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i love ordinals so much, and due to my inability to explain ordinals well to other people, i love when a good video about ordinals drops. thank you for applying the successor function to the number of awesome videos on youtube. in fact, maybe i would even say that for every video x that was on youtube before this one, you applied the successor function to |{y in the set of youtube videos : y is better than x}|. and you definitely turned my happiness into an ordinal (which is a significant improvement because negative ordinals don't exist) :)))))
An alternative proof: if your ordinal number α cannot be represented as a finite sum of powers of 10, then there would be an infinte, strictly descending sequence of ordinals below α, which contradicts the fact that ordinals are well ordered by definition.
This proof doesn't work because the assumption that a strictly descending sequence of ordinals implies they are not well ordered is wrong, for example it's easy to create a sequence of real numbers that's strictly descending, but with the axiom of choice it's possible to prove that any set, and in particular the reals, has a well order: en.wikipedia.org/wiki/Well-order#Reals
@@Midori179 You cannot use that argument, because the standard ordering or reals is not a well-order. Well-orders cannot have an infinitely descending sequence, because a set of elements of such a sequence does not have the least element (which is required by the definition of a well-order). Any well-order on reals will look nothing like the standard order.
Those are different orderings of the real numbers. Under choice, the real numbers are only "well-orderable," not necessarily well-ordered. The argument of descending ordinals works so long as you assume *some* decimal expansion exists for any ordinal, but you'd still have to prove this claim.
@zanderhill9842: I agree, my proof sketch has some gaps. I need to show that there is such a representation first, and I also need to show that the partial sums would be descending, which are both nontrivial claims.
One thing I've never seen addressed is that the standard definition of ordinal exponentiation is that 0 raised to a limit ordinal equals 1. For example, 0 to the omega is sup{0⁰,0¹,0²,0³,...} = sup{1,0,0,0,...} = 1.
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.)
Great video. Quick question, do we need something extra in your definition to guarantee some limit ordinals exist? Given 0 exists, the 'there's always a bigger ordinal' generates all the finite numbers, but don't we need the existence of at least one infinite ordinal to be able to conclude that omega exists by the second (well ordering) property? (And likewise for larger limit ordinals).
Yeah, I tried to keep it simple or else I'd be opening a can of worms, but you're right. The "rules" I provided are satisfied by natural numbers. A quick save would be to say that there is always a bigger ordinal than any *set* of existing ordinals. ... Well, now that I think about it, was that all I needed to do?
@@SheafificationOfG Definitely not my area of expertise, but does that lead to problems with being able to create a set of all ordinals? It would seem to at least imply that the collection of all ordinals are too big to be a set (does this automatically happen with the equivalence class of well ordered sets definition?)
The collection of ordinals is indeed too large to be a set! The reason is basically a well-ordered version of Russell's paradox: if all ordinals formed a set, then that set would be well-ordered (by my second rule), and therefore correspond to another ordinal (by the actual precise definition). If you believe the axiom of choice, it also would follow from the fact that the collection of all sets is too large to form a set.
@@SheafificationOfG Interesting. I still wonder if having supremums of arbitrary sets introduces ordinals that wouldn't otherwise exist. I guess that's the same as asking if there is an unbounded set of ordinals. Obviously there can't be if we use Von Neumann ordinals, just take the union, but if we reject choice, is every well ordered set still equivalent to a Von Neumann ordinal? Anyway, fun rabbit hole, thanks again for the video!
Every well-ordered set is indeed order-isomorphic to a unique (von-Neumann) ordinal! The idea is to construct the isomorphism by well-founded induction on your well-ordered set, mapping successive elements to successive ordinals.
I watched this because it's a suggested prerequisite to a video which is a prerequisite to a video that I want to watch. The first time around I felt like I didn't get it. Then halfway through a rewatch I realised I was completely skipping over the "(nonzero)" in "infinitely many (nonzero) digits". Now I'm pretty sure I get it - you can write an infinite ordinal by writing the base-ω expansion of the infinite part of the ordinal, followed by an infinite number of zeroes, followed by the decimal expansion of the finite part of the ordinal. That's simplifying, obviously there might be more strings separated by infinitely many zeroes. Let me make sure I have something else straight - if we had an ordinal like, say (ω+ω)•25+ω•17+309, we would write 25, then an infinite number of zeroes, then 17, then an infinite number of zeroes, then 309. 25000...00017000...000309, if you like. Isn't there some way to repeat this process in turn until we have ω finite strings of digits separated by ω infinite strings of zeroes? I know I'm just missing some intuition because I've known this concept for like half an hour, tops, so what am I missing?
Seems like you've got the right idea! [Edited from earlier] Numbers past omega^omega would have at least omega many chunks of omega-length segments of digits (but only finitely many of these will have nonzero digits). A caveat of this decimal expansion (used in the video for instructional purposes) is that they get a lot less informative once ordinals get *really big*, but the same is true for Cantor normal form.
@@SheafificationOfG thanks for the reply! I guess the bit of my question that I forgot to ask was: doesn't the fact that ordinals larger than ω^ω would have an infinite number of substrings of nonzero digits contradict that any original can be written with finitely many nonzero digits?
@Tinybabyfishy shoot sorry! While you can definitely write down numbers with infinitely many chunks of finitely many digits, it will always be equal to a different representation with finitely many zeroes (an infinite sequence of finitely many chunks of nonzero digits, each separated by an infinite sequence of zeroes is likely going to be equal to omega^omega). I edited my last comment. When things get big, it's better to use a more rigorous mathematical representation for the numbers you're referring to (i.e. write it out as a possibly infinite sum of digits, indexed by ordinals), and it'd be easier to compare them with Cantor Normal Form. The intuitive decimal expansion I used in the intro is awful in practice, and was just there as a hook/motivation.
