Technically you define this exponential as a *limit* (namely "limsup" would work) to ensure that the result gives you zero, but you're right that the naïve definition with a supremum doesn't quite work!

@@SheafificationOfG Thank you. That's what I'd thought; I'd just never seen it addressed. Also, subscribed! (As of your Goodstein Sequence video, which I watched before this one.)

I hope this helps: omega is the smallest ordinal bigger than any finite ordinal, so omega - 1 must be finite *if it exists*. In other words: IF there is an ordinal alpha such that alpha + 1 = omega, then [by the argument you summarised] omega would have to satisfy omega < omega, which is nonsense. By contradiction, we reject the premise, but t he premise is *not* that omega exists; the premise is that *alpha* exists... and therefore alpha (i.e., "omega - 1") doesn't exist. The existence of omega itself is fine (it's the order type of the natural numbers).

​@@SheafificationOfG But at 6:20 next turn is 1 less then the previous one, shouldn't we be able to subtract 1 from w in order for this to work? How else did we get w³•3+w²•3+w•3+3 from w^w?

Right, we're not actually subtracting one from the ordinal shape directly! Rather, we're subtracting one from the corresponding finite number! Rather than subtracting 1 from w^w, we are subtracting 1 from something like 3^3, which gives 3^2*2 + 3*2 + 2 (and the shape of that is w^2*2 + w*2 + 2).

@@SheafificationOfG I think I get it, thank you

... if you can formalize your argument in Peano arithmetic (which it appears you're essentially trying to do), then it is wrong. (it's not a proof to look at the sequences and notice the pattern of decreasing constants. how do you know this pattern continues forever? for instance, in the sequence for 5, the third term has a sudden • 3. why couldn't this happen later in the sequence too?)

I want to build on @myca9322's comment. You have the right idea about the general approach (there is some sense in which the "shape" of a pure base-b expansion decreases as you progress along any Goodstein sequence). However, the devil's in the details: how do we make your argument precise? These shapes can be quite elaborate due to their highly recursive nature, so how do we accurately "measure" a shape's complexity (so that, in particular, we can say that successive terms of the Goodstein sequence are less and less complex)? As you say, the shape is really a kind of formal expression (where the base is replaced by some abstract symbol "b"), and a rough heuristic about complexity is "if the constants are smaller, the shape is less complex". However, we need to be more careful: the shape b^{b + 1} is "more complex" than b^{b*100} + b*1000, as you're probably already aware. But if we want to *prove* Goodstein's theorem, we need to spell out an actual way of comparing two of these "shapes" appropriately! Let's look at the shape "b" by itself (e.g., the shape of 5 in base 5). This shape is more complex than constants, so "b > 0", "b > 1", "b > 2", ... and so on. Inevitably, this makes "b" some kind of analogue of omega, even if you were trying to avoid infinite ordinals. Similarly, trying to explain the complexity of expressions like "b * 5" or "b^2" or "b^b" will inevitably lead to analogues of ordinal arithmetic in the same sense. Finally, there's one more question: why do these shapes that "decrease in complexity" *actually* run out? There are infinitely many shapes that are "less" complex than b^b (for example, b, b^2, b^3, b^4, b^5, ... are all less complex than b^b), so at least something needs to be said about this, too. As other comments remark, the finiteness of any decreasing sequence is automatic if we use ordinals; however, if you avoid ordinals, you would have to prove the inevitable finiteness of these decreasing-complexity sequences by structural induction on your shape! In any case, you'll find that formalising your proof may very well put you back in the position of the proof in this video. That's not to say that ordinals are the *only* way to do this (as you can use other recursive structures, such as tree-like structures), but a result of Kirby and Paris ensures that any proof is going to be as "strong" (in a proof-theoretic sense) as one using infinite ordinals.

@@myca9322 Did you actually read my comment or were you just yapping? the sudden 3 is because on the third iteration, it goes from 3^3 to 4^4 - 1, but we can't have negatives in pure base notation, so we need to break the 4^4 into 4^3 * 3 +..... This is *literally* the base being depleted like I was talking about. the 4^4 had to have its exponent turned into a 3(which will no longer grow) in order to replenish the +3 you talked about. eventually the +3 is depleted, and then the 7 * 3 must have its constant depleted when it becomes 8 * 3 - 1 and become 8 * 2 + 7 to replenish the standalone constant. formally, each -1 eats away at every term that doesn't include the base, until the base itself needs to be depleted.

