An uncountable countable set
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From: stephen on 3 Oct 2006 13:48 Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > stephen(a)nomail.com wrote: >> Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: >> >>>stephen(a)nomail.com wrote: >> >>>>Han.deBruijn(a)dto.tudelft.nl wrote: >>>> >>>>>Worse. I have fundamentally changed the mathematics. Such that it shall >>>>>no longer claim to have the "right" answer to an ill posed question. >>>> >>>>Changed the mathematics? What does that mean? >>>> >>>>The mathematics used in the balls and vase problem >>>>is trivial. Each ball is put into the vase at a specific >>>>time before noon, and each ball is removed from the vase at >>>>a specific time before noon. Pick any arbitrary ball, >>>>and we know exactly when it was added, and exactly when it >>>>was removed, and every ball is removed. >>>> >>>>Consider this rephrasing of the question: >>>> >>>> you have a set of n balls labelled 0...n-1. >>>> >>>> ball #m is added to the vase at time 1/2^(m/10) minutes >>>> before noon. >>>> >>>> ball #m is removed from the vase at time 1/2^m minutes >>>> before noon. >>>> >>>> how many balls are in the vase at noon? >>>> >>>>What does your "mathematics" say the answer to this >>>>question is, in the "limit" as n approaches infinity? >> >> >>>My mathematics says that it is an ill-posed question. And it doesn't >>>give an answer to ill-posed questions. >> >> That is a perfectly reasonable answer. But you do agree that >> for this problem, the vase is empty at noon for any finite n. >> So one wonders what criteria you used to determine that >> this infinity cannot be approached via limits. > We can say that the number of balls Bk at step k = 1,2,3,4, ... is: > Bk = 9 + 9.ln(-1/tk)/ln(2) where tk = - 1/2^(k-1) for all k in N . > And that's ALL we can say. The version of the problem used here is > the first experiment in: > http://groups.google.nl/group/sci.math/msg/d2573fcb63cbf1f0?hl=en& > Han de Bruijn Why can't we say that every ball that is added is also removed? ball #m is added to the vase at time 1/2^(floor(m/10)) minutes before noon. ball #m is removed from the vase at time 1/2^m minutes Every ball is removed before noon, no matter how many balls there are. Stephen
From: Lester Zick on 3 Oct 2006 14:20 On 3 Oct 2006 10:13:54 -0700, imaginatorium(a)despammed.com wrote: >Lester Zick wrote: > ><snip lesser maundering...> > >> But if the limit concept is the only sensible way to approach the >> infinite, one is left to ponder whether the limit concept is infinite >> and if not how a limits to the limit concept can be approached? > >Beautifully put, Lester, but surely the only way would be by >regression? Naturally by finite tautological regression to self contradictory alternatives. Or subdivision. >(Try interplanetary travel, Lester - the sky's the limit there!) Fraid not, Brian. Planetary travel is about as far as we're likely to get. Besides the sky is falling anyway. ~v~~
From: Virgil on 3 Oct 2006 14:58 In article <452279ce(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > >>> David R Tribble wrote: > >>>>> So I ask again, where are those infinite naturals and reals you keep > >>>>> talking about? It's obvious they are not in N. > >>> Tony Orlow wrote: > >>>> [No] it's not. > >>> Every member of N has a finite successor. Can you prove that your > >>> "infinite naturals" are members of N? > >>> > >> Yes, if "finite successor" is the only criterion. > > > > But it is not. In addition, N must be minimal with respect to closure > > under successorship, which excludes TO's "infinite naturals". > > Which of Peano's axioms states such a requirement? The last one: If a property holds for the first natural, and holds for the successor of every natural number for which it holds, then the property holds for all natural numbers. (This axiom of induction is also called "mathematical induction". Its point is to limit the natural numbers to just those which are required by the other axioms.) ERGO: If the first natural is finite and the successor of a finite natural is a finite then every natural is finite. > > Von Neumann's N is such that > > (1) {} is a member of N > > {} is the empty set, not a natural number. > > > (2) If x is a member of N then so is x \union {x} > > (3) If S is an subset of N such that > > (3.1) {} is a member of S, and > > (3.2) if x is a member of S then so is x \union {x} > > then S = N. > > > > And that is, by general consent, a satisfactory model for N. > > You may be satisfied with it. Von Neumann was, and the vast majority of those who followed have been. That TO is not only bolsters my confidence in it. > > > > > But that model cannot contain any of TO's allegedly infinite naturals. > > Well, it can. Let S be the all finite naturals in vN , the set of von Neumann naturals, then (1) {} is a member of S (2) if x is a member of S then so is its successor Thus S = vN, and there is no room in vN for TO's illusionaries. > > > > > (1) there cannot be a first infinite natural, and, > > No, there can't, but you have a fallacious first infinite ordinal, so > why complain? Because TO is trying to put objects into vN that cannot be there. > > > (2) since this N can be proved to be well ordered, > > Therefore, every subset of N with no first element must be empty. > > My N is not "well ordered" any more than the reals. Then it is not vN, and does not satisfy the Peano properties, and is irrelevant in any discussion of natural numbers.
