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From: Daryl McCullough on 25 Oct 2005 06:53 albstorz(a)gmx.de says... >David R Tribble wrote: >> Obviously, you do not understand. >> >> My set is: >> S = {0, 2^0, 2^2^0, 2^2^2^0, 2^2^2^2^0, ...} >> S = {0, 1, 2, 4, 16, 65536, ...} >I don't know what kind of math you apply here. Tribble-O-Math? It's quite ordinary math, except for the fact that perhaps you are unfamiliar with the use of ^ to mean exponentiation? 2^0 = 1 2^1 = 2 2^2 = 2*2 = 4 2^3 = 2*2*2 = 8 2^4 = 2*2*2*2 = 16 etc. David's sequence is 0, 2^0, 2^(2^0), 2^(2^(2^0)), etc 2^0 = 1 2^(2^0) = 2^1 = 2 2^(2^(2^0)) = 2^2 = 4 2^(2^(2^(2^0))) = 2^4 = 16 2^(2^(2^(2^(2^0)))) = 2^16 = 65536 etc. -- Daryl McCullough Ithaca, NY
From: albstorz on 25 Oct 2005 08:53 David R Tribble wrote: > Albrecht S. Storz wrote: > > But you are not able to give respond to my argumentation. You correct > > mistakes which have no connection to my argumentation. You just ignore > > it. You speak with yourself. Why do you post in this thread if you are > > not willing to debate my argumentation? > > > > This is not only adressed to you, David, but also to all the other > > participants of this thread. > > Okay, I'll respond to your original post, one more time. Thanks. > > > > Let's start with a representation of the natural numbers in unitary > > (1-adic) system as follows: > > > > O O O O O O O O O ... > > O O O O O O O O ... > > O O O O O O O ... > > O O O O O O ... > > O O O O O ... > > O O O O ... > > O O O ... > > O O ... > > O ... > > . > > . > > . > > > > 1 2 3 4 5 6 7 8 9 ... > > > > Each vertical row shows a natural number. Horizontally, it is the > > sequence of the natural numbers. Since the Os, the elements, are local > > distinguished from each other, we can also look at the rows as sets. A > > set may contain the coordinates of the elements as their representation > > or may look like this: S3 = {O1, O2, O3} e.g. . > > > > From the Peano axiomes follows that the set of all naturals is > > infinite. So, the set of the elements of the first horizontal row is > > infinite. Actually the set of the elements of every horizontal row is > > infinite. And the set of all the elements in this representation is > > also infinite. > > > > But there is no vertical row with infinite many elements since there is > > no infinite natural. > > That is correct. The set of natural numbers (N) derived from the > Peano axioms is an infinite set containing only finite naturals. OK > > > > Now let's fill the horizontal rows or lines with other symbols. We have > > to take into account that only lines should be filled with #s which > > containes Os. > > > > # O O O O O O O O O ... 1 > > # # O O O O O O O O ... 2 > > # # # O O O O O O O ... 3 > > # # # # O O O O O O ... 4 > > # # # # # O O O O O ... 5 > > # # # # # # O O O O ... 6 > > # # # # # # # O O O ... 7 > > # # # # # # # # O O ... 8 > > # # # # # # # # # O ... 9 > > . . . . . . . . . . . > > . . . . . . . . . . . > > . . . . . . . . . . . > > > > The vertical sequence of sets of #s fullfill the peano axiomes exactly > > as the horizontal sequence of the sets of Os does. > > But there is a slight difference. Since there is no infinite natural in > > form of a set of Os and since after every set of #s there should be a > > O, the size of the set of the naturals as sets of #s could not extend > > the "biggest" number of the naturals in form of sets of Os. > > Since there is no biggest number and since there is no infinite number, > > the size of the set of numbers in form of sets of #s is undefined as > > the biggest natural number is undefined. > > > > But the sequence of the sets of # fullfill the peano axiomes. So this > > set must be infinite. > > > > The cardinality of a set is not able to be infinite and "not defined" > > at the same time. > > > > This is the contradiction. > > There is no contradiction. The set contains only finite elements, > and the set is infinite (i.e., contains an infinite number of > elements). There are many sets. What do you are talking from? Let's have names: The first horizontal row or line of #s: L#1 = {#1} The first horizontal row or line of Os: LO1 = {O1, O2, O3, ...} The second horizontal row or line of #s: L#2 = {#1, #2} and so on The first vertical row or column of #s: C#1 = {#1, #2, #3, ...} The first vertical row or column of Os: CO1 = {O1} and so on Now, as defined, CO1, CO2, CO3, ... is the sequence of the natural numbers in 1-adic representation. So the cardinality of the set LO1 is equivalent to the cardinality of the set of natural numbers N. Now we had found, that the sequence L#1, L#2, L#3, ... also follows the peano axioms. But by construction, the cardinality of the set of L#x, which is equivalent to card(C#1), could not be bigger than any finite natural number. Ergo: The size of the set C#1 must be infinite and undefined at the same time. > > The size (cardinality) of the set is infinite, but this set measure > is not a natural number, and is not a member of the set itself. > Infinite cardinal numbers are not natural numbers, so they are not > members of sets of naturals. > > The "largest natural number" is undefined because it does not exist; > there is no such number. On the other hand, the size of the set is > well-defined as the smallest infinite set size (Aleph_0). But the > size is not a natural number. > > It is reasonable