Georg Cantor - Contributions to the Founding of the Theory of Transfinite Numbers - Section 19

This is video number 19 in a series of videos on Cantor's 1895 and 1897 articles on the theory of aggregates and transfinite numbers. Section 19 is titled The Normal Form of the Numbers of the Second Number Class. Personally, I found some parts of this section particularly confusing for the first couple of attempts at reading through it. There's a lot of ground to cover in this section, and Cantor, I would imagine, in order to keep the section to a reasonable length, omits many of the details of this section. So I'll try to clarify as many points as I can, even at the risk of seemingly going into a pedantic level of detail, but that said, there may still be some minor details which I've overlooked here and there. Please let me know if there's anything that you spot that I've missed. It might be noticed in this section that sometimes less than or equal is written with the less sign above the equals sign, and sometimes with the equals sign above the less than sign. I've not been able to determine why this is, or whether this is supposed to convey some significant distinction. I haven't found any need to understand them as meaning different things, so I'll treat them as symbolising the same relationship in this video. By Theorem F of Section 18, it follows that Omega to the power of Alpha is greater than or equal to Alpha. And by Theorem B of Section 18, then if Alpha is less than XI, we have that Omega to the power of Alpha is less than Omega to the power of XI, giving Omega to the power of XI is greater than Alpha for XI greater than Alpha. On the assumption that Beta is a number of the second kind, and since we know then that Omega to the power of Beta Nu is less than or equal to Alpha for all Nu, then if it were the case that the limit of Omega to the power of Beta Nu is greater than Alpha, then there would necessarily be a Nu-naught for which Omega to the power of Beta Nu is greater than Alpha for all Nu greater than or equal to Nu-naught, which contradicts the minimality of Beta, since we know that Beta Nu is less than Beta for all Nu. Therefore, Beta is not a number of the second kind. There's a small typo coming up on this page as indicated on the screen now. So Beta is a number of the first kind, but the number immediately preceding Beta, denoted Alpha-naught, may be a number of either the first kind or the second kind. Point 1 follows, since we have Alpha-naught is less than Beta, we must also have that Omega to the power of Alpha-naught is less than Alpha, but Omega to the power of Alpha-naught times Omega is equal to Omega to the power of Alpha-naught plus 1, by theorem D of section 18. And since Alpha-naught plus 1 is equal to Beta, we have Omega to the power of Beta is greater than Alpha, where Beta is the smallest such power of Omega. And so Omega to the power of Alpha-naught times Omega is greater than Alpha. That Kappa-naught is greater than 0 can be seen as follows. Since by 1 of this section, Omega to the power of Alpha-naught is less than or equal to Alpha, then Omega to the power of Alpha-naught times 1 is less than or equal to Alpha. And so if Omega to the power of Alpha-naught times Nu is greater than Alpha, then Nu must be greater than or equal to 2. Thus Kappa-naught plus 1 is greater than or equal to 2. And so Kappa-naught is greater than or equal to 1. And so Kappa-naught is greater than 0. 0. The inequalities of point 2 follow from the definition of kappa 0, since kappa 0 was defined to be the least value of nu, for which omega to the power of alpha 0 times nu is greater than alpha. And point 3 follows from point 10 on page 155. For the inequalities of 4, we have the argument as shown on the screen. By point 4, we have 0 is less than power of alpha 0. And by point 1, we have omega to the power of alpha 0 is less than alpha. Therefore we have 0 is less than or equal to alpha primed is less than omega to the power of alpha 0 is less than alpha, which gives alpha primed is less than alpha. The number alpha 0 was chosen to be a number less than or equal to alpha. The theorem A has effectively been proved by the arguments presented on the previous couple of pages. Most of this page is self-explanatory, so jumping to the bottom of page 186 then, alpha greater than alpha primed, greater than alpha primed primed, and so on, is a decreasing sequence of transfinite numbers. If alpha sigma is of the second number class, then the process continues, so eventually alpha sigma must be of the first number class. If alpha sigma never became finite, then the sequence would continue to decrease, and there would be no smallest number, which is a contradiction, since the aggregate of alphas is a part of a well-ordered aggregate. If alpha sigma is finite, but not equal to 0, then alpha sigma can be written as alpha sigma equals omega to the power of 0 times kappa sigma plus alpha sigma plus 1, where, by