@@SheafificationOfG okay. there are no negative digits in any ordinary base. any pure base representation has some term which corresponds to B^0 * N where N is strictly less than B. At 4:04 in the video, N is 1, 0, 3, 2, 1, 0, 7, 6, 5, 4, 3. What we'll notice that on each iteration, this value is unaffected by the B increases by 1 each time. This is because N is strictly less than B at all times. What we'll notice is that when N reaches 0 as in the 3rd and 7th iterations, on the very next iteration, it *would* become negative because this part of the value is unaffected by the increasing nature of our series. However, there is no such thing as a negative representation, so we must find the B^1*M term. this corresponds to 7 * 3 in iteration number 7. When we increase B and subtract 1, part of our equation must then be B^1 * M - 1, as M is strictly less than B. To reconcile, we must turn this into B^1 * (M-1) + B^0 * (B - 1). Notice that our coefficient associated with the B^0 term is strictly less than B and will no longer increase. Our B^1 coefficient will also not increase. On our next iteration, our B^0 coefficient has been frozen in place and will eventually be depleted because its a descending series of finite ordinals. Once that coefficient reaches 0, we run into the same problem where we must then do B^1 * (M-1) then becomes B^1 * (M-1) - 1, and to reconcile the fact that we cannot have a negative constant, we must break open the B^1 * (M-1) into B^1 * (M-2) + (B-1). This continues until the B^1 coefficient, denoted as M will also reach 0 because its a descending series of ordinals. Then we have to break open the B^2 thing. I hope that you see that this term's coefficient is also going to eventually reach 0 for the same reasons as listed above. Then, we must do B^3, and so on until we either run out or reach a special B^K term where K is greater than or equal to B. If we have no B^K term, it's pretty obvious through induction that our series will terminate just because all coefficients are decreasing and there are finitely many exponents that can be applied to B. Our inductive reasoning will eventually work its way through and deplete evry single exponent. If we have a B^K term, we know that K itself is some ordinal and even if its increasing, its only finite, and its composed of some assortment of Bs and constants strictly less than B as per our pure base requirement. The parts strictly less than B will decrease once all terms besides the B^K term are depleted in order to replenish them. Once all the constants strictly less than B are also depleted, the parts that include B must be depleted themselves. This will continue until K is less than B, and then all our previous induction will apply. This explanation is very similar to your infinite ordinal explanation but in my opinion, all this talk about shape and complexity ends up obfuscating away from the point to the point that its not even clear what you're talking about. I'm not saying that you're wrong, just that your explanation is very confusing and unnecessarily wordy whereas I think that the one I've presented here is something that could be explained in only a minute or two given visuals. Basically, our goodstein increasing of the base only affects the total quantity, but not the number of terms. The number of terms must eventually reach 0 because the -1 affects constants which are not affected by the goodstein increasing, and each time that these constants are depleted, the number of terms must come closer to decreasing, until it does, and it eats away the entire value.

@ethos8863 while your explanation works for a specific example, the issue is when trying to handle the theorem in full generality. The intuition is right, like I said, but the ordinals (or some equally powerful substitute) is necessary to explain why your intuitive process works (or even makes sense). Like was mentioned earlier, Goodstein's theorem is undecidable in first order Peano arithmetic, so an argument purely in terms of natural numbers is, in some weak sense, impossible. I would recommend consulting the paper by Kirby and Paris (or starting a discussion on this somewhere like MathStackExchange) where you might be able to get more insight about the matter. I won't say my explanation is the best or cleanest, but delving into ordinals is necessary for a complete argument. Anyway, I hope this is somewhat helpful, and I'm glad you've engaged critically with the content, and I hope you can find a satisfactory explanation elsewhere if not here 🙂

You're right (as was mentioned by someone else, too), but since the result is pretty counterintuitive the first time you see it, I figured it wouldn't hurt to "demonstrate" it instead of just going with a "trust me; it's by definition!" In hindsight, maybe a more direct argument from well-foundedness would've been more appropriate, but what's done is done!

This is a nonsense comment. All mathematical theorems are true by definition. He was illustrating how the basic definition that omega is the smallest ordinal greater than all finite ordinals implies that all decreasing sequences of ordinals must terminate.

@@ethos8863 No, only the axioms are true by definition. Also he was not just talking about omega, he was talking about all ordinals.

@@MichaelDarrow-tr1mn the theorems use the definition of the axioms to be true by definition themselves. You prove a theorem by taking the definition of the axiom and showing that the axiom being true requires the theorem to also be true. So by the definition of the axiom, the theorem is also true, hence the "true by definition". Also, ordinals being well ordered is required because they are well-ordered which how all nonempty subsets have a smallest element. what this means is that its required that as you start at any element, there is some subset of all elments less than that element, and this subset must terminate somewhere because it is a nonempty subset of the main set. this *requires* that given the set of infinite ordinals, there must be some smallest element which is omega. By the definition of there being infinite ordinals, it is required that all descending series of ordinals terminate somewhere by the reasoning given in the video.

Damn, that would really upset (in a happy way?) the 10 set theorists that are still alive

I would argue that my last video did that 😤😤😤 (do you have an infinite problem in mind?)