@@SheafificationOfG word! Well it definitely served its purpose as a hook. I studied linguistics and took a couple of math electives, but I'm definitely approaching this from an undergraduate perspective, at best, and I'm totally on board for the ride and thinking more about sets than I likely ever have. (Even the undergrad stuff I did was like intro to vector calc and a number theory/cryptography course, I haven't touched sets since high school)
I'm a little confused about how ordinal arithmetic is defined for limit ordinals. If a + omega = sup (a + b | b < omega), where a is a finite ordinal, then wouldn't a + omega just be the same as omega? The set { a + b | b < omega} would be {a, a+1, a+2, a+3, ...} which is a proper subset of {1, 2, 3, 4 ....}, so if omega is defined as the supremum of {1, 2, 3, 4 ....} then how can omega + a be the supremum of a set contained within {1, 2, 3 ,4 ....} ? Don't they have to be the same?
Yes it would be like that, that's how ordinals work. Addition and multiplication of ordinals are not commutative. For a finite a, ω+a = {0, 1, 2, ..., ω, ω+1, ..., ω+(a-1) }, which is obviously not a subset of ω.
I am new to your channel and i like the style of video You know what u are doing, and the gen z style math memes keep it interesting And ofc i like the rigour you provide, after oversimplification of something
The last proof only works because of well-ordering of ordinals, which prohibits infinite descending sequences. Without that, this kind of induction can loop back on itself to infinite depth, and while it would still prove the existence of an expansion, it would not prove its finiteness.
I like how it wasn't "you can't just take the remainder because it's always 0", instead it's "you can't just take the remainder because then the math is a headache" lol
It doesn't make sense, w to be divisible by every finite number. let's assume w is divisible by 2, then exists a number x such that w = 2*x, so x < w, but because w is the smallest number bigger than all finite numbers, that means x must be a finite number, but there is no finite number that doubled we get w. For that reason w is not divisible by 2, 3, 5, etc. Which makes it a prime number? Because of this, I do not agree that 1...000 = ...999. EDIT: My assumption of: if a>1, then forall b. b < a*b is wrong.
"exists a number x such that w = 2*x, so x < w" is a false statement even in naturals, as 0 = 2×0. What you missed is that multiplication of ordinals is not commutative: 2×ω equals ω, ω×2 does not. Same with addition: 1+ω = ω.
@@vytah when I wrote 2*x I meant x+x (which is written as x*2, my bad, and thanks for the warning :) ), with that change the reasoning holds x < x+x = x*2 = w
@@arturjorge5607 But the definition of divisibility used in this video puts the divisor on the left. So to check if ω is divisible by 2, you should find x that satisfies ω = 2×x, not ω = x×2. And similarly, the division splits the number into d×q+r (d=divisor, q=quotient, r=remainder) in this precise order. So ω/2 = ω remainder 0, because 2×ω+0 = ω. A lot of standard ring algebra stuff that you're used to doesn't work with ordinals, for example distributivity doesn't works from the left: (1+1)×ω = 2×ω = ω, but 1×ω+1×ω = ω+ω = ω×2
12121212.......121212....121212 can be written like: V = 12 + 100 V It's better to completely get rid of constructs like "+ ...", and use recursion, so that you can trivially solve for their values. (1 - 100)V = 12 V = -12/99. V = 12 + 100(12 + 100 V) = ...
Being a computer guy, I prefer taking the limit of n's complement in base n as the number of digits goes to infinity (for n=2, 2's complement and ones' complement become equivalent in this case, which is why it's interesting to a computer guy). In other words, we can't define that ...9999 = some infinite ordinal because I need ...9999 (or equivalently ...1111 in binary) for -1!
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?
@@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
This proof seemed initially unintuitive as it's an immediate reflex to consider numbers like 0(...999) as clearly not being representable with finitely many digits, but it falls right out of the definition of limit ordinals that any such expansion is equal to omega, being larger than every finite ordinal. The fact that you can change the digits and they need not be 9 is also unintuitive without this understanding. The hook of the introduction did its job! The induction is also fairly obvious once all of the definitions are nailed down. I wish I had been introduced to ordinals in my discrete mathematics or number theory classes, I'm not sure we have a lower division class specifically for set theory.
I'm also pretty sure that with a bit of effort this can cause the decimal expansion of the surreals to fall out, where it is possible for non-ordinal values like omega minus 1 to show up.
Does this work for ordinals larger than 𝜀_0? 𝜀_0 has the property to be smallest ordinal that satisfies 𝜀_0 = 𝜔^(𝜀_0). The Wikipedia article for Cantor normal form mentions this. Also, I have hard time thinking a union of continuum number of continuums could be represented as a single decimal expansion, so there should be more ordinals packed there for this to work?
Despite the weirdness of epsilon_0, it does have a decimal expansion! Perhaps unsurprisingly, the decimal expansion of epsilon_0 is 10^(epsilon_0). When you deal with really big ordinals, this is going to happen a lot. Cantor normal form (base omega expansion) also exists for any ordinal. However, it's only particularly *useful* if the ordinal is less than epsilon_0, because in this case, the exponents of Cantor normal form will be strictly smaller than the original ordinal. In general, you won't have that guarantee. The Cantor normal form for epsilon_0 is just omega^(epsilon_0), after all!
It does work, but it's a bit of a cop-out. omega^epsilon_0 is still in Cantor normal form if we allow each exponent to be _at most_ (rather than strictly less than) the ordinal being represented.
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.
@@SheafificationOfG I have no idea, that's the point. It seems there is no way to, but there is clearly a way to define the number itself, for example as a limit of the sum from n=1 to n=N of d_n*10^(N-n) as N tends to infinity, with d_n being the n-th digit of pi. What am I missing?
