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From: Tony Orlow on 25 Oct 2005 16:06 David R Tribble said: > David R Tribble: > >> 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, ...} > > Albrecht Stortz: > > I don't know what kind of math you apply here. Tribble-O-Math? > > I don't want discuss your very interesting system. > > Look at my starting posting if you want to know in what I'm interested > > to discuss. > > Start a new thread if you search for people which want talk about > > Tribble-O-Math. > > It's obvious that you don't understand simple arithmetic. 2^x is 2 > raised to the x power. 2^2^x is 2^(2^x), using ordinary arithmetic. > > I'm giving you set S so that you can tell us whether it is an > infinite set or not. You said: > > >> either there are infinite natural numbers or there is no infinite set. > > So I'm giving you set S, which obviously does not contain any > infinite numbers. So by your rule, the set is finite, right? > > If it doesn't contain any infinite members, it's not infinite. Those terms differ by more than a constant finite amount, but rather a rapidly growing amount greater than 1. There is no way you have an infinite number of them without achieving infinite values within the set. -- Smiles, Tony
From: Randy Poe on 25 Oct 2005 16:06 Tony Orlow wrote: > Virgil said: > > What "last element" is talking about? > The one that completes the set and corresponds to the last bit in the mapping > natural. The set is complete. It has no last element. - Randy
From: Tony Orlow on 25 Oct 2005 16:08 David R Tribble said: > David R Tribble wrote: > >> 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. > > > > Albrecht Storz wrote: > > 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. > > This may be obvious to you, but it is not to anyone who understands > infinite sets. > > > David R Tribble wrote: > >> Your contradiction is wrong. The contradiction comes from saying > >> that the size of the set is also a member of the set. > > > > Albrecht Storz wrote: > > 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. > > You are claiming that one of the members of the infinite set is > equal to the size of the set. This is not obvious to us because it > contradicts proven set theory. It is your responsibility to prove > that your claim is true, and that standard set theory is wrong. > Saying that your claim is "obvious" is not a proof. > > His claim is obvious based on the diagram he offered, which graphically shows the element values along one side of a growing square, and the element count along the other side. They are obviously, graphically, inductively, always equal. One is not infinite while the other is finite. It's impossible. It is these kinds of pictures the axiomatikers need to practice seeing. -- Smiles, Tony
From: William Hughes on 25 Oct 2005 16:09 Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > Virgil said: > > > > <snip> > > > > > > > So, which element of the power set does not have a natural mapped to it? > > > > > > > > {x in S:x not in f(x)} > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > No, just the entire set. The set does not have a last element. > > > > - William Hughes > > > > > Then there is no last bit to the natural that maps to it, so it cannot be truly > identified, can it? Oh, I see, the natural that maps to the entire set is another one of the things that exist but cannot be specified. - William Hughes
From: David Kastrup on 25 Oct 2005 17:53
Tony Orlow <aeo6(a)cornell.edu> writes: > David Kastrup said: > >> 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. > > A decree does not make it correct either. Well, one knows that there is no natural number depicting the size. > The first option is better, to admit that the set size is equal to > the largest member, but that there is no identifiable largest member > and therefore no idnentifiable set size, which makes sense given the > lack of endpoint for measure. This has, of course, the advantage that one need not deal with number-like entities which behave rather peculiarly when doing arithmetic with them. It has the disadvantage that there happens to be a hierarchy of surjectability even between infinite sets. And those sets can be compared in the context of surjectability, and it turns out that one can group them in equavalence classes. And that makes it convenient to also have a _name_ for the cardinality of the naturals and other infinite sets. This name can be used for sorting; and a few rules concerning the arithmetic can be derived, too, when deriving those rules from what happens to cardinality when you form the union of disjoint sets. It _is_ a valid stance to leave the cardinality unnamed or unspecified or unidentifiable or whatever else. But it is stopping short of what _can_ be achieved in a consistent manner. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |