infinity ...
in [Math]
|
Prev: math
Next: The proof of mass vector.
From: imaginatorium on 26 Oct 2005 12:01 William Hughes wrote: > Tony Orlow wrote: > > William Hughes said: > > > > > > Tony Orlow wrote: > > > > David R Tribble said: > > > > > Tony Orlow wrote: > > > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > > > > > > > Virgil said: > > > > > >> {x in S:x not in f(x)} > > > > > > > > > > > > > > > > Tony Orlow wrote: > > > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > > > > > Your mapping does not include any natural that maps to the entire set > > > > > *N, which it must do in order to be called a bijection, since *N is one > > > > > of the members of P(*N). Show us that one single mapping, please. > > > > It would obviously be the natural with an infinite unending strings of 1's, > > > > right? Ah, but you want to know, how MANY 1's? > > > > > > No, this is irrelevent. No natural has a binary representation that > > > consists only of 1's. > > I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit > > the bill. Why do you say I cannot declare that the 1's go on indefinitely? We > > are talking about *N, the set including infinite naturals, with infinite > > numbers of bits. When will I run out of bits for my subset elements? Remember, > > there is no last element, as you remind me again, below. > > > You do not "run out of bits". > The reason that no natural number can > have a binary representation composed only of ones > is that every natural has a successor which also > has a binary representation. > > Suppose that a natural n > had a binary representation that is composed only of ones. > What could the representation of n+1 be? It cannot consist > only of ones (this is the binary representation of n). It cannot > have a 0, as a binary representation containing a 0 is the binary > representation of a number smaller than the number with a binary > representation consisting only of ones. Thus n+1 cannot have a binary > representation. Contradiction. Hmm, doesn't this just mean it must be the last one? After all, the principle that you can always add one to an integer and get another integer must be applied with care, lest some previous declaration be rendered invalid. Brian Chandler http://imaginatorium.org
From: Robert J. Kolker on 26 Oct 2005 13:08 Tony Orlow wrote: >> > > Perhaps. Cardinality is better than nothing. And yet, I think that noting that > the size really is not a number with an absolute value but needs to have some > value applied to it in order to measure it, You continually confuse "how many" with "how much". The former is denominated by cardinal numbers. That later by some form of measure. Google <measure theory>. The simplest sorts of measures are length, area, volume etc. and generalizations thereof. Mathematicians do not confuse these two kinds of quantities, but you do, therefore you are not a mathematician. Your inability to learn from your betters indicates strongly that you will never learn to be one. Before you criticize well established mathematical theories you should learn what they are first. You have not done this and you show no indication of ever doing so. Bob Kolker
From: Virgil on 26 Oct 2005 14:40 In article <MPG.1dc956a36ff838c098a571(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > David Kastrup said: > > 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. > > > > > Perhaps. Cardinality is better than nothing. And 'nothing' is what TO's set sizes amount to. Ther are a number of ways of measuring different aspects of sets, any of which can be clled "size". None of them, except for cardinality, allow comparison of arbitrary sets. For example, set inclusion cannot compare sets neither of which is a subset of the other. similarly, non-numeric sets cannot be compared bases on "value ranges" since non-numeric objects cannot be subtracted one from another to get "ranges". > And yet, I think that > noting that the size really is not a number with an absolute value > but needs to have some value applied to it in order to measure it, > leads to more fruitful ways of dealing with the set. Their is no such thing as 'the' size of the set, there are many sizes, depending on which proprties one is looking at. But only the cardinality size can be applied to every set. all other types of size are limited in what sets they can compare. > I hope that's okay, even if it sounds like babbling > hogwash. Does deep hogwash run still? Hmmm.... If TO is content to wallow in his hogwash, ...
From: David R Tribble on 26 Oct 2005 14:52 David R Tribble said: >> [Alrecht] 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. > Tony Orlow wrote: > 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. His second diagram is an infinite list of member sets, where each set contains a finite number of leading #s and an infinite nubmer of trailing 0s. Each member set is associated with a finite natural, so the entire set is the same size as the set of finite naturals (N). There's not much to say about his diagram, though, because it doesn't show any contradictions in standard set theory. It also does not show that the size of the set is a member of the set.
From: David R Tribble on 26 Oct 2005 14:52
David R Tribble said: >> So I'm giving you set S, which obviously does not contain any >> infinite numbers. So by your rule, the set is finite, right? > Tony Orlow wrote: > 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. Yes, you and Albrecht keep saying that repeatedly. Please demonstrate why it must be so, because it's not. |