Uncountability of the Rationals?
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From: Brian Chandler on 14 Jun 2010 12:15 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 I > Next.... > > > > > But I suspect that our Leader will determine that this is a Trick Set, > > to which special rules apply. We may have to declare a new Unit > > Infinity, or Recalibrate the naturals, or something. But you'll have > > to wait for the answer till he comes back from lunch. > > > > Brian Chandler
From: Michael Stemper on 14 Jun 2010 13:14 In article <389b7d8c-cd2d-4351-b7e5-7ea4834aab5c(a)y11g2000yqm.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> writes: >On Jun 10, 5:06=A0pm, MoeBlee <jazzm...(a)hotmail.com> wrote: >> > > You retreat every time you fail to define your terminology in a non- >> > > circular way, every time you fail to answer the substantive questions >> I said circular TERMINOLOGY. You have no primitives, so your >> terminology is never grounded. It either goes circularly (and I showed >> you instances of that a few years ago), or it just dangles at terms >> themselves not defined not primitive. >> Or, please, I'd love for you to list your exact primitives: constants, >> predicate symbols, function symbols (or words that signal such >> formalisms). >Moe - You seem to think that coming up with a complete alternative to >transfinite set theory from the ground up is a simple matter to be >easily spewed out. I doubt that he thinks that. But, until you have primitives and axioms, you have nothing. You start with them, and work out their consequences. If you don't like the consequences, you change them and try again. But, until you have some primitives and axioms, you don't have anything at all. -- Michael F. Stemper #include <Standard_Disclaimer> The FAQ for rec.arts.sf.written is at: http://www.geocities.com/evelynleeper/sf-written Please read it before posting.
From: David R Tribble on 14 Jun 2010 14:19 David R Tribble wrote: >> You need to stop beating around the bush and just go ahead >> and state your primary axiom: >> Tony's Axiom. [TA1] >> Given property P(x) for all real x, >> P(w) for any non-finite ordinal w. > Brian Chandler wrote: > No. Even Tony isn't quite that stupid. See, there's a property F > called "finite". And everyone knows that even though F(x) is true for > all (finite) real x, F(x) is false for all non-finite x. Serves me right for trying to think like Tony. Oh well, this is easily solved! The problem is that TA1 is too general, so we need to limit its scope. Instead of any property P, let's just consider arithmetic properties and operators. So, perhaps something like this: Tony's Axiom [TA2] Given arithmetic function f(x) = y, where x and y are reals, then f(w) = u, where w and u are infinite ordinals. Yeah, I know this really isn't any better when taken any further, but Tony really needs to put his money where his mouth is and just go ahead and write his assumptions down already. He can't proceed with any of his other statements about bigulosity, ICI, IFR, or whatever until he starts with an axiom or two. And he knows it. > Tony's answer > to this conundrum is a piece of verbiage I can never remember exactly, > something about "limit", "zero", and "tending", but in practice it > means that TA1 is qualified by a condition equivalent to "provided > Tony can't see any immediately derived contradiction". > See the long rambling thread about the "staircase limit". Tony's > intuition says that a staircase is always a staircase, therefore the > limit of the staircases is not a normal point set (which would have > the expected length 1/root(2)) but an "infinite staircase", a new sort > of point set in which alternate points are 90 degrees apart in > orientation. Yeah, I remember some of those discussions. Another million brain cells wasted. > So it is obvious there will never be a coherent version of TA1, since > it's just part of Tony's Belief System, which is indefinitely elastic > with respect to new discoveries. Oh, I think Tony's still working under the same belief structure he's had since he started posting on sci.math. He seems quite creative when it comes to shoe-horning little eccentricities and "discoveries" into his private theory of sets. Just a few days ago, for example, he posited his tired old statement that N+, being of size omega, must therefore contain an omega-th element. That particular piece of coal goes back a number of years. So I don't think he's really actually changed any of his original ideas. > **Tony does not understand the > concept of axiomatic mathematics, so you are wasting your time.** Tony's has never given us any reason to doubt this.
From: David R Tribble on 14 Jun 2010 14:48 Tony Orlow wrote: >> However omega cannot exist within the set. Consider this. >> Are the following two statements logically equivalent? >> AneN n<omega >> ~EneN n>=omega >> >> Does either imply, actually, that such a size exists? No. Both assert >> that no natural number will suffice. However, this is not to say that >> there exists ANY number which does. Logically, you must concede, >> neither statement proves that omega exists as a number. > David R Tribble wrote: >> Yes, that's correct. Neither of your two statements implies that >> there exists any number equal to card(N). >> Neither statement implies card(N) does not exist, either. > Tony Orlow wrote: > Nope. They simply say that IF N has a size, it cannot be a member of N. Neither statement says anything about the size of N, unless you're taking omega to indeed be the definition of the size of N. In which case you're attempting to disprove the definition of omega. But then, definitions are not "proved" or "disproved" in logic. But you already knew that. Also, neither statement says anything about "If" N has a size. 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 ...". Do you have a test in BO for "size" that determines when a set has a size and when it does not? Tony Orlow wrote: > I never said the two statements I gave above prove the nonexistence of > |N|, but that the statements also do not state that it exists, only > that it cannot be any natural number, if it exists. Yes, omega is not a natural. We already knew that. I suppose then that you had some bigger point to make about omega not being a natural? > Bigulosity of countably infinite sets is always in functional relation > to omega, but omega is more of a limit than a number. Now you're saying that omega is not a number. How then can you say (in a previous post) that there must be an omega-th (indexed) member of N+? That appears to be two contradictory points of view about how "omega" acts. Tony Orlow wrote: > With IFR, N is hardly the smallest caountably infinite set, consider > the evens, squares of naturals, natural powers of 2, etc. Yes, in BO that may be the case. It's hard to tell right now, though, because we still don't really know what you mean by "size" and "less than" in BO. And does the old bijection cardinality still work in BO or not? Maybe only sometimes?
From: David R Tribble on 14 Jun 2010 14:56
Tony Orlow wrote: >> ELiminating proper subset equinumerosity is worthwhile. > David R Tribble wrote: >> That kind of statement hints at a deep misunderstanding of >> the way mathematics works. >> >> You seem to be suggesting that by inventing an alternative way >> of measuring set sizes (a more "intuitive" way), that the existence >> of bijections between infinite sets and their proper infinite subsets >> will disappear or stop working. > Tony Orlow wrote: > You mean like your ability to read and retain? Remember that IFR is > EXACTLY dependent of the establishment of a bijection. The definition > of an infinite set as one wherein a bijection may be established > between it and a proper subset works for both the countably and > uncountably infinite. No one said bijections would disappear, unlike > that infinite number of balls in the vase when you change their > labels. Then likewise, bijections between infinite sets and their infinite proper subsets, what you call "proper subset equinumerosity", will not disappear either. Yet you specifically said you wanted to eliminate it (see quoted text above). Obviously IFR can't do this. David R Tribble wrote: >> Unless you invent a system of "sets" that flat out does not allow >> relations like injections and surjections to be formulated at all, >> you can't do this. Either that or come up with a system having >> completely different meanings for "subset" and "mapping". > Tony Orlow wrote: > I don't know what you think you're blathering about. I'm responding to your desire to "eliminate proper subset equinumerosity", simply pointing out that inventing a new kind of "set size" cannot accomplish this. |