@@SpaghettiToaster That's definitely one way to try and formalise your value, and if you do it that way, the limit you get will be omega (as the limit will be the smallest ordinal larger than each of the terms in your limit's sequence), which is equal to the decimal expansion 1[...]0000. The point is that even if there is *some* expansion where infinitely many digits are nonzero, the expression is inevitably equal to another decimal expansion where only *finitely* many of them are nonzero.
@@SheafificationOfG Oh, I understand, thanks! But that means that an infinitely large circle constructed in this manner would have a radius-to-circumference ratio of 1, doesn't it? Shouldn't it instead be some indeterminate form that you can somehow beat into converging to pi?
@@SpaghettiToaster I think it's important to emphasize that ordinals are only concerned with ordering (rather than "length" or "magnitude" or "size"), so having a circle of radius "omega" doesn't necessarily make sense.
5:33 - I can see how direct limits are filtered colimits, but is the converse necessarily true? That is, given a filtered diagram D, is there always a directed subdiagram D' such that colim D = colim D'?
At what point does the series for a = 0 to k, 9*10^-a == 1, given that for any natural value of k, the series is less than 1 and that the derivative is always a nonzero positive?
Intersting, just dont get the point @9:17 How is it possible, can you give an example to find the digits of Omega? - Is there only one Omega or many differents?- I don't mean w+1,w+2..etc oder do you mean that with different "w"(Omegas) here? When the remainder is always "w"(Omega) and Omega has all Numbers as possible divisors that give back "w"(Omega) then the digits would be all "w"(Omega) right - or what? Best regards
Look closer at the algorithm we use to generate digits. The digit d_n is the remainder, not the quotient of each operation. The remainder when dividing omega by 10 happens to be 0. We get an infinite string of zeroes, but the operation can't ever terminate or produce a limit, so it doesn't give us a well defined output. When operating the other way, we get the same infinite string of zeroes but there's an extra 1 at the omega'th place.
omega does refer to a specific ordinal (namely, the smallest infinite ordinal). Counterintuitively, omega = n * omega for any finite n (and, notably, n * omega is *not* the same as omega * n), so omega is (left) divisible by every finite number. Nonetheless, following the proof, the decimal expansion of omega is 1...0000. This is because when you find the least ordinal xi such that 10^xi
@@pudy2487 @SheafificationOfG So its pretty much the same to divide an infinite number by a number (maybe only a finite one) repetively that would give you endless strings? Like divide 1000...000... (a 1 with infinite Zeros) by 10 repetively and it would also give you a 1 with infinite infinite zeros - or divide it by 5 repetively to give you a 2 with infinite zeros etc. Also it kinda sounds like the operations with the infinite symbol ♾ Like x times (♾) infinte = (♾) infinite. What is the difference to Omega compared to the (♾) infinity symbol?
@@Lokalgott If you divide omega by 5, you just get omega again, so some of your intuition about division from the land of the finite doesn't transfer to the infinite setting.
Actually, the proof shows that the expansion is necessarily finite, too! Granted, proofs by (strong) induction obfuscate this, but I only ever work with finite sums.
It follows from the (strong) induction hypothesis: we assumed that any ordinal smaller than alpha had a finite decimal expansion, and so in particular, the remainder rho after dividing by 10^(xi_1), being strictly smaller than alpha, had a finite decimal expansion. Say that expansion had length N, then adding 10^(xi_1)d_1 to the sum gives a length-(N+1) decimal expansion of alpha.
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").
Hypernaturals, brought to you by the guy who also mentioned nonstandard analysis in a previous video ;) I don't actually know much about hypernatural numbers. Do they have a "finite digit representation" analogue, or is the H-many 1's a counterexample?
@@SheafificationOfG😂feelin like a youth pastor from that comment about advocacy. The point of the comment is a counterexample, since it’s a related concept of infinite number. Like if ordinals had square roots and stuff (all of arithmetic by transfer principle). I rec looking up keislers infinite microscope and telescope for a good viz, or the numberphile The Infinitesimal Monad with Carol Wood
@@mzg147no, but as a refinement of that, they ARE internally well ordered, which covers a lot actually (all standard formulae for one and allows “internal induction“ which is like a more user-friendly transfinite induction. Same for completeness of hyperreals. Incomplete but internally complete.
It seems to me that placing a number in the infinite place is meaningless. It is defining into existence something that has no meaning or actual value since infinity with an extra digit is still just infinity. Put an infinite to the infinite number of digits in front of infinity and it's still just infinity. It also seems that omega is a meaningless value as if you actually try to consider what it would equal, you would quickly find that there is no first number that is greater than all finite ordinals. It is either a finite value, which contradicts its own definition or it's an infinite value which makes all further addition to it meaningless since they're all just infinity. It's like little kids yelling "I have infinite power!" and another yelling "I have infinite times infinite power." Omega = Infinity. Omega + Omega = Infinity.
Yes, if you select the first number bigger than all numbers you are effectively selecting the empty set. Everything else is nonsense stemming from applying this empty object to random situations, which gives different broken results depending on the direction from which you approach the task, leading to all the paradoxes such as sums of infinite positive series giving -1/12.
Ordinals are used to classify well-ordered sets. A set corresponding to omega has a different structure to a set corresponding to omega×2, even though they have the same size. In other words, you can create a bijection between them, but that bijection will never preserve the ordering. An example pair of such sets are N and {0,1}×N.
