinals and Transfinite Arithmetic

[@BeMedic]

Correction: the continuum hypothesis (CH) is not "unknown", it has been solved. By solved, I mean Gödel and Cohen showed some time ago that in the context of ZFC the CH is independent of ZFC. In other words, it has been proven that CH is unprovable (independent) wrt ZFC. In other words still, ZFC+CH and ZFC+not(CH) are both consistent; more accurately, relatively as consistent as ZFC. In some other type of set theory, maybe, the problem could be open.

[@jaca2899]

I personally believe that Beth_1 should be the same as Aleph_1

[@Erotemic]

While the second part of your comment is true, Woodin would disagree with the statement that it's been solved. Perhaps ZFC is missing some fundamental axioms. You allude to this, but solved within ZFC may not be equivalent to solved.

[@kennyearthling7965]

Ok I was not ready for this, need to rewatch. Heavy stuff, very interesting

[@pabloagsutinnavavieyra2308]

Awesome. I am glad to found this gem of channel :) You definitely have a new subscriber. Cheerio!

[@pabloagsutinnavavieyra2308]

Is so nice to find this type of content available in video form.

[@wdfusroy8463]

Like every other video I have ever seen on the topic of transfinite ordinals we are told that [1 + ω] is not = to [ω +1] because the latter expression adds the 1 to the end of the already specified ordinal ω, while the former merely throws another non-transfinitely ordered number, [i.e. 1] into the already ordered number ω. O.K., that makes sense. But no one ever goes on to explain the analogous difference between [1 - ω] and [ω 1 exactly that is supposed to mean! Worse still, if we include extensions of the integers into the hyperreals of Non-standard Analysis, or the Surreal Numbers, we can easily generate new numbers like ω/2, or sqrt[ω], or even lnω and other such odd beasties. But what is never made clear, at least that I have ever seen, is why these new and smaller infinite ordinals do not contradict the original definition of ω which specifies that ω represents the "first transfinite integer. I hope someone can explain why such a contradiction does not occur, if indeed it does not. Thanks!

[@tomkerruish2982]

I don't know enough about the hyoerreals to answer your question concerning them. As for the surreals, omega is still the smallest infinite ordinal, where ordinals are defined to be surreal numbers with no right options.

[@linuxp00]

Well, in hyperreal setting. It's really a relabeling of lemniscata symbol ∞ for ω, given that a concrete definition, that is its the ordinal representing the cardinality of ℕ. Said that, they are just different relative to others transfinite numbers of the same order, that is, ω + 1 < ω + 2 < ω², or their reciprocal ε, such that ε = 1/ω. But compared to the finite numbers, is just as ∞ in the standard analysis. Example: (ω²+2ω)/ω = ω+1 -ω/√ω = -√ω (3ε²+2ε)/ε = 3ε+2 5ε/ε = 5 When taken the "standard part" that will be simplified to a finite number or a regular old infinity. The examples above have std part equal to: ∞, -∞, 2, and 5, respectively.

[@masonmireles9295]

Can anybody please dummify this down? I really want to understand, but this is difficult. What are the implications to the reals world? Where can I use this?

[@boubakersoltani4566]

This is just pure math, no real world implications at all. Basically, mathematicians said: what if we pack all numbers in one black box and name this box "omega"? One can object: but numbers are infinite, so how can one fit them all in a box? But we should not forget that this is just math, it is all only about being consistent with your first assumptions no matter whether they are from reality (like biology and physics) or just from mere mind. So, by using this box "omega" as a new unit, we can now count the boxes: first omega box, second omega box, ... millionth omega box and go on. Here again you can see that the series of boxes is endless, so why do not pack all these boxes in a new bigger box? this new box has "omega" number of smaller boxes, which by turn has "omega" numbers each. The new box is called then "Omega times Omega". And we continue counting this way, and getting more bigger embedding boxes. This recursive process is what we call "ordinal generation".

[@masonmireles9295]

@@boubakersoltani4566 Thank you so much for helping me to better understand this! It truly means a lot! I am currently pursuing a major in mathematics. Is there any chance I could email you to ask about this further please?

