Georg Cantor - Contributions to the Founding of the Theory of Transfinite Numbers - Section 9

Welcome to this video, the ninth video in my series of videos on George Cantor's 1895 and 1897 articles on the theory of aggregates and transfinite numbers. Section 9 is titled the ordinal type eta of the aggregate r of all rational numbers which are greater than zero and smaller than one in their natural order of precedence. Page 123 outlines three properties which characterize the ordinal type eta of the aggregate r. The second and third characteristic properties namely that r has no lowest rank element and no highest rank element and that between any two elements of r there are other elements of r are straightforward enough to show and furthermore since it's been shown that the aggregate r can be rearranged into the aggregate r naught and r and r naught are equivalent then r has cardinal number aleph naught. Page 124 formalizes the three characteristic properties of the ordinal type eta into a theorem that the aggregate m can be brought into the form of a well-ordered aggregate of type omega is based on the idea that m and omega have the same cardinal number aleph naught. Therefore from section 1 we know that m and omega must be equivalent though of course they're not similar and therefore there must be a one-to-one correspondence between m and omega. Therefore the element m which is coordinated with the element nu of omega will be denoted m sub nu and m sub nu precedes m sigma in m naught whenever nu precedes sigma in omega. Note that the one-to-one correspondence between m and omega does not need to be explicitly stated at least in Cantor's view. The correspondence can be assumed to exist simply because m and omega have the same cardinal number and are therefore equivalent aggregates. Also note that well-ordered aggregates will be properly introduced in section 12 although they've already been mentioned on page 115. The ordinal type and indeed ordinal number omega has also been mentioned on page 115 as well. There's a comment in the introduction which is relevant to the claim that m can be brought into the form of a well-ordered aggregate. On page 62 Cantor is quoted as saying every aggregate of the power of the first class that is aleph naught is innumerable by numbers of the second class and only by these and the aggregate can always be so ordered that it is enumerated by any prescribed number of the second number class and analogously for the higher classes. An aggregate this is enumerable is understood to be an aggregate that can be put into a one-to-one correspondence with the aggregate of positive integers. Note that r1 in r is to be correlated with the element m1 in m and not in m naught. The correlation that we require is between r and m. The aggregate m naught contains all of the elements of m. m naught is used merely to help form the correspondence between r and m. Since m naught contains all of the elements of m and since there must be elements of m which are lower in rank or higher in rank than m1 then there must be at least index i2 for which m sub i2 is lower in rank or higher in rank than m1. If not then m naught would only contain the elements of m which were either higher in rank than m1 or lower in rank than m1. The elements r1 r2 etc are not being randomly chosen from r. They are in fact the elements of the aggregate r naught and in the order in which they appear in r naught. The r1 chosen to start things off is the lowest rank element of r naught. r2 is the second lowest rank element of r naught. So just like m naught is being used to help form the relevant correspondence between r and m so r naught is also being used for this purpose. Note also that r naught contains all of the elements of r without exception. Thus if and when a suitable correspondence is made between each r new and an element of m then there is a correspondence between each element of r and m. The process by which each r new is correlated to an element of m is simple enough at least from a theoretical perspective. So a correlation is obtained whereby r1 is correlated to m1 and each r new is correlated to m sub i new for new greater than 1. Note that the sequence of subscripts 1 i2 i3 and so on is not necessarily and in fact probably won't be an increasing sequence. The claim is that we have correlated definite elements m i new of m to all the elements r new in r and the elements m i new have in m the same order of precedence as the corresponding elements r new in r. This is easy enough to confirm since each r new has been correlated in the way described to an element of m but the r new account for all elements of r. Recall that r naught contains all elements of r without exception and so all elements of r have been correlated to an element of m. Moreover by the way that the m i new have been chosen they have the same order of precedence in m as their corresponding r new in r. What about the other direction? This is a bit trickier to deal with and I'll just clarify some of the points mentioned in the proof. The proof follows