to say that the number of elements in the set is > greater than any member in the set. Therefore, the size cannot be > a member of the set, and it cannot be a natural number. It's not reasonable in the case of the natural numbers. The size of a set could not be greater than any member in the case if the members count themself. This is so obvious. Since it is obvious and you and many other mathematics can't see it, ther is no rational discussion possible about this issue. > > > > Or let's say it in another form: The first vertical row of #s could not > > exceed the biggest vertical row of Os (and could not be smaller). So, > > the cardinality of this set is undefined like the biggest natural > > number. But the set of the elements of the first vertical row of #s has > > the same cardinality like the set of the natural numbers. > > --> Contradiction. > > > > Or did I construct a monster set which cardinality is subtransfinite? > > Neither. If you consider the second set to be a set of subsets (where > each subset is represented by a row of #s and Os), the argument is > the same - the set is infinite, but no member of the set is infinite. > > Your contradiction is wrong. The contradiction comes from saying > that the size of the set is also a member of the set. Since it is constructed like this it is like this: the size of the set is a member of the set. If you just deny the obvious facts, there is no further communication possible. > > There is not much else to say about it. Yes. Regards AS
From: David Kastrup on 25 Oct 2005 09:05 albstorz(a)gmx.de writes: > David R Tribble wrote: > >> The size (cardinality) of the set is infinite, but this set measure >> is not a natural number, and is not a member of the set itself. >> Infinite cardinal numbers are not natural numbers, so they are not >> members of sets of naturals. >> >> The "largest natural number" is undefined because it does not >> exist; there is no such number. On the other hand, the size of the >> set is well-defined as the smallest infinite set size (Aleph_0). >> But the size is not a natural number. >> >> It is reasonable to say that the number of elements in the set is >> greater than any member in the set. Therefore, the size cannot be >> a member of the set, and it cannot be a natural number. > > It's not reasonable in the case of the natural numbers. The size of > a set could not be greater than any member in the case if the > members count themself. This is so obvious. You did not read _any_ of what David wrote above, right? Or at least you did not understand a word of it. Every member "counts" itself. And every member "counts" a set which ends with itself. The set of natural numbers does not end with any member, and so it is not "counted" by any of its members. Only the last member of such a set "counts" the set, and there is no last natural number to "count" it. So you have the options of declaring that the set can't be counted, or you have to invent an unnatural number which, while not counting the set in the customary way, is supposed to represent the count of the members of a set obeying the Peano axioms. Not by actually doing the counting, but by checking whether the axioms hold, and then declaring the count of the set to be aleph_0, by decree. This does not make aleph_0 a member of the set of naturals. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: albstorz on 25 Oct 2005 10:01 Daryl McCullough wrote: > albstorz(a)gmx.de says... > > >David R Tribble wrote: > > >> Obviously, you do not understand. > >> > >> My set is: > >> S = {0, 2^0, 2^2^0, 2^2^2^0, 2^2^2^2^0, ...} > >> S = {0, 1, 2, 4, 16, 65536, ...} > > >I don't know what kind of math you apply here. Tribble-O-Math? > > It's quite ordinary math, except for the fact that > perhaps you are unfamiliar with the use of ^ to mean > exponentiation? > > 2^0 = 1 > 2^1 = 2 > 2^2 = 2*2 = 4 > 2^3 = 2*2*2 = 8 > 2^4 = 2*2*2*2 = 16 > etc. > > David's sequence is 0, 2^0, 2^(2^0), 2^(2^(2^0)), etc I had read 0, 2^0, (2^2)^0, (2^2^2)^0, etc since he has written: "I've got a set S = {0, 2^0, 2^2^0, 2^2^2^0, ...}, which contains all the powers of 2 of the form 2^p, where p=0 or 2^q." Maybe I had misunderstood. > > 2^0 = 1 > 2^(2^0) = 2^1 = 2 > 2^(2^(2^0)) = 2^2 = 4 > 2^(2^(2^(2^0))) = 2^4 = 16 > 2^(2^(2^(2^(2^0)))) = 2^16 = 65536 > etc. > > -- > Daryl McCullough > Ithaca, NY Yes, there are many possible sequences to build. Regards AS
From: Randy Poe on 25 Oct 2005 10:09
albst...(a)gmx.de wrote: > > David's sequence is 0, 2^0, 2^(2^0), 2^(2^(2^0)), etc > > I had read 0, 2^0, (2^2)^0, (2^2^2)^0, etc > since he has written: > "I've got a set S = {0, 2^0, 2^2^0, 2^2^2^0, ...}, which contains > all the powers of 2 of the form 2^p, where p=0 or 2^q." Which means the elements are 0, and 2 raised to the power of q where q is another element. Thus, since 2^0 is an element, 2^(2^0) is an element, and so is 2^(2^(2^0))), etc. > > Maybe I had misunderstood. Yes. I don't see how to interpret p = (2^2)^0 as 2^q where q is another element of the set. It's the same idea as the Peano definition of the naturals, defining elements as 0, or as the successor to other elements. But here the successor operation is succ(q) = 2^q, instead of succ(q) = q + 1. Peano doesn't specify what the successor operation is. This is a perfectly good Peano set. You could, for instance, do induction on this set. - Randy |