Uncountability of the Rationals?
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From: Virgil on 17 Jun 2010 15:31 In article <a2159aab-1d41-43bc-860a-e0fca1539617(a)i31g2000yqm.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 7:45�pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > >> Sometimes there are logical arguments themselves which don't appeal to > > >> everyone equally. What I have been told is that the size of omega must > > >> be larger than every natural since no natural is large enough to > > >> express it, and so it is some infinite number, aleph_0. However, that > > >> logical argument, as I pointed out, really just proves that aleph_0 > > >> cannot be finite. Along with the argument that any initial segment of N > > >> + of size x contains an xth element whose value is x, which would > > >> imply that aleph_0 or omega is a member of N+, we arrive at a > > >> contradiction implying that aleph_0 cannot actually exist. Which axiom > > >> contradicts this logic? > > > > David R Tribble wrote: > > >> The problem is that there is no logic there. > > >> You consider finite sets of the form {1, 2, 3, ..., k}, all of which > > >> have a least and a greatest member, to have size k and to also > > >> have k as a member. Then you take an unjustified leap and > > >> claim that this implies that sets of the form {1, 2, 3, ...}, > > >> which do not have a largest member, to also have a size equal > > >> to one of their members. > > > > Tony Orlow wrote: > > > No, they all have a size, and if that size is n, then there exists an > > > nth. If there does not exist an nth, then there are not n elements in > > > the sequence, which this set also is. > > > > But what if a set has an nth member, for every n in N+, but has > > no largest member? Does the set still have a size? Is that size > > a member of the set? If so, is it the largest member or some > > other member? > > Is that supposed to answer my question? Is that the Axiom of > Avoidance? Which axiom did I violate? > > If the set is an initial segment of N+ with no largest member it has > no size. I have said this about a zillion times (or at least tav). Since we did not believe it the first max(zillion,Tav) times you said it, why should we believe it now? > > > > > David R Tribble wrote: > > >> So the question becomes, what axiom justifies your logical > > >> leap, applying a property of finite sets to infinite sets without > > >> largest members? > > > > Tony Orlow wrote: > > > There is no leap. Then there can be no difference between finite sets and infinite sets, thus everything true of finite sets must be true of infinite sets, thus all infinite sets must be finite since that is true of finite sets. At least following TO's logic. > > > > By your logic, you state that all sets of the form {1,2,3,...,k} > > have a largest member, therefore sets of the form {1,2,3,...} > > also have a largest member. I assume you have an axiom > > or theorem in mind to help you reach this conclusion. > > No, read carefully. I said that every initial segment of N+ with size > x has an xth element with value x. Not for the set of naturals most of us use.
From: David R Tribble on 17 Jun 2010 16:44 Tony Orlow wrote: >> Sometimes there are logical arguments themselves which don't appeal to >> everyone equally. What I have been told is that the size of omega must >> be larger than every natural since no natural is large enough to >> express it, and so it is some infinite number, aleph_0. However, that >> logical argument, as I pointed out, really just proves that aleph_0 >> cannot be finite. Along with the argument that any initial segment of N+ >> of size x contains an xth element whose value is x, which would >> imply that aleph_0 or omega is a member of N+, we arrive at a >> contradiction implying that aleph_0 cannot actually exist. Which axiom >> contradicts this logic? > David R Tribble wrote: >> The problem is that there is no logic there. >> You consider finite sets of the form {1, 2, 3, ..., k}, all of which >> have a least and a greatest member, to have size k and to also >> have k as a member. Then you take an unjustified leap and >> claim that this implies that sets of the form {1, 2, 3, ...}, >> which do not have a largest member, to also have a size equal >> to one of their members. > Tony Orlow wrote: >> No, they all have a size, and if that size is n, then there exists an >> nth. If there does not exist an nth, then there are not n elements in >> the sequence, which this set also is. > David R Tribble wrote: >> But what if a set has an nth member, for every n in N+, but has >> no largest member? Does the set still have a size? Is that size >> a member of the set? If so, is it the largest member or some >> other member? > Tony Orlow wrote: > Is that supposed to answer my question? Is that the Axiom of > Avoidance? Which axiom did I violate? No, it points out the error in logic that you made. You didn't violate any specific axiom. There is no "axiom that contradicts your logic". However, you didn't use any of the existing axioms when you made the logical jump from stating that a property of sets of the form {1,2,3,...,k} also applies to sets of the form {1,2,3,...}. Therefore the second half of your argument is unfounded (not based on the axioms and theