An uncountable countable set
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From: Lester Zick on 8 Oct 2006 14:40 On Sat, 07 Oct 2006 23:45:23 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >mueckenh(a)rz.fh-augsburg.de wrote: >> Tony Orlow schrieb: >> >>> mueckenh(a)rz.fh-augsburg.de wrote: >>>> Tony Orlow schrieb: >>>> >>>> >>>>>>> Why not? Each and every number of the list terminates. That one is a number >>>>>>> that does *not* terminate. >>>>>>> >>>>>>> > If you think that 0.111... is a number, but not in the list, >>>>>> It is me who insists that it is not a representation of a number. >>>>> Well, Wolfgang, that sets us apart, though I agree it's not a "specific" >>>>> number. It's still some kind of quantitative expression, even if it's >>>>> unbounded. Would you agree that ...333>...111, given a digital number >>>>> system where 3>1? >>>> That is the similar to 0.333... > 0.111.... But all these >>>> representations exist only potentially, in my opinion. The difference >>>> is, that 0.333... can be shown to lie between two existing numbers, so >>>> we can calculate with it, while for ...333 this cannot be shown. >>> I think it can be shown to lie between ...111 and ...555, given that >>> each digit is greater than the corresponding digit in the first, and >>> less than the corresponding digit in the second. >> >> Yes, but only if we define, for instance, >> >> A n eps |N : 111...1 < 333...3 where n digits are symbolized in both >> cases. >> >> This approach would be comparable with the "measure" which gives >> >> A n eps |N : |{1,2,3,...,2n}| = 2*|{2,4,6,...,2n}|. >> >> I don't know whether these definitions are of any use, but I am sure >> that they are not less useful than Cantor's cardinality. >> >> Regards, WM >> >> . >> > >My opinion about that is, if one wants to talk about what happens "at >infinity", that's the way that makes sense, not the measureless way of >abstract set theory. I trust limit concepts, but not limit ordinals. Tony, would it be fair to characterize what you're trying to say as that there is some kind of positive/negative crossover at infinity such that {-00, . . .,-1, 0, +1, . . . +00}? I haven't really been following this thread too closely so I'm trying to understand what you're after here in basic terms instead of the exact arguments involved. ~v~~
From: Tony Orlow on 8 Oct 2006 14:51 Virgil wrote: > In article <45287184(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> David R Tribble wrote: >>> 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. >>> >> Strings with only finite bit positions. > > Wrong!!! Strings with only finite bit positions can still have > infinitely many bit positions as there are infinitely many finite > naturals. Finite naturals always have a finite most significant bit > position and only finitely many non-zero digits. Incorrect. If every bit is in finite position, then there is no location in the string where it can be said to have an infinite value. > >>> 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 will not produce 1 bits in infinite positions without an infinite >> number of successions. > > Which the naturals forbid. Correct. Given the restriction of finiteness on the values of the elements and the constant finite difference between successors, it is impossible for N to have "an infinite number of successions", even if the bound on the successions is undefinable. >>>> 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. >>> >> 1) ....00000 is a number. > > It is a digit string, and might be a numeral, but without a suitable > context it is not a number. > That's the equivalent of Peano Axiom 1. I declared it a number, and a member of the set. So, it's a number. > >> 2) If x is a number, then the successive number, formed by inverting the >> rightmost 0 and all 1's to the right of it, is also a number. > > And if it is NaN then neither are any of those other things. I declared it a number. What makes Peano's '0' a number? Consider it a member of the complete binary language, which set may be interpreted as numbers.
From: Lester Zick on 8 Oct 2006 14:53 On 8 Oct 2006 05:11:25 -0700, mueckenh(a)rz.fh-augsburg.de wrote: > >Virgil schrieb: > >> In article <ac6c7$45260f70$82a1e228$27946(a)news2.tudelft.nl>, >> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >> >> >> > Sure. Simply "define" something that's undefined. And create the self >> > fulfilling prophecy that suits you best. >> >> Beats hell out of defining things that are already defined. >> >> Every definition worth having defines something that would be undefined >> without that definition. > >A definition is an abbreviation. Nonsense. The name assigned to a definition is an abbreviation for the series of predicates or properties constituting that definition in sequence and in relation to one another. ~v~~
From: Tony Orlow on 8 Oct 2006 14:55 Virgil wrote: > In article <45286ce5$1(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> David R Tribble wrote: > >>> What about: >>> sum{n=0 to oo} (10n+1 + ... + 10n+10) - sum{n=1 to oo} (n) >>> The left half specifies the number of balls added to the vase, and >>> the right half specifies those that are removed. >>> >> Do you mean: >> sum{n=0 to oo} (10) - sum{n=0 to oo} (1)? >> That sounds like what you re describing, and termwise the difference is >> sum(n=0 to oo) (9). That's infinite, eh? > > But the sums are not given termwise in the question, but sumwise, so > cannot be calculated termwise in your answer, but must be done sumwise. > > And sumwise they are no different. That's bass-ackwards. The gedanken specifically states that the insertion of 10 and the removal of 1 are coupled as an iteration in the process under discussion. Between every removal of 1 and the successive removal of 1 is an insertion of 10. So, each term '+10' must be coupled with a term '-1', which together make a '+9' per iteration. Where you violate the condition of the experiment that specifies this coupling of events into an iteration, you create your "discontinuity" at noon, and the monster under your bed.
From: Lester Zick on 8 Oct 2006 14:58
On 8 Oct 2006 05:20:24 -0700, mueckenh(a)rz.fh-augsburg.de wrote: > >Lester Zick schrieb: > >> On Fri, 06 Oct 2006 12:52:43 -0600, Virgil <virgil(a)comcast.net> wrote: >> >> >In article <1160122708.108138.77770(a)e3g2000cwe.googlegroups.com>, >> > mueckenh(a)rz.fh-augsburg.de wrote: >> > >> > >> >> I am sure, the results "all balls in B" and "all balls not in B" are >> >> not to be interpreted as an actual contradiction of set theory. It is >> >> just counter intuitive. >> >> >> >> Regards, WM >> > >> >It seems to be intuitive in "Mueckenh" 's world, but it is not only not >> >intuitive, it is not true in any set theory of my acquaintance. >> >> So it's counter intuitive. What's the problem? Surely it isn't the >> first counter intuitive suggestion you've ever run across. > >I did not know that the simultaneous existence of A and ~A is not a >contradiction but only counter intuitive. Now I learned it, and I find >ZFC is not very attractive to me. Well technically the simultaneous existence of A and not A is only contradictory and not counter inutitive because that relation defines the existence of space. If you said "A is not A" that would be both self contradictory and counter intuitive. Doesn't mean it can't happen mechanically in intuitive terms just that it's mechanically useless. ~v~~ |