an induction argument. Suppose that lambda is chosen sufficiently large so that all of the elements m1 m2 up to m new are amongst the elements m1 m i2 up to m i lambda of the correspondence. Note that m1 is accounted for that is m1 is amongst the m1 This is the base case for the induction. The elements m1 m2 up to m new are the first new elements of m naught and recall that m naught contains all elements of m without exception. So if we show that all m new are amongst the m1 m i2 m i3 and so on then we can see that every element of m is accounted for in the correspondence derived on the previous couple of pages. We consider the two possible cases either m sub new plus one is amongst the m i lambda or m sub new plus one is not amongst the m i lambda. In the first case there's nothing to do we only have to really consider the second case. The argument works in a similar way to the construction of the correspondence on the previous couple of pages where we hunted through the elements of the aggregate m naught for a suitable element to correlate to each of the ri by choosing at each stage the smallest value index amongst the elements of m naught. Here though we look for a suitable element of r naught to correlate to m new plus one. This is obtained by again choosing the least value index lambda plus sigma for which r sub lambda plus sigma has the same ordinal position in r to the r1 r2 up to r lambda as m new plus one has in m to m1 m i2 up to m i lambda. At the risk of stating the obvious I'll just clarify some of the details of the induction proof which I feel are glossed over a bit. Since lambda plus sigma is greater than lambda and r lambda plus sigma corresponds to m i lambda plus sigma then m i lambda plus sigma cannot be equal to any of the m1 m i2 up to m i lambda and so cannot be equal to any of the m1 m2 up to m new. In the diagram on the screen the upper row of dots represents the elements m1 m i2 up to m i lambda and the lower row of dots represents r1 r2 up to r lambda. A possible position of m new plus one is shown. Consider r lambda plus one and assume that r lambda plus one does not have the same ordinal position with respect to r1 r2 up to r lambda as m new plus one has to m1 m i2 up to m i lambda. We see that r lambda plus one cannot be in the indicated position since then it would have the same ordinal position with respect to r1 r2 up to r lambda as m new plus one has to m1 m i2 up to m i lambda which would contradict the minimality of lambda plus sigma. So suppose for example that r lambda plus one is in the indicated position then by the correspondence r lambda plus one corresponds to m i lambda plus one and that m i lambda plus one has the same ordinal position to m1 m i2 up to m i lambda that r lambda plus one has to r1 r2 up to r lambda. If we include both r lambda plus one and m i lambda plus one in the diagram we see that in fact m new plus one has the same ordinal position to m1 m i2 up to m i lambda plus one as r lambda plus sigma has to r1 r2 up to r lambda plus one. We can continue this for all mu such that one is less than or equal to We see this indeed m new plus one has the same ordinal position to m1 m i2 up to m i lambda plus mu that's r lambda plus sigma has to r1 r2 up to r lambda plus mu for each one less than or equal to mu less than or equal to sigma minus one and in particular m new plus one has the same ordinal position to m1 m i2 up to m i lambda plus sigma minus one that Rlambda plus Sigma has to R1, R2 up to Rlambda plus Sigma minus 1 as claimed. Then since by the correspondence I lambda plus Sigma is the smallest index in M0 for which Mi lambda plus Sigma has the same ordinal position in M to M1, Mi2 up to Mi lambda plus Sigma minus 1 as Rlambda plus Sigma has to R1, R2 up to Rlambda plus Sigma minus 1 in R and I lambda plus Sigma cannot be equal to any of the 1, 2 up to nu then I lambda plus Sigma is equal to nu plus 1 and so Mi lambda plus Sigma is equal to M nu plus 1. Therefore Mi lambda plus Sigma is equal to M nu plus 1. And so we've shown that if I lambda is chosen sufficiently large so that M1, M2 up to M nu are amongst the M1, Mi2 up to Mi lambda then we're able to, if necessary, choose lambda plus Sigma sufficiently large so that M1, M2 up to M nu plus 1 are amongst the M1, Mi2 up to Mi lambda plus Sigma and this completes the induction. This confirms that the correspondence found is in fact an imaging of M and R and thus M is similar to R. Some of the various ordinal types that are mentioned which involve eta have some nice visualizations such as 1 plus eta can be visualized as all rationals in the open interval 0, 1 with 0. 1 plus eta plus 1 can be visualized as all rationals in the closed interval 0, 1. Eta plus 1, all rationals in the open interval 0, 1 with 1. Eta plus eta, all rationals in the open interval 0, 2 except 1. Eta plus 1 plus eta, all rationals in the open interval 0, 2. Eta times eta, all points on the interior of the unit square with rational coordinates. Thank you for watching. 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.