An uncountable countable set
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From: David R Tribble on 4 Oct 2006 19:22 Tony Orlow wrote: >> For the sake of this argument, we can talk about infinite reals, of >> which infinite whole numbers are a subset. > David R Tribble wrote: >> Every member of N has a finite successor. Can you prove that your >> "infinite naturals" are members of N? > Tony Orlow wrote: > Yes, if "finite successor" is the only criterion. > > To prove finiteness of such a string: > > The bits over each sequence are indexed by natural numbers, which are > all finite, yes? > > For any finite bit position, the string up to and including that bit > position can only represent a finite value, yes? > > Therefore, there is no bit position where the string can have > represented anything but a finite value, see? If the length is > potentially, but not actually, infinite, so with the value. So you're saying that finite bitstrings can only represent finite naturals. > To prove successorship of such a string: > > The rule for successorship for finite values is > 1. Find the rightmost (least significant) 0 > 2. Invert from that 0 rightwards > > This works for all values where there is a rightmost 0. That excludes > ...111, which can only have successor given ignored overflow, allowable > in some cases. So obviously this rule, given a starting point of 0, a finite natural and a finite-length bitstring, can never produce anything but another finite-length bitstring as a successor. So you've proven that N can contain only finite naturals. Unless you think that your rule allows an infinite bitstring successor to be formed from some finite bitstring? > You don't really question why the successor to ...11110000 is equal to > ...11110001, do you? Again, I can't answer that until you define those numbers in a meaningful way. As you proved above, they are obviously not members of N.
From: Dik T. Winter on 4 Oct 2006 20:49 In article <1159976673.345698.261310(a)h48g2000cwc.googlegroups.com> "MoeBlee" <jazzmobe(a)hotmail.com> writes: > Tony Orlow wrote: .... > > It doesn't have primitive operators 'e'and 'succ'? > > It has the primitive 'S' (read as 'successor'), yes. It does NOT have > 'e'. > > The usual primitives are: > > 0, S, +, *, as well as = from identity theory. As far as I understand, you can even dispense with "+" and "*". But you get recursive definitions of "+" and "*". As I am not very far in logic, I do not know whether that fits entirely. On the other hand, it would be great if Tony (who is a computer scientist) wrote up the computer routines that would do arithmetic on naturals based on the existence of 0 and the S and = operation only. It is possible. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 4 Oct 2006 20:59 In article <1159978513.826507.125470(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > You can read it above. There is the order preserving union of a set of > > > k negative numbers and a set of omega natural numbers including zero. > > > > But that has ordinal k + omega, > > that is what I said! No. > > not -k + omega. A set of k negative numbers > > has ordinal k; not -k. > > But subtraction of a set of positive numbers from the set omega is > expressed, as I did, by -k + omega. Ah, apparently you are defining something new here. > > > > Now try to > > > > apply that definition with -k, where k is positive. What is a set > > > > with ordinal "-k"? Do you know how ordinal addition works? > > > > > > You can find it in Cantor's collected works, e.g., p. 201: "Die > > > Subtraktion kann nach zwei Seiten hin betrachtet werden." And you can > > > read above how it works in this special case. The first k elements are > > > cut off. > > > > And that, again, is quite different from what you did write. But I will > > read and see how Cantor does define omega - 1. A left-handed subtraction > > is indeed possible, but that is not the same as a left-handed addition by > > a negative number. > > No? Who decides that? You see, I knew already that, according to > Cantor, subtraction is possible. If I express this as addition of > negative, what do you think did I meant? I did not know because there is no definition presented nor available. Addition is defined between ordinal numbers. Ordinal numbers are (by their very definition) larger than or equal to 0. But you want to define addition between ordinal numbers and non-ordinal numbers. Pray go ahead, and supply your definitions. (I still have not found the time to see how Cantor defined omega - 1.) > > That is just opinion. > > Like everything else. In that case there is no room for discussion. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Ross A. Finlayson on 5 Oct 2006 00:41 Lester Zick wrote: > On Wed, 04 Oct 2006 10:47:02 -0400, Tony Orlow <tony(a)lightlink.com> > wrote: > > >David R Tribble wrote: > >> Tony Orlow wrote: > >>>> On the other hand > >>>> I don't know why I said "neither can the reals". In any case, the only > >>>> way the ordinals manage to be "well ordered" is because they're defined > >>>> with predecessor discontinuities at the limit ordinals, including 0. > >>>> That doesn't seem "real" > >> > >> Virgil wrote: > >>>> In what sense of "real". There are subsets of the reals which are order > >>>> isomorphic to every countable ordinal, including those with limit > >>>> ordinals, so until one posits uncountable ordinals there are no problems. > >> > >> Tony Orlow wrote: > >>> In the sense that the real world is continuous, and you don't just have > >>> these beginnings with nothing before them. The real line is a line, with > >>> each point touching two others. > >> > >> That's a neat trick, considering that between any two points there is > >> always another point. An infinite number of points between any two, > >> in fact. So how do you choose two points in the real number line > >> that "touch"? > >> > > > >They have to be infinitely close, so actually, they have an > >infinitesimal segment between them. :) > > There you go, Tony. Now between points you're beginning to see some > light. > > ~v~~ Ha ha ha! Virgil's uncountable is unnatural. That is the first word that Virgil ever tried to make a pun about: unnatural. Then, he started trying to talk as if what he said about TO was his. Don't get me wrong, I don't hold that against Virgil. You see, unnatural ordinals would just count backwards from infinity, towards zero. It just reverses the natural ordering. Also, don't get me wrong, I like Virgil. Thankfully, Cantor is talking about infinity, minus one, or so our translators tell us. See also: less than infinity. Definition: everything, including infinity. All a scientist ever wants is to have a research institution named after them. Ross -- Buddy, was a good dog, president Clinton's dog Buddy. He was hit by a car.
From: Han de Bruijn on 5 Oct 2006 03:20
Randy Poe wrote: > Math has to be logical. It doesn't have to be physically > realizable. Wrong. Math has to be an idealization which can be materialized again into something that is physically realizable. Quote: Physicists also > realize that things can exist in mathematics that aren't even > approximations of a physical realizable. That aren't physically > sensible in other words. That's only true for non-disciplinary mathematics. Han de Bruijn |