Uncountability of the Rationals?
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From: Virgil on 15 Jun 2010 20:19 In article <3e8a10bc-7035-492b-9af6-8617c5b2ff62(a)f6g2000vbl.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Any of the copuntably infinite collection of countable > infinities is tied to omega=|N+|, with an algebraic function on |N+|. > The countable set has zero measure with respect to the uncountable. In > that I agree with Lebesgue. "Copuntably"? Besides which, for a set A, |A| is its cardinal, not its ordinal, so |N| = |omega|, but since omega is an rdinal but not a cardinal one does not have omega = |A|, for any set A whatsoever. > > > > > > I guess that to you an n-dimensional space (let n=2 for simplicity) > > consisting of the pairs of naturals > > {a, b} is a "square" of side Tav. I further guess that the "number" of > > these pairs of naturals {a,b} (a and b being independent naturals) is > > Tav^2. > > > > Tav us a bad name, but whether you are talking about omega or a > zillion, the equivalence relation holds. Not with omega! > > > Well, I asked about the set of pairs {p,p}, where (groan) p=p. In > > other words typical members of this set are {1, 1}, {72, 72} and so > > on. {23, 190} is _not_ a member of this set. Since this set is a > > proper subset of the set of _all_ pairs of naturals {a,b}, I believe > > its Bigulosity should be smaller. If Big({(p,p)}) = Tav^2, then > > Big({(a,b)}) must be more than Tav^2. > > No that set has exactly one member for each memeber of N+. By what bijection? It's > basically the actual diagonal of the set, as long as it is wide, with > one element on each row and column, therefore making the height and > width equal. But it is its "area" which is relevant. Consider the cardinalities of Np x Nq, where Nr = {x in N: x <= r}
From: Transfer Principle on 15 Jun 2010 20:22 On Jun 15, 3:15 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 15, 4:40 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > According to Tribble, there are many questions which TO > > can't and won't answer about his theory. One of these > > questions (asked IIRC by MoeBlee) is to which of the axioms > > of ZFC does TO object? But to me, the answer to this > > question is obvious. If a poster disagrees with how the > > infinite sets work under ZFC, then they reject the axiom > > which guarantees their existence -- and that axiom is, of > > course, the Axiom of Infinity. > You must have only skimmed the relevant exchanges in this thread on > the subject of axioms. In that case, let me go back to the references to axioms in order to see TO's opinion on this subject: TO: That it doesn't necessarily exist according to those elementally logical statements, and that I am not obliged by the axioms or that argument to believe that any such absolute size exists for such a set. TO: It does not follow from the axioms, as the mention nothing about set size or cardinality. Aatu's response to TO: Just as PA proves nothing about primes since they aren't mentioned in any of its axioms? MoeBlee: > The formulas you mentioned don't entail that there is such an ordinal > number, true. But we do prove from our AXIOMS that there exists such > an ordinal number. So we can see what's going on here. TO doesn't accept that ZFC proves the existence of (standard) cardinality or ordinality because "card" isn't a primitive symbol that's mentioned in any of the axioms. But MoeBlee points out how that ZFC proves the existence of ordinals despite not being mentioned in the axioms, and Aatu adds that PA does prove the existence of primes even though the word "prime" doesn't appear in the language or axioms of PA. This, of course, goes back to the question that comes up often in these threads. Is it possible for one to accept all of the axioms of a theory, without accepting all of the consequents of those axioms? And if one does reject theorems proved from axioms that he finds unobjectionable, does he deserve to be called by a five-letter insult? It is often mentioned that those called by such insults want more control over what results can be proved, while those who aren't so insulted are delighted when they can prove unexpected results from axioms. (Of course, on sci.math less than a year ago there was an adherent of ZFC+AD who refused to accept that his chosen theory is inconsistent. In other words, he accepts all of the axioms of ZFC+AD, but not the theorem "1=0" which is provable in ZFC+AD. This poster defies classification.) If TO accepts all of the axioms of ZFC, but rejects the theorem "there exists a cardinal (or ordinal) number," then I'll agree to call TO "wrong." Still, I believe that if we can show him a theory which does satisfy his intuitions, he'll have less of a reason to criticize the adherents of ZFC. We notice that TO isn't the first sci.math poster to reject ordinals. The poster tommy1729 also has a low opinion of ordinals as well. Of course, I doubt that TO or tommy1729 rejects the existence of finite ordinals such as 0 or 1 -- evidently, it's the infinite ordinals to which they object. So I believe that the best we can do with regards to preventing as many ordinals as possible from existing would be to switch from ZF to Z. Without the Replacement Schema, we can't prove that omega+omega exists (assuming that the theory is consistent), though we can still prove the existence of omega+n for finite n (as well all of the finite ordinals, of course). We can't do better unless we reject Infinity altogether. But TO goes a step further than tommy1729, for the latter accepts the existence of all standard cardinals up to aleph_omega (or aleph_aleph_0), since aleph_(omega+1) has the objectionable omega+1 as its index. TO, on the other hand, rejects all infinite cardinalities and proposes that Bigulosities be used instead of cardinalities. Of course, one can still attempt to assign every set whose existence is guaranteed in ZFC a Bigulosity. And if so, then we might have the following theorem: If two sets have the same Bigulosity, then they have the same cardinality as well.
From: Virgil on 15 Jun 2010 20:25 In article <ab487c14-ebbd-400f-b626-4aaf288adf03(a)b35g2000yqi.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Just because an infinite set is defined by the fact that it can be > bijected with a proper subset doesn't mean that this bijection implies > the counterintuitive equinumerosity that one would NOT expect from a > "correct" theory. Where, besides in your own mind, do you find anything declaring YOUR theory "correct" and any other theory "incorrect. Certainly intuitions, particularly those of sloppy amateurs like TO, are not binding on mathematics as a whole.
From: Virgil on 15 Jun 2010 20:29 In article <f614aae3-8fa0-4214-b9bf-86d0aab5cbf4(a)j4g2000yqh.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Please step back, and consider the finite case. In -> the finite case <- for a set , purely as a set, its "size" is the number of members it has.
From: Virgil on 15 Jun 2010 20:34
In article <91171084-ab38-448e-9a78-3101085da24d(a)u7g2000yqm.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 15, 6:07�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <0cef0e10-6e81-4031-9673-2df1cad85...(a)k39g2000yqb.googlegroups.com>, > > as in 2*N^2 + 1? > > But is such an N an ordinal (omega) or a cardinal (aleph_0)? > > > > It makes considerable difference! > > > > And why (ordinal vs cardinal)? > > Those terms are meaningless in Bigulosity. It's a "count". It's not even a "knight". Since both ordinality and cardinality already exist, and differ, one must be careful to use them correctly, even in "Bigulosity". |