Uncountability of the Rationals?
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From: Tim Little on 4 Jun 2010 00:17 On 2010-06-03, Tony Orlow <tony(a)lightlink.com> wrote: > This is partly because mathematicians misrepresent the theory, > claiming that every conclusion they draw "follows logicallly from > the axioms." The conclusions *do* follow logically from the axioms. > This claim is simply not true, as cardinality is not mentioned in > the axioms Calling a given symbol "cardinality" is just a convenient notational shorthand for a set of logical propositions expressible in terms of the foundation symbols of the theory. That's what is meant by a mathematical definition. > much less anything about omega or the alephs Actually the symbol omega is frequently expressed directly in the Axiom of Infinity of ZF: "There exists omega such that ...". In expressions of the axioms that do not directly use the symbol omega, it is a notational shorthand for such a set. Likewise the aleph symbols are notational shorthands for properties expressed in the language of the theory. > except for a declaration that something homomorphic to the naturals > exists, which is ultimately not a set, but a sequence. In the theory of ZF, it is provable that there exist sets that model the naturals. One example is the set of von Neumann ordinals. > By demanding to know with which axioms a non-transfinitologist > disagrees, mathematicians obscure the question. After some years of > this discussion have arisen some obvious culprits: the von Neumann > ordinals as some complete set The existence of a set satisfying the properties defining the von Neumann ordinals follows as a purely logical deduction from the axioms. You appear to be objecting to nothing more than giving that set a name. > and the simplistic assumption of equivalence of "size" based on > bijection alone. If you don't like the notion of size being determined by cardinality, feel free to come up with a better formalization that doesn't depend on mere labelling of elements of the set. Zuhair did try for a while in this newsgroup, but didn't really succeed. - Tim
From: Tim Little on 4 Jun 2010 00:20 On 2010-06-03, Tony Orlow <tony(a)lightlink.com> wrote: > One might think there were something like aleph_0^2 rationals, but > that's not standard theory. That's perfectly standard theory. It's just that aleph_0^2 = aleph_0. - Tim
From: Tim Little on 4 Jun 2010 00:23 On 2010-06-04, Tim Little <tim(a)little-possums.net> wrote: >> except for a declaration that something homomorphic to the naturals >> exists, which is ultimately not a set, but a sequence. > > In the theory of ZF, it is provable that there exist sets that model > the naturals. One example is the set of von Neumann ordinals. My apologies, "the set of *finite* von Neumann ordinals". > The existence of a set satisfying the properties defining the von > Neumann ordinals follows as a purely logical deduction from the > axioms. Once again, *finite* ordinals. There is no set of all von Neumann ordinals. - Tim
From: Transfer Principle on 4 Jun 2010 02:05 On Jun 3, 3:51 pm, "porky_pig...(a)my-deja.com" <porky_pig...(a)my- deja.com> wrote: > On Jun 3, 1:10 pm, kunzmilan <kunzmi...(a)atlas.cz> wrote: > > Since in both triangles are more different elements than in the first > > row of the matrix R, there are more rational numbers than in the set > > of the natural numbers and the rational numbers are uncoutable. > So we had Herc claiming that reals are countable. now we have > kunzmilan claiming the rationals are uncountable. The plot thickens. At one point, I'd wanted to introduce phrases such as "substandard theorists" and "superstandard theorists." The former would refer to posters like Herc/Cooper who believe in _fewer_ types of infinity than the standard (sub- means below -- Herc believes that the infinity of the reals and the infinity of the naturals are the same.) The latter would refer to posters such as TO and kunzmilan who believe in _more_ typpes of infinity than the standard (super- means above -- these two posters believe that the infinity of rationals and the infinity of the reals are in fact distinct). This would have emphasized that there are posters who oppose ZFC from two different sides -- those who believe that ZFC has too many set sizes and those who believe that ZFC doesn't have enough set sizes. But I've said that I would avoid calling different types of posters as so-and-so "theorists," and so I will not be using those labels at all. I only point out what I once had in mind.
From: Michael Stemper on 4 Jun 2010 09:09
In article <b65b2a33-7335-410a-8da7-91026e63755e(a)q13g2000vbm.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> writes: >Looking at some of Herc's recent threads about Cantor's diagonal proof >of the uncountability of the reals, a question occurs to me which I >actually don't know how a subscriber to transfinite set theory would >answer. I'm curious. > >In Cantor's list of reals, for every digit added, the list doubles in >length. In order to follow his logic, we need not consider any real >numbers which are not rational. In any such list of rationals, it is >always true that we can concoct another rational which is not on the As others have pointed out, the number that you construct is not known to be rational. >BTW, I understand that the rationals are considered countable because >of a rather artificial bijection with N, Better to say "the rationals *are* countable because they biject with N". As a matter of fact, there are many possible bijections between Q and N. The use of the word "artificial" is kind of silly, because most of what we talk about in math is "artificial". I'm not sure how one could define the difference between an "artificial" function and a "natural" one. But, it's not only possible to establish bijections between Q and N, it's even possible to define injections from Q to N that are not onto. In other words, you can map every rational to a unique natural, while still having an infinite set of naturals left over. Express any rational, q, in the form a*(m/n), where a = +/- 1 and m,n are naturals with gcd(m,n) = 1. Define f(q) to be: (2^m)*(3^n) if a == +1 (5^m)*(7^n) if a == -1 Do you see that this maps every rational to a different natural? So, it's an injection from Q to N. Do you also see that no rational gets mapped to 11 or 17 or 42? -- Michael F. Stemper #include <Standard_Disclaimer> Indians scattered on dawn's highway bleeding; Ghosts crowd the young child's fragile eggshell mind. |