But I guess what is more important is that you're not aware how overloaded the term "infinity" is in maths. Here are some meanings I can name out of top of my head: - infinity as a lack of bounds in a particular direction on the real number line, it has two variants: -∞ and +∞ --- it's also used in contexts of integers, as they are a subset of reals - infinity as a symbol of divergence of a sequence, -∞ if the sequence is always decreasing, +∞ is it's always increasing --- note that there are other non-numbers used to describe sequence limit behaviour, like for example 0⁺ meaning "the limit is 0 and it's approached from the right" - infinity as a symbol of divergence of a limit of a function - the complex equivalents of all those above: a lack of bounds, a limit of a sequence, or a limit of a function on a complex plane - infinity as a symbol representing distance between two elements that cannot reach reach other - infinity as an element of a projective plane or a wheel; in this context, 1/0=∞ is not a sloppy shorthand for a limit lim x→0 1/x, but an actual algebraic expression - infinity as a size (cardinality) of a set - cardinality of sets is measured using cardinal numbers, and there are infinitely many infinite cardinal numbers - infinity as a description of an order of an infinite well-ordered set - you use ordinal numbers, which are the subject of this video: --- two sets with the same ordinal number have a bijection between them that preserves ordering, two sets with different ordinal numbers cannot have such a bijection --- ordinal numbers have some nice properties they share with natural numbers, like the ability to do induction --- there's a connection between ordinal numbers and cardinal numbers, as all well-ordered sets with a given ordinal number have the same cardinality, so every ordinal number can be mapped to some cardinal number (obviously not vice versa though) --- infinite cardinal numbers are usually indexed with the ordinal numbers, starting from ℵ₀ - infinity as any surreal number larger than any finite number - surreal numbers contain all ordinal numbers, so that's obvious There are more, but I think that's enough for starters.
I thought we agreed everyone is actually (if secretly) an ultrafinitist, as is right and proper. :P I enjoy your videos a lot. Keep up the good work. (At whatever finite pace you wish to set for yourself and your fan.)
They're finite, I swear! Just count the digits! It would hardly be a secret if I only ever spoke about small finite numbers like a good and proper pure mathematician.
8:00 So you're saying with Brilliant, even girls can learn math? 😝 Seriously, though, as a lifelong handyman/DIYer who dabbles in math by way of youtube, I know what it's like from both their points of view. Kurei's tool-bag is almost as cool as the chill teaching attitude she inherited, and I laughed _inordinately_ hard at Serufu's unexpected yet inevitable display of power-drill finesse. Yours is the first _DIY!!_ meme I've seen, and it's perfect. Subbed.
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.
Where did this guy learn math? He is basiclally saynig that one infinite number is equal to another infinite number, but infinity cannot compare with infinity of the same sign. Yes, +inf is bigger than -inf, but +inf and +inf do not cmopare, you cannot say that any of them is equal to or greaer than the other. Hence his argument falls apart completely. And no, infinity does not have finite number of non-zero digits unless you talk about a finite string of digits followed by an infintie number of 0's in which case it is infinity and my comment above applies. 1 followed by infinitely many zeroes is neither larger nor smaller than 2 followed by infintely many zeroes, they are both infinity and so cannot be compared. Definitely not subscribing.
I think you got some things mixed up there. This video talks about ordinal numbers and not about the concept of infinity as an extension/direction of the real number line. There is a long answer on a comment from thelordz33 that lists some different meanings of the word "infinity"
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Dang, brilliant really do be scanning _all_ of the YT edusphere to sponsor anyone they can shoot an email at xD
Gotta say, they have good taste ;)
My man's already made it to the big leagues!
I did not understand anything in this video. Subscribed.
Absolutely cursed premise, Amazing video
i love ordinals so much, and due to my inability to explain ordinals well to other people, i love when a good video about ordinals drops. thank you for applying the successor function to the number of awesome videos on youtube. in fact, maybe i would even say that for every video x that was on youtube before this one, you applied the successor function to |{y in the set of youtube videos : y is better than x}|. and you definitely turned my happiness into an ordinal (which is a significant improvement because negative ordinals don't exist) :)))))
omg it's you yto
@@metachirality omg hi!!! :3
An alternative proof: if your ordinal number α cannot be represented as a finite sum of powers of 10, then there would be an infinte, strictly descending sequence of ordinals below α, which contradicts the fact that ordinals are well ordered by definition.
This proof doesn't work because the assumption that a strictly descending sequence of ordinals implies they are not well ordered is wrong, for example it's easy to create a sequence of real numbers that's strictly descending, but with the axiom of choice it's possible to prove that any set, and in particular the reals, has a well order: en.wikipedia.org/wiki/Well-order#Reals
@@Midori179 You cannot use that argument, because the standard ordering or reals is not a well-order. Well-orders cannot have an infinitely descending sequence, because a set of elements of such a sequence does not have the least element (which is required by the definition of a well-order). Any well-order on reals will look nothing like the standard order.
Those are different orderings of the real numbers. Under choice, the real numbers are only "well-orderable," not necessarily well-ordered. The argument of descending ordinals works so long as you assume *some* decimal expansion exists for any ordinal, but you'd still have to prove this claim.
@@Midori179 the first sentence of the page you linked says "the standard ordering of the reals is not a well ordering"
@zanderhill9842: I agree, my proof sketch has some gaps. I need to show that there is such a representation first, and I also need to show that the partial sums would be descending, which are both nontrivial claims.
what are mathematicians smoking
Only crystal math
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One thing I've never seen addressed is that the standard definition of ordinal exponentiation is that 0 raised to a limit ordinal equals 1. For example, 0 to the omega is sup{0⁰,0¹,0²,0³,...} = sup{1,0,0,0,...} = 1.
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.)
he has the same property as i do, but minimally. so simple yet so powerful. definitely to be feared.
insanely cool oh my god
wtf is this THE azali ??