section 18, theorem b part a, omega to the power of 0 is equal to 1, kappa sigma is equal to alpha sigma, and alpha sigma plus 1 is equal to 0, so alpha tau plus 1 is equal to 0 for tau plus 1 greater than or equal to sigma plus 1. Note that star is not an application of theorem a from section 19, since alpha sigma is not of the second number class, but nevertheless star still holds. Even if alpha is of the form omega to the power of alpha 0 times kappa 0, then alpha primed is equal to 0, and in this case alpha tau plus 1 is equal to 0 for all tau plus 1 greater than or equal to 1. In any case where it is necessary to apply theorem a to alpha primed also, then tau plus 1 will necessarily be greater than 1, so tau may be equal to 0, evidently only in the case where alpha is of the form omega to the power of alpha 0 times kappa 0, but at any rate tau plus 1 will necessarily be other than 0. That tau plus 1 be finite ensures that the form of alpha given on page 187, the normal form, has only a finite number of terms. Therefore the normal form of a number of the second number class shows us that any number of the second number class can be represented as a finite number of terms, and each term is a finite multiple of a possibly transfinite power of omega. Notice that in section 17 we were limited to finite powers of omega, and thus section 17 only dealt with a certain subclass of numbers from the second number class. The normal form is however completely general. That is, every number of the second number class has a normal form representation. Point 9 follows from point 8 of section 14, and point 10 can be seen as shown on the screen. Thus we see from point 10 that any power of omega which is followed on the right by a term which is of higher degree in omega may be omitted. Point 11 follows by repeated application 13 we have as shown on the screen now, and the argument for point 14 is shown on the screen now. It can be seen from point 14 that the degree of alpha to the power of lambda is alpha-naught times lambda. Thus for any given term, omega to the power of alpha-naught times lambda-naught of the fundamental series omega to the power of alpha-naught times lambda, there will be terms of the fundamental series alpha to the power of lambda which are greater than omega to the power of alpha-naught times lambda-naught. Namely, whenever lambda is greater than or equal to lambda-naught. Similarly, for any given term alpha to the power of mu-naught of the fundamental series alpha to the power of lambda, there will always be terms of the fundamental series omega to the power of alpha-naught times lambda which will be greater than alpha to the power of mu-naught. Namely, when lambda is greater than mu-naught. Therefore the two fundamental series are coherent. And since the fundamental series alpha to the power and Omega to the power of Alpha nought times Lambda are a coherent series, then they have equal limits. The limit of Alpha to the power of Lambda is Alpha to the power of Omega, and the limit of Omega to the power of Alpha nought times Lambda is Omega to the power of Alpha nought times Omega. Therefore we have that Alpha to the power of Omega is equal to Omega to the power of Alpha nought times Omega, which is 15. And so 16 can be deduced as shown on the screen. Theorem C can be approached in a similar way to page 175, that's section 17, and the proofs of the results of theorem D are very similar in nature to the proofs of A, B and C at the bottom of page 175 and top of page 176 of section 17. And so, I'll omit these here. The result of point 10 of this section will also be useful in the proofs of the results of theorem D. Theorem E can be proved as shown on the screen. And now a proof of theorem F. Theorem G, the derivation of the product form of alpha, is virtually identical to the derivation of the product form of phi in section 17. A little more care has to be taken when dealing with the powers of omega, but the argument is roughly the same and I won't repeat it here. The kappas which appear in the product form of alpha are the same kappas as in the normal form of alpha. Notice that the kappas in the normal form are kappa naught to kappa tau from left to right, but in the product form the kappas are kappa tau to kappa naught from left to right. Thus the order is reversed. Due to issues with commutativity, it's not possible to reorder the factors of the product form. Each factor is where it is and has to remain in that position. The proof of theorem H can be shown as on the screen. Theorem I simply states that for two numbers of the second class to commute under addition, then they must be, possibly different, finite multiples of the same number gamma. Theorem K states that for two numbers of the second number class, and which are both numbers of the first kind, to commute under multiplication, then they must be, possibly different, finite powers of the same number gamma. The proofs of these