From: imaginatorium on 3 Oct 2006 16:46 Lester Zick wrote: > On 3 Oct 2006 10:13:54 -0700, imaginatorium(a)despammed.com wrote: > > >Lester Zick wrote: > > > ><snip lesser maundering...> > > > >> But if the limit concept is the only sensible way to approach the > >> infinite, one is left to ponder whether the limit concept is infinite > >> and if not how a limits to the limit concept can be approached? > > > >Beautifully put, Lester, but surely the only way would be by > >regression? > > Naturally by finite tautological regression to self contradictory > alternatives. Or subdivision. Hmm. Not sure what self-contradictory alternatives are. How can something be alternative to something that's self-contradictory? Oh, but I see, we could just use subdivision. Wow - that really would _never_ have occurred to me. But I'm sorry, I don't quite see what we would be subdividing; is there any chance you could clarify? (Hope the answer won't entail any of these self-contradictory alternatives...) Brian Chandler http://imaginatorium.org
From: Tony Orlow on 3 Oct 2006 18:17
Virgil wrote: > In article <452279ce(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > >>>>> David R Tribble wrote: >>>>>>> So I ask again, where are those infinite naturals and reals you keep >>>>>>> talking about? It's obvious they are not in N. >>>>> Tony Orlow wrote: >>>>>> [No] it's not. >>>>> Every member of N has a finite successor. Can you prove that your >>>>> "infinite naturals" are members of N? >>>>> >>>> Yes, if "finite successor" is the only criterion. >>> But it is not. In addition, N must be minimal with respect to closure >>> under successorship, which excludes TO's "infinite naturals". >> Which of Peano's axioms states such a requirement? > > The last one: > If a property holds for the first natural, and holds for the successor > of every natural number for which it holds, then the property holds for > all natural numbers. (This axiom of induction is also called > "mathematical induction". Its point is to limit the natural numbers to > just those which are required by the other axioms.) Its point is to define the inductive set. There is no indication there that an uncountably infinite number of successions can't occur. > > ERGO: If the first natural is finite and the successor of a finite > natural is a finite then every natural is finite. > Only for a finite number of successive increments. > > >>> Von Neumann's N is such that >>> (1) {} is a member of N >> {} is the empty set, not a natural number. >> >>> (2) If x is a member of N then so is x \union {x} >>> (3) If S is an subset of N such that >>> (3.1) {} is a member of S, and >>> (3.2) if x is a member of S then so is x \union {x} >>> then S = N. >>> >>> And that is, by general consent, a satisfactory model for N. >> You may be satisfied with it. > > Von Neumann was, and the vast majority of those who followed have been. > > That TO is not only bolsters my confidence in it. Good. >>> But that model cannot contain any of TO's allegedly infinite naturals. >> Well, it can. > > Let S be the all finite naturals in vN , the set of von Neumann > naturals, then > (1) {} is a member of S > (2) if x is a member of S then so is its successor > Thus S = vN, and there is no room in vN for TO's illusionaries. >>> (1) there cannot be a first infinite natural, and, >> No, there can't, but you have a fallacious first infinite ordinal, so >> why complain? > > Because TO is trying to put objects into vN that cannot be there. I reject the von Neumann ordinals as an inadequate model of the naturals. >>> (2) since this N can be proved to be well ordered, >>> Therefore, every subset of N with no first element must be empty. >> My N is not "well ordered" any more than the reals. > > Then it is not vN, and does not satisfy the Peano properties, and is > irrelevant in any discussion of natural numbers. Not really. |