@@LostExcaliburyea unexpected
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definitely going to have to rewatch this one a few times
wake up babe, new big g just dropped
When I see a comment like this I know to sub
Great video. Quick question, do we need something extra in your definition to guarantee some limit ordinals exist? Given 0 exists, the 'there's always a bigger ordinal' generates all the finite numbers, but don't we need the existence of at least one infinite ordinal to be able to conclude that omega exists by the second (well ordering) property? (And likewise for larger limit ordinals).
Yeah, I tried to keep it simple or else I'd be opening a can of worms, but you're right. The "rules" I provided are satisfied by natural numbers. A quick save would be to say that there is always a bigger ordinal than any *set* of existing ordinals.
... Well, now that I think about it, was that all I needed to do?
@@SheafificationOfG Definitely not my area of expertise, but does that lead to problems with being able to create a set of all ordinals? It would seem to at least imply that the collection of all ordinals are too big to be a set (does this automatically happen with the equivalence class of well ordered sets definition?)
The collection of ordinals is indeed too large to be a set! The reason is basically a well-ordered version of Russell's paradox: if all ordinals formed a set, then that set would be well-ordered (by my second rule), and therefore correspond to another ordinal (by the actual precise definition).
If you believe the axiom of choice, it also would follow from the fact that the collection of all sets is too large to form a set.
@@SheafificationOfG Interesting. I still wonder if having supremums of arbitrary sets introduces ordinals that wouldn't otherwise exist. I guess that's the same as asking if there is an unbounded set of ordinals. Obviously there can't be if we use Von Neumann ordinals, just take the union, but if we reject choice, is every well ordered set still equivalent to a Von Neumann ordinal? Anyway, fun rabbit hole, thanks again for the video!
Every well-ordered set is indeed order-isomorphic to a unique (von-Neumann) ordinal! The idea is to construct the isomorphism by well-founded induction on your well-ordered set, mapping successive elements to successive ordinals.
I really like the way the ordinals and cardinals both extend N to infinity, but are pretty wildly different
I watched this because it's a suggested prerequisite to a video which is a prerequisite to a video that I want to watch. The first time around I felt like I didn't get it. Then halfway through a rewatch I realised I was completely skipping over the "(nonzero)" in "infinitely many (nonzero) digits". Now I'm pretty sure I get it - you can write an infinite ordinal by writing the base-ω expansion of the infinite part of the ordinal, followed by an infinite number of zeroes, followed by the decimal expansion of the finite part of the ordinal. That's simplifying, obviously there might be more strings separated by infinitely many zeroes.
Let me make sure I have something else straight - if we had an ordinal like, say (ω+ω)•25+ω•17+309, we would write 25, then an infinite number of zeroes, then 17, then an infinite number of zeroes, then 309. 25000...00017000...000309, if you like. Isn't there some way to repeat this process in turn until we have ω finite strings of digits separated by ω infinite strings of zeroes? I know I'm just missing some intuition because I've known this concept for like half an hour, tops, so what am I missing?
Seems like you've got the right idea!
[Edited from earlier] Numbers past omega^omega would have at least omega many chunks of omega-length segments of digits (but only finitely many of these will have nonzero digits).
A caveat of this decimal expansion (used in the video for instructional purposes) is that they get a lot less informative once ordinals get *really big*, but the same is true for Cantor normal form.
@@SheafificationOfG thanks for the reply! I guess the bit of my question that I forgot to ask was: doesn't the fact that ordinals larger than ω^ω would have an infinite number of substrings of nonzero digits contradict that any original can be written with finitely many nonzero digits?
@Tinybabyfishy shoot sorry! While you can definitely write down numbers with infinitely many chunks of finitely many digits, it will always be equal to a different representation with finitely many zeroes (an infinite sequence of finitely many chunks of nonzero digits, each separated by an infinite sequence of zeroes is likely going to be equal to omega^omega). I edited my last comment.
When things get big, it's better to use a more rigorous mathematical representation for the numbers you're referring to (i.e. write it out as a possibly infinite sum of digits, indexed by ordinals), and it'd be easier to compare them with Cantor Normal Form. The intuitive decimal expansion I used in the intro is awful in practice, and was just there as a hook/motivation.
@@SheafificationOfG word! Well it definitely served its purpose as a hook. I studied linguistics and took a couple of math electives, but I'm definitely approaching this from an undergraduate perspective, at best, and I'm totally on board for the ride and thinking more about sets than I likely ever have. (Even the undergrad stuff I did was like intro to vector calc and a number theory/cryptography course, I haven't touched sets since high school)
merci d'avoir posté
What’s the source (I’m guessing a book) you used as basis for the more technical definitions?
I mostly worked off memory (horrible idea), but Jech's set theory book is a great place to find things like this done carefully.
@@SheafificationOfGthanks
For technical detail of ordinal i recommend you to read one of the trinity Ordinal book,the Holy W.sierpinski Ordinal and Cardinal
I'm a little confused about how ordinal arithmetic is defined for limit ordinals. If a + omega = sup (a + b | b < omega), where a is a finite ordinal, then wouldn't a + omega just be the same as omega? The set { a + b | b < omega} would be {a, a+1, a+2, a+3, ...} which is a proper subset of {1, 2, 3, 4 ....}, so if omega is defined as the supremum of {1, 2, 3, 4 ....} then how can omega + a be the supremum of a set contained within {1, 2, 3 ,4 ....} ? Don't they have to be the same?
Yes it would be like that, that's how ordinals work. Addition and multiplication of ordinals are not commutative. For a finite a, ω+a = {0, 1, 2, ..., ω, ω+1, ..., ω+(a-1) }, which is obviously not a subset of ω.
@@vytah oh that's very interesting
@@vytah So to reiterate, a+ω = ω < ω+a?
@@farklegriffen2624 Yes, precisely, for any finite a.
