Uncountability of the Rationals?
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From: Virgil on 14 Jun 2010 20:03 In article <1959a4fe-5e75-4ae6-a9e0-df23d18e470b(a)y4g2000yqy.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > I'm of course assuming we're talking about ZFC and standard > > arithmetic and so forth. In your BO ("Bigulosity Ordering"), I > > guess not all sets have a size, so it might very well be possible > > for statement in BO to say "if set N has a size, then ...". > > If it's a countably infinite set then it is expressed with relation to > standard N+. Then you are claiming that EVERY subset of N+ has a well defined bigulosity? If it is an uncountable number of points it begins to > have Lebesgue measure, and can start to be quantified with respect to > zillions. > > > > > Do you have a test in BO for "size" that determines when a > > set has a size and when it does not? > > Countably infinite sets as normally defined have only parametric size > in relation to N+. DO they all have bigulotic "sizes"? > >> Therefore, omega as a definition is self contradictory, and connot > "particularly" exist. But omega as a set is merely N, so that N cannot exist either in Bigulosity.
From: Virgil on 14 Jun 2010 20:08 In article <3b89977b-eb55-4134-803c-2d2b32bd7f64(a)g19g2000yqc.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Don't you make any connection between a set of pairs and, say, spatial > coordinates? When you talk about points in n dimensional space, do you > not define them as unique n-tuples in the spatial set of points? > Hmmmm..... Not necessarily. One can easily have a space without having to have a coordinatized space. What bits of geometry are so dependent on coordinates that they cannot be done without it?
From: Virgil on 14 Jun 2010 20:11 In article <88dcc5f6-04d5-4da9-a4f4-67bc92cca388(a)z10g2000yqb.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 14, 12:04�pm, Brian Chandler <imaginator...(a)despammed.com> > > This is nonsense, of course. "Standard Cantorian cardinality" (SCC) > > doesn't "reject" anything -- but then TP is totally hung up on > > "standard theorists" (or whatever they're called this week) > > "rejecting" things. Anyway, yes, cardinality is a natural extension of > > the "size" comparison of finite sets which preserves the first notion. > > But the simplest measure that preserves the second notion is the > > subset relation: perfectly sound "standard theorists'" maths, but > > inevitably only a partial ordering. Bigulosity simply isn't a theory > > in any normal sense -- just a braindump of Tony's musings. > Eat me, Brian. Is that supposed to be an effective rebuttal, TO?
From: Jesse F. Hughes on 14 Jun 2010 21:49 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 14, 12:15 pm, Brian Chandler <imaginator...(a)despammed.com> > wrote: >> Tony Orlow wrote: >> > On Jun 14, 1:21 am, Brian Chandler <imaginator...(a)despammed.com> >> > wrote: >> > > David R Tribble wrote: >> > > > Tony Orlow wrote: >> > > > > Actually, in Bigulosity, there are much larger countably infinite >> > > > > sets. The square roots of the naturals have a higher "density" than >> > > > > the naturals. >> >> > > > How is that? The two sets should be exactly the same size >> > > > (your size(), not just cardinality). Every natural has a square root, >> > > > and every natural square root has a corresponding natural. >> >> > > Can't you see? You're just using "cardinality" to point out a >> > > bijection between naturals and their square roots. Bigulosity Theory, >> > > or BT, simply rejects such arguments, in its search for the Answer. >> >> > > Consider any large number Q (perhaps a quillion): it is clear that in >> > > the range 1-Q there are Q naturals, but there are (roughly, at least) >> > > Q^2 real values x such that x^2 is a natural. Now appealing to ICI, >> > > PCT, QD, and any other TLAs to hand, declare Tav* to be our unit >> > > infinity, let Q tend to Tav, and lo! the answer we wanted. >> >> > > * Last letter of the Hebrew alphabet according to Wikipedia. >> >> > > It is true that BT, of which I am but a beginning student, leaves >> > > nagging doubts. Suppose, for example we thought about the bigulosity >> > > of the set NQ >> >> > > NQ = { {p,q} | p^2 = q e N} (set of pairs of naturals with their >> > > roots) >> >> > Done. You have an omega X omega^2 matix. You omega^3 elements. >> >> Is this the answer, Dear Leader? "You omega^3 elements"? This sentence >> no verb? Explication possible? >> >> I would have thought that since their is one pair {p,q} for each >> natural q, that there might be omega of these pairs. Or possibly, >> since their is one pair {p, q} for each root of a natural p (of which, >> DL, you say there are omega^2), then there might be omega^2 of these >> pairs. But you tell us there are omega^3?? >> >> I see no matrix, by the way. I also wonder what the bigulosity is of >> the set of pairs {p, p} where p is a natural? >> >> Brian Chandler >> > > Don't you make any connection between a set of pairs and, say, spatial > coordinates? When you talk about points in n dimensional space, do you > not define them as unique n-tuples in the spatial set of points? > Hmmmm..... You didn't answer the question. Do you think that there are omega sets of pairs (p,p) or omega^2? Note that we're not talking about all sets of pairs of naturals, but only those sets where the first and second coordinates are equal. -- Jesse F. Hughes "If you are a consumer that's taking advantage of the technologies that exist ... then the spam problem for you is solved." --MS spokesman verifying that the spam problem has been solved.
From: Brian Chandler on 14 Jun 2010 23:37
David R Tribble wrote: > Tony Orlow wrote: > >> The square roots of the naturals have a higher "density" than > >> the naturals. > > > > David R Tribble wrote: > >> How is that? The two sets should be exactly the same size > >> (your size(), not just cardinality). Every natural has a square root, > >> and every natural square root has a corresponding natural. > >> (I assume you're only dealing with the positive roots.) > > > > Tony Orlow wrote: > > Holy cannoli! There exists a bijection, f:N->S and g:S->N, so indeed > > they have the same cardinality. They do not share the same bigulosity, > > since f and g are not the identity function. The number of square > > roots over omega is omega^2. > > Sorry, but it's not clear what you mean by "the number of square > roots over omega is omega^2". OK, let me try to guess. let Qn be the set of (real) values p<=n, such that p^2 is a natural. (the "roots up to n") Then for any n, |Qn| (call it q(n)) equals (approximately) n^2. So lim (n->oo) q(n) = n^2 Therefore, by a general principle with many names (QD basically), declaring 'omega' to be our unit infinity we have q(omega) = omega^2 Isn't that it? Brian Chandler |