@@SheafificationOfG Idk (It's a good meme though). I guess something like proving induction is sound for natural numbers as we know them, or a similar philosophical meme along those lines

Guess that depends on if the problem is provable to begin with 🙃

Absolutely (I even defined them to be well-ordered in my last video)! I figured since it's such a counterintuitive consequence, it was worth "reproving" with examples. But yeah, the argument isn't terribly *well-founded* 😏

I came across Naviary's video on the longest infinite chess game. It was a real pleasure to watch (and I suck at chess).

@@SheafificationOfG Thanks for the rec! Love that video, it's actually the one that got me into infinite ordinals and how I came across your videos when searching for them!

Not sure if my comment got eaten, but I recommend the "Essence of Set Theory" series

Statements made by the utterly deranged. A certain meme I saw a while ago

Nah, calculus is "doing finite things and then taking limits." 🙃

you mean CS channel?

He already gave an example of what would happen in that case. He first started off by giving an example of a_0 being a finite number and then a_1 being 1 less than it. Obviously, this would result in a finite sequence. What's more interesting is if a_0 had been an infinite ordinal. We already know that any descending sequence of finite ordinals is going to be finite.

just because the sequence can be arbitrarily large doesnt mean that it can be infinite.

Each term in the sequence just has to be less than the previous term, it doesn't have to be exactly 1 less

> i actually kinda like how we just ignore a search for truth and just use overpowered paradoxical machinery to prove normal theorems ahh but you see mathematics isn't about the search for truth. Its about the discovery and use of semantic constructions of objects and documenting the properties those objects have in relation to each other. Truth is but a particular type of pattern in certain contexts of semantic objects! (i think)

you should not be using that tone for someone barely acquainted with the construction of basic number sets my guy

i wonder if i will ever see it :(

i don't get it lol

@@TheArtOfBeingANerd you gotta -sell your soul- to a rhythm game to understand that reference

If you write an ordinal out as a finite decimal expansion, then yes; the sum of its digits will indeed be finite.

lmaoo thanks fam

Ordinals are only concerned with counting, and we can only really count by (repeated) 1's.

Actually, that's not an error: 10^(xi_1) is less than or equal to alpha, but since rho is strictly less than 10^(xi_1), this means that rho is also strictly less than alpha.

Thanks fam!

I expected a p-adics comment, but I didn't expect it from a "computer guy" :^) Isn't it enough that you have Z/2^{64} with a sign convention? 😉

I'm not sure if what I'm describing is exactly the p-adics. It's certainly closely related in its construction, but the p-adics seem to have some weird notion of absolute value attached to them with zero having an infinite absolute value. If you represent your numbers the p-adic way but treat them as having the normal absolute value, is it still the p-adic's? Does it work to do that?

@@SheafificationOfG Also, "isn't it enough?" is what was said at various times about Z/2^20 and Z/2^32. It wasn't enough. 😂

​@@JonBrase If you're only interested in the p-adic integers (which, of course, contains all integers), then you can construct it purely algebraically with out mentioning the p-adic norm. (Also, just to clarify, the p-adic *valuation* of zero is infinity, but the p-adic *absolute value* of zero is just zero!) If you're only interested in *ordinary* integers, they embed into the p-adic integers, after which you can use whatever absolute value you're comfortable with!

@@SheafificationOfG Now we get to the evil part: what I said about 2s complement and 1s complement being equivalent in the limit of an infinite number of bits isn't actually quite true if we only have bits to the left of the radix point as in the strict 2-adics. In 1s complement you negate a number by flipping all the bits, while in 2s complement you negate it by flipping all the bits and adding one to the least significant bit, so for them to be equivalent the least significant bit needs to be infinitely far to the right. If we have an infinite number of bits on either side of the radix point, we get: ...1111.0000... = -1 Flip every bit: ...0000.1111... = 1 And add 1 to the least significant bit: ...0000.1111... + ...0.000...1 = ...0001.0000... Just as in 1s complement, we have a positive zero that's all zeroes and a "negative zero" that's all ones: ...0000.0000... = 0 Flip every bit: ...1111.1111... = 0 Add one to the LSB: ....1111.1111.... + 0.000...1 = ...0000.0000... = 0

omega-1 is actually not well-defined (because omega is not the successor of anything). The reason is actually exactly because if omega-1 were to exist, it would have to be finite.

omega + 1 is indeed not divisible by anything (besides 1 and itself, of course). There is a concept of "prime ordinals" (and a canonical factorisation of all ordinals into primes); see en.wikipedia.org/wiki/Ordinal_arithmetic#Factorization_into_primes You might notice, though, that omega is mentioned as prime here, which might seem strange giving that omega = n * omega for any finite n. This is because if omega divides a product a*b, then omega must divide a or b (which is the more proper notion of "prime").

@@SheafificationOfG i see, thank you

Your proof works, but I think a simpler argument is: 1. D is downward-closed (if y is in D, and x < y, then x is in D), which falls out of the definition. 2. xi < sup D (because xi + 1 = sup D by assumption), so xi must be less than something in D, and therefore must fall in D by step 1.

Oh, I see. Step2 is the contraposition of the definition of supremum.