theorems follow on the remaining pages of this section. A typographical error on page 191 is corrected on the screen now. And note that finite numbers commute under addition, so in theorem D is applied to find alpha plus beta and beta plus alpha respectively. For alpha plus beta we would have mu plus nu as the multiplier of omega to the power of alpha naught, and for beta plus alpha we would have nu plus mu as the multiplier. But mu plus nu is equal to nu plus mu. The proof of I presents no real difficulties. I'll spend the remainder of this video going through the proof of theorem K, since there's a few points that are worth clarifying in the proof. Notice that in the product form describing theorem G of this section, that if a number is of the first kind, then gamma naught is equal to zero. And hence, this omega to the power of gamma naught is equal to one, by theorem B of section 18. Therefore, since both alpha and beta are numbers of the first kind, by the assumptions in the statement of theorem K, the product forms can be written as shown on the screen. It may be that both of these product forms agree for some number of terms at the left end of each expression. This is the delta which is used to express alpha and beta as delta alpha primed and delta beta primed respectively. In the expressions for alpha primed and beta primed, there are no common factors, by assumption, at the very left end of each expression other than one. This doesn't mean that there are not common factors elsewhere in the product forms of alpha primed and beta primed, but simply that they do not appear at the leftmost end. That alpha primed is greater than beta primed is shown on the screen. And that alpha primed delta beta primed is equal to beta primed delta alpha primed is also shown on the screen. Note also that alpha primed, beta primed and delta are all numbers of the first kind, since none of them have a product form which begins omega to the power of xi, for some xi greater than or equal to one, as required by theorem H of the current section, for any number of the second kind. Each of the product forms will have rho mu times omega to the power of xi one plus one, at the left end, where rho mu is possibly equal to one. Theorem G of the current section says that the product form of a number is unique. Therefore, since alpha primed delta beta primed and beta primed delta alpha primed are equal, then they have the same unique product form. In particular, the left end of the product form of alpha primed delta beta primed will be identical to the left end of the product form of beta primed delta alpha primed. Cantor does a bit of case analysis, the details of which I'll fill out here, considering various possible combinations of alpha primed and beta primed being finite or transfinite, and going through a process of elimination until eventually a single possibility is arrived at. Suppose the product form of both alpha primed delta beta primed and beta primed delta alpha primed has rho mu times omega to the power of xi one plus one at the left end. Then if both alpha primed and beta primed are transfinite, we must have that both alpha primed and beta primed have rho mu, omega to the power of xi one plus one, at the left end of their individual product forms. Which is a contradiction since alpha primed and beta primed we assumed to have no common factors at the left end of their product forms other than one. If alpha primed and beta primed are both finite, then the only way that both alpha primed delta beta primed and beta primed delta alpha primed can be transfinite is if delta is transfinite. The product form of delta then starts with kappa times omega to the power of xi one plus one. Then the product form for alpha primed delta beta primed has a finite factor alpha primed kappa at the left end of the expression, and beta primed delta alpha primed has a finite factor of beta primed kappa at the left end. These must be equal since the product forms for alpha primed delta beta primed and beta primed delta alpha primed are identical, and so alpha primed kappa is equal to beta primed kappa. Since these are finite terms then there's no problem cancelling the common factor of kappa giving alpha primed equals beta primed, which is a contradiction since we know that alpha primed is greater than beta primed. Therefore one of alpha primed and beta primed must be finite and the other transfinite. Since alpha primed is greater than beta primed and any transfinite number is greater than any finite number, then it must be that alpha primed is greater than omega and beta primed is less than omega. Note that alpha primed cannot be equal to omega since alpha primed is a number of the first kind, but omega is a number of the second kind. Now suppose that alpha primed has product form starting kappa times omega to the power of psi 1 plus 1 and beta primed equals nu naught and assume that nu naught is greater than 1. Then since the