[@boubakersoltani4566]

@@masonmireles9295Hi Mason! I wanted to really thank you for your trust, but indeed I am not that expert in set theory. Maybe you won't gain much from me.

[@masonmireles9295]

@@boubakersoltani4566 Okay. Thank you anyways for your help! I truly appreciate it! Not a lot of people go out of their way to help others out anymore. You helping me truly made my day.

[@boubakersoltani4566]

@@masonmireles9295 glad to here that ^^ I wish you all success.

[@biggerthaninfinity7604]

5:13 w1 is the first uncountable ordinal. What is e0?

[@hypercosmologyexplanations5178]

e0 is omega to the power omega to the power omega...[repeat omega times]. e0 is much much smaller than w1

[@arkeusalexander9054]

@@hypercosmologyexplanations5178 You mean greater as it is an infinite tetration/Hyper"w" ?

[@MkhanyisiMadlavana]

4:45 weren't you supposed to say 'not necessarily bigger' instead of just 'not bigger' because it depends on the order type?

[@LarghettoCantabile]

I'm not sure how the continuum hypothesis might not be true. Represent [0, 1] by the set of subsets of N, each represented by an infinite bitmap, which can also be interpreted as a list of binary decimals. There will be collisions in the interpretation, with for instance .10+ being the same real number as .01+ (i.e. a zero followed by an unbounded list of 1's); but we will still have an onto function; and we can presumably build a bijection from it. Our number of bitmaps is 2 to the aleph null; so this is presumably the cardinality of [0, 1]. Now, define a bijective function between [0,1] and R, for example by considering a semicircle of size 1, with a tangent at midpoint, and mapping each semicircle point to a tangent point by drawing a radius. This proves [0, 1] and R have the same cardinality, which is presumably the cardinality of our list of decimal numbers, i.e. beth_1.

[@HPTopoG]

That doesn’t prove that the continuum hypothesis is true because it doesn’t say anything about the actual value of the continuum. What’s important to understand when dealing with CH is that we don’t know which ℵ number 2^ℵ0 actually evaluates to. It’s a bit like if somebody asked you to calculate 2×3 but you didn’t know what 3 meant. The reason we know CH is unprovable is that we can describe mathematical universes in which the continuum is exactly ℵ1 and other universes in which the continuum is provably larger than that. The former is true in Gödel’s constructible universe L and the latter is true in Cohen’s forcing extension. Both are somewhat difficult to understand.

[@lukostello]

how would this definition handle multiplying by fractions? How do you get fractions cardinally?

[@HPTopoG]

For honest fractions, you don’t. You can define cardinal division, but it’s relatively boring. For a pair of cardinals κ≤λ, you can define λ/κ as the maximum cardinal μ satisfying κ·μ≤λ. But since the product of cardinals is given by κ·μ=max(κ,μ), this just means that at least one of them is actually equal to λ. But then the result of the above division is just λ every time. It doesn’t work for honest fractions, i.e. λ/κ for λ

[@philosopherhobbs]

Great video. What about division?

[@BarnabyFWNightingale]

What is set A at 2:52?

[@arkeusalexander9054]

First time i encountered this set (learned maths in french). So i know : N, Z, D, Q, R, C

[@Xnoob545]

Algebraic numbers

[@BarnabyFWNightingale]

@@Xnoob545 Thanks!

[@christophersedlak1147]

thanks

[@Fircasice]

I do not understand why m^lambda = n^lambda = 2^lambda.

[@HPTopoG]

As an easier example, let m=3 and n=2. Then an element of m^λ is a ternary sequence f of length λ and an element of n^λ is a binary sequence g of length λ. Take any ternary f and reinterpret it in binary to get a binary g. Go from binary to ternary to get the other map direction. If you can show these are both injective or both surjective, then the Cantor-Bernstein theorem implies they have the same cardinality. Alternatively if n

[@epicchrist2941]

is there any chance I contact you through email?

[@christophersedlak1147]

thanks