New favorite channel. Can’t wait till you get big
Thanks fam!
Great video, love your channel.
pi * omega
proud of you getting that sponsorship. great videos!
The goat is back with another banger 🚨
I am new to your channel and i like the style of video
You know what u are doing, and the gen z style math memes keep it interesting
And ofc i like the rigour you provide, after oversimplification of something
Do you recommend any specific math materials? I want to practice this to better understand it.
The last proof only works because of well-ordering of ordinals, which prohibits infinite descending sequences. Without that, this kind of induction can loop back on itself to infinite depth, and while it would still prove the existence of an expansion, it would not prove its finiteness.
I like how it wasn't "you can't just take the remainder because it's always 0", instead it's "you can't just take the remainder because then the math is a headache" lol
It doesn't make sense, w to be divisible by every finite number. let's assume w is divisible by 2, then exists a number x such that w = 2*x, so x < w, but because w is the smallest number bigger than all finite numbers, that means x must be a finite number, but there is no finite number that doubled we get w. For that reason w is not divisible by 2, 3, 5, etc. Which makes it a prime number?
Because of this, I do not agree that 1...000 = ...999.
EDIT:
My assumption of:
if a>1, then forall b. b < a*b
is wrong.
.
"exists a number x such that w = 2*x, so x < w" is a false statement even in naturals, as 0 = 2×0. What you missed is that multiplication of ordinals is not commutative: 2×ω equals ω, ω×2 does not. Same with addition: 1+ω = ω.
@@vytah when I wrote 2*x I meant x+x (which is written as x*2, my bad, and thanks for the warning :) ), with that change the reasoning holds x < x+x = x*2 = w
@@arturjorge5607 But the definition of divisibility used in this video puts the divisor on the left. So to check if ω is divisible by 2, you should find x that satisfies ω = 2×x, not ω = x×2. And similarly, the division splits the number into d×q+r (d=divisor, q=quotient, r=remainder) in this precise order. So ω/2 = ω remainder 0, because 2×ω+0 = ω.
A lot of standard ring algebra stuff that you're used to doesn't work with ordinals, for example distributivity doesn't works from the left: (1+1)×ω = 2×ω = ω, but 1×ω+1×ω = ω+ω = ω×2
@@arturjorge5607 Aside from the reasons vytah mentioned, it remains a nitpick that x + x is still not always greater than x (with x = 0)
12121212.......121212....121212
can be written like:
V = 12 + 100 V
It's better to completely get rid of constructs like "+ ...", and use recursion, so that you can trivially solve for their values.
(1 - 100)V = 12
V = -12/99.
V = 12 + 100(12 + 100 V)
= ...
Being a computer guy, I prefer taking the limit of n's complement in base n as the number of digits goes to infinity (for n=2, 2's complement and ones' complement become equivalent in this case, which is why it's interesting to a computer guy). In other words, we can't define that ...9999 = some infinite ordinal because I need ...9999 (or equivalently ...1111 in binary) for -1!
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
This proof seemed initially unintuitive as it's an immediate reflex to consider numbers like 0(...999) as clearly not being representable with finitely many digits, but it falls right out of the definition of limit ordinals that any such expansion is equal to omega, being larger than every finite ordinal. The fact that you can change the digits and they need not be 9 is also unintuitive without this understanding. The hook of the introduction did its job! The induction is also fairly obvious once all of the definitions are nailed down. I wish I had been introduced to ordinals in my discrete mathematics or number theory classes, I'm not sure we have a lower division class specifically for set theory.
I'm also pretty sure that with a bit of effort this can cause the decimal expansion of the surreals to fall out, where it is possible for non-ordinal values like omega minus 1 to show up.
me knowing that the set including all decimal numbers between 1 and 0 is higher than all full numbers but failing to understand this video:
Does this work for ordinals larger than 𝜀_0? 𝜀_0 has the property to be smallest ordinal that satisfies 𝜀_0 = 𝜔^(𝜀_0). The Wikipedia article for Cantor normal form mentions this. Also, I have hard time thinking a union of continuum number of continuums could be represented as a single decimal expansion, so there should be more ordinals packed there for this to work?
Despite the weirdness of epsilon_0, it does have a decimal expansion! Perhaps unsurprisingly, the decimal expansion of epsilon_0 is 10^(epsilon_0). When you deal with really big ordinals, this is going to happen a lot.
Cantor normal form (base omega expansion) also exists for any ordinal. However, it's only particularly *useful* if the ordinal is less than epsilon_0, because in this case, the exponents of Cantor normal form will be strictly smaller than the original ordinal. In general, you won't have that guarantee. The Cantor normal form for epsilon_0 is just omega^(epsilon_0), after all!
It does work, but it's a bit of a cop-out. omega^epsilon_0 is still in Cantor normal form if we allow each exponent to be _at most_ (rather than strictly less than) the ordinal being represented.
Minor nitpick at 14:01
You say strictly less than alpha but the graphics show less than or equal
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.
5:54 but limit of something like {5+1, 5+2, 5+3, .......} is same as limit of {1, 2, 3, 4, ......}.
Yes, that's why 1 + omega is not the same as omega + 1.
9:43 was very confusing until I realized ordinal multiplication is not commutative
Question: is this base dependent?
you are a shadow
also its not
Nope, there's nothing special about 10; the base can even be infinite!
@@SheafificationOfG If the base is the variable x, does this generalize to polynomials?
I'm not entirely sure what this would mean... is there an analogue of this over the ordinary naturals?
Is there a book you could recommend to learn more about this kind of stuff?
It's been ages since I learned this stuff, but the place I started was Jech's book on set theory. I think it's very well-written!
How much aderoll is required to watch this?
What about number a defined as "pi without the decimal point" or the decimal expansion of any other constructive real?
How would you formalise the ordinal "pi without the decimal point"?
@@SheafificationOfG I have no idea, that's the point. It seems there is no way to, but there is clearly a way to define the number itself, for example as a limit of the sum from n=1 to n=N of d_n*10^(N-n) as N tends to infinity, with d_n being the n-th digit of pi. What am I missing?
@@SpaghettiToaster That's definitely one way to try and formalise your value, and if you do it that way, the limit you get will be omega (as the limit will be the smallest ordinal larger than each of the terms in your limit's sequence), which is equal to the decimal expansion 1[...]0000.
The point is that even if there is *some* expansion where infinitely many digits are nonzero, the expression is inevitably equal to another decimal expansion where only *finitely* many of them are nonzero.
@@SheafificationOfG Oh, I understand, thanks!
But that means that an infinitely large circle constructed in this manner would have a radius-to-circumference ratio of 1, doesn't it? Shouldn't it instead be some indeterminate form that you can somehow beat into converging to pi?
@@SpaghettiToaster I think it's important to emphasize that ordinals are only concerned with ordering (rather than "length" or "magnitude" or "size"), so having a circle of radius "omega" doesn't necessarily make sense.
5:33 - I can see how direct limits are filtered colimits, but is the converse necessarily true? That is, given a filtered diagram D, is there always a directed subdiagram D' such that colim D = colim D'?
Every filtered category admits a cofinal functor from a directed category, so these notions of colimit are (in a strong sense) the same.
@@SheafificationOfG Awesome, thanks!
0:23 this really is the kind of complicated math that makes you forget how to count
At what point does the series for a = 0 to k, 9*10^-a == 1, given that for any natural value of k, the series is less than 1 and that the derivative is always a nonzero positive?
Keep up the good work!
Intersting, just dont get the point @9:17
How is it possible, can you give an example to find the digits of Omega? - Is there only one Omega or many differents?- I don't mean w+1,w+2..etc oder do you mean that with different "w"(Omegas) here?
When the remainder is always "w"(Omega) and Omega has all Numbers as possible divisors that give back "w"(Omega) then the digits would be all "w"(Omega) right - or what?
Best regards
Look closer at the algorithm we use to generate digits. The digit d_n is the remainder, not the quotient of each operation. The remainder when dividing omega by 10 happens to be 0. We get an infinite string of zeroes, but the operation can't ever terminate or produce a limit, so it doesn't give us a well defined output. When operating the other way, we get the same infinite string of zeroes but there's an extra 1 at the omega'th place.
omega does refer to a specific ordinal (namely, the smallest infinite ordinal).
Counterintuitively, omega = n * omega for any finite n (and, notably, n * omega is *not* the same as omega * n), so omega is (left) divisible by every finite number.
Nonetheless, following the proof, the decimal expansion of omega is 1...0000.
This is because when you find the least ordinal xi such that 10^xi
@@pudy2487 @SheafificationOfG So its pretty much the same to divide an infinite number by a number (maybe only a finite one) repetively that would give you endless strings?
Like divide 1000...000... (a 1 with infinite Zeros) by 10 repetively and it would also give you a 1 with infinite infinite zeros
- or divide it by 5 repetively to give you a 2 with infinite zeros etc.
Also it kinda sounds like the operations with the infinite symbol ♾
Like x times (♾) infinte = (♾) infinite.
What is the difference to Omega compared to the (♾) infinity symbol?
what is the differnce to the infinite symbol ♾ here to omega?
@@Lokalgott If you divide omega by 5, you just get omega again, so some of your intuition about division from the land of the finite doesn't transfer to the infinite setting.
You didn't prove the claim in the title though. Just proved that you can get a decimal expansion, not that it is finite.
Actually, the proof shows that the expansion is necessarily finite, too! Granted, proofs by (strong) induction obfuscate this, but I only ever work with finite sums.
Wouldnt adding by 0.1 be smaller by 1? Or do we only use integers?
Ordinals are only concerned with counting, and we can only really count by (repeated) 1's.
At 14:20, how do we know the decimal expansion is finite?
It follows from the (strong) induction hypothesis: we assumed that any ordinal smaller than alpha had a finite decimal expansion, and so in particular, the remainder rho after dividing by 10^(xi_1), being strictly smaller than alpha, had a finite decimal expansion. Say that expansion had length N, then adding 10^(xi_1)d_1 to the sum gives a length-(N+1) decimal expansion of alpha.
Great video :D
If w (omega) is divisible by all natural numbers, then is w+1 not divisible by anything? Are there no primes after w?
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
just watched vsauce how to count past infinity video again and got this lol
Math is just counting an infinite amount of zeros.
a twist on this perspective: represent an hypernatural number by using a fixed hypernatural H many 1’s
ordinal number addition and hypernatural addition don't match, hypernaturals are not well ordered
Hypernaturals, brought to you by the guy who also mentioned nonstandard analysis in a previous video ;)
I don't actually know much about hypernatural numbers. Do they have a "finite digit representation" analogue, or is the H-many 1's a counterexample?
@@SheafificationOfGthe infinite series of ones would be a counter example thanks to the transfer principle
@@SheafificationOfG😂feelin like a youth pastor from that comment about advocacy.
The point of the comment is a counterexample, since it’s a related concept of infinite number. Like if ordinals had square roots and stuff (all of arithmetic by transfer principle).
I rec looking up keislers infinite microscope and telescope for a good viz, or the numberphile The Infinitesimal Monad with Carol Wood
@@mzg147no, but as a refinement of that, they ARE internally well ordered, which covers a lot actually (all standard formulae for one and allows “internal induction“ which is like a more user-friendly transfinite induction.
Same for completeness of hyperreals. Incomplete but internally complete.
goated video just for "there's always a bigger fish"
Anakin had omega^omega-many midichlorians
I don't even do a math degree what am I doing here
We ought to sacrifice our souls to homotopy type theory
How many Latin _and_ Greek alphabet can you fit in one math video
Reminds me of vsauce's video about infinities
infinit numbers with infinit type of symbols!!!!!!!!!!!!! yes the biggest infinity of all!!!!!!!!!!!!!!!!!!!
why do you have a filthy Brilliant sponsor?
It seems to me that placing a number in the infinite place is meaningless. It is defining into existence something that has no meaning or actual value since infinity with an extra digit is still just infinity. Put an infinite to the infinite number of digits in front of infinity and it's still just infinity. It also seems that omega is a meaningless value as if you actually try to consider what it would equal, you would quickly find that there is no first number that is greater than all finite ordinals. It is either a finite value, which contradicts its own definition or it's an infinite value which makes all further addition to it meaningless since they're all just infinity. It's like little kids yelling "I have infinite power!" and another yelling "I have infinite times infinite power."
Omega = Infinity. Omega + Omega = Infinity.
I think you're mixing up ordinals and cardinals
Yes, if you select the first number bigger than all numbers you are effectively selecting the empty set. Everything else is nonsense stemming from applying this empty object to random situations, which gives different broken results depending on the direction from which you approach the task, leading to all the paradoxes such as sums of infinite positive series giving -1/12.
Ordinals are used to classify well-ordered sets. A set corresponding to omega has a different structure to a set corresponding to omega×2, even though they have the same size. In other words, you can create a bijection between them, but that bijection will never preserve the ordering. An example pair of such sets are N and {0,1}×N.
But I guess what is more important is that you're not aware how overloaded the term "infinity" is in maths. Here are some meanings I can name out of top of my head:
- infinity as a lack of bounds in a particular direction on the real number line, it has two variants: -∞ and +∞
--- it's also used in contexts of integers, as they are a subset of reals
- infinity as a symbol of divergence of a sequence, -∞ if the sequence is always decreasing, +∞ is it's always increasing
--- note that there are other non-numbers used to describe sequence limit behaviour, like for example 0⁺ meaning "the limit is 0 and it's approached from the right"
- infinity as a symbol of divergence of a limit of a function
- the complex equivalents of all those above: a lack of bounds, a limit of a sequence, or a limit of a function on a complex plane
- infinity as a symbol representing distance between two elements that cannot reach reach other
- infinity as an element of a projective plane or a wheel; in this context, 1/0=∞ is not a sloppy shorthand for a limit lim x→0 1/x, but an actual algebraic expression
- infinity as a size (cardinality) of a set - cardinality of sets is measured using cardinal numbers, and there are infinitely many infinite cardinal numbers
- infinity as a description of an order of an infinite well-ordered set - you use ordinal numbers, which are the subject of this video:
--- two sets with the same ordinal number have a bijection between them that preserves ordering, two sets with different ordinal numbers cannot have such a bijection
--- ordinal numbers have some nice properties they share with natural numbers, like the ability to do induction
--- there's a connection between ordinal numbers and cardinal numbers, as all well-ordered sets with a given ordinal number have the same cardinality, so every ordinal number can be mapped to some cardinal number (obviously not vice versa though)
--- infinite cardinal numbers are usually indexed with the ordinal numbers, starting from ℵ₀
- infinity as any surreal number larger than any finite number - surreal numbers contain all ordinal numbers, so that's obvious
There are more, but I think that's enough for starters.
So, does that mean that infinite numbers can have a finite digit sum?
If you write an ordinal out as a finite decimal expansion, then yes; the sum of its digits will indeed be finite.
I thought we agreed everyone is actually (if secretly) an ultrafinitist, as is right and proper. :P
I enjoy your videos a lot. Keep up the good work. (At whatever finite pace you wish to set for yourself and your fan.)
They're finite, I swear! Just count the digits!
It would hardly be a secret if I only ever spoke about small finite numbers like a good and proper pure mathematician.
Counting is easy until you get to the one larger than three
SUCC
This seems awfully similar to the real numbers somehow.. 🤨
8:00 So you're saying with Brilliant, even girls can learn math? 😝
Seriously, though, as a lifelong handyman/DIYer who dabbles in math by way of youtube, I know what it's like from both their points of view. Kurei's tool-bag is almost as cool as the chill teaching attitude she inherited, and I laughed _inordinately_ hard at Serufu's unexpected yet inevitable display of power-drill finesse.
Yours is the first _DIY!!_ meme I've seen, and it's perfect. Subbed.
if the digit d is in the eth position 😂
Is ω-1 finite
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.
The intro seems like ω
Wow
W
I ruined the 69 likes in hope of 420
Where did this guy learn math? He is basiclally saynig that one infinite number is equal to another infinite number, but infinity cannot compare with infinity of the same sign. Yes, +inf is bigger than -inf, but +inf and +inf do not cmopare, you cannot say that any of them is equal to or greaer than the other. Hence his argument falls apart completely.
And no, infinity does not have finite number of non-zero digits unless you talk about a finite string of digits followed by an infintie number of 0's in which case it is infinity and my comment above applies. 1 followed by infinitely many zeroes is neither larger nor smaller than 2 followed by infintely many zeroes, they are both infinity and so cannot be compared.
Definitely not subscribing.
I think you got some things mixed up there. This video talks about ordinal numbers and not about the concept of infinity as an extension/direction of the real number line.
There is a long answer on a comment from thelordz33 that lists some different meanings of the word "infinity"
*Please stop using memes.*
They are so annoying and *will lose you likes* from anyone who actually knows and enjoys the subject.
stop having fun