product forms of alpha primed delta beta primed and beta primed delta alpha primed are identical, then this means that nu naught must be a non-trivial factor of kappa. This is what Cantor means when he says that it, that is beta primed, would be contained in the finite factor at the left end of alpha primed. And this contradicts that alpha primed and beta primed have no common factors at their respective left ends other than 1. Therefore, beta primed must be equal to 1. And so, since beta is equal to delta times beta primed, then beta equals delta times 1, which by point 9 of section 14 gives beta equals delta. Since alpha primed delta beta primed is equal to beta primed delta alpha primed and beta primed is equal to 1, then alpha primed delta times 1 is equal to 1 times delta alpha primed. And so, by point 9 of section 14, we have alpha primed delta is equal to delta alpha primed. And also, since beta equals delta, we therefore conclude that alpha primed beta is equal to beta alpha primed. Since alpha equals beta alpha primed, then from theorem H from section 14, we can conclude that alpha primed is less than or equal to beta alpha primed, which equals alpha, and hence alpha primed is less than or equal to alpha. But Cantor states that alpha primed is less than alpha. This is because we also have that alpha primed beta is equal to beta alpha primed. And so also by theorem H of section 14, we have alpha primed is less than alpha primed beta, which equals alpha, and hence alpha primed is less than alpha as required. We keep applying the process just described to generate a series of numbers, alpha, alpha primed, alpha primed and so on. And this can be summarised in the following array, where each row represents an iteration of the process. This process continues until the stage is reached, which must happen by theorem B of section 16, where alpha row naught is less than or equal to beta. Note that the row naught in brackets does not represent a power, but represents an index. A small typographical error on this page is shown on the screen now. Suppose that alpha row naught is equal to beta. Then note that alpha equals beta alpha primed equals beta beta alpha primed primed and so on. Then alpha is equal to beta to the power of row naught plus one. Thus theorem K is proved in the case that alpha row naught is equal to beta, if we set gamma equal to beta. However, if alpha row naught is less than beta, then on the assumption that alpha row naught is equal to beta one, then beta one is less than beta. Alpha equals beta to the power of row naught times beta one, as shown. Since we already know from the array previously shown that alpha row naught times beta is equal to beta alpha row naught, then we have beta one times beta is equal to beta times beta one. Therefore the condition beta one times beta equals beta times beta one and beta one less than beta means we can repeat the process described on page 193 to 194 for beta and beta one instead of alpha and beta. Again this process can be summarised in the following array, which is similar in nature to the previous array which was shown. Once again the stage must be reached where beta row one is less than or equal to beta one. Suppose beta row one is equal to beta one. Then we're in the same situation as at the top of page 194 for alpha row naught equals beta. So assume beta row one is less than beta one and set beta row one equal to beta two. Then the process can be repeated until we obtain beta one row two is less than or equal to beta two. At which point either beta one row two is equal to beta two and the process terminates or beta one row two is less than beta two, in which case the process continues. A sequence is generated as shown. Additional details also are given on the screen. We see then that ultimately both alpha and beta can be written as a power of beta kappa with respective powers mu and nu. The values of the powers can be written in terms of the numbers row naught, row one up to row kappa, but they will be incredibly complicated expressions. So rather than attempt to give expressions for the values of mu and nu, Cantor gives instead an expression for the ratio mu divided by nu. The derivation of this expression which Cantor doesn't give goes as shown. Note that in general alpha, beta, beta one up to beta kappa minus one can all be written as a power of gamma which is equal to beta kappa. Where I've written power of such and such, then it's referring to the exponent of that number when written as a power of gamma. Thank you for watching this video. If you found this video useful then please hit the like button and subscribe to my channel. You can see more videos on various topics on my channel and if you have any suggestions for topics for future videos then please feel free to let me know and I'll try my best to put something together. Don't forget to hit that subscribe button and keep an eye out for new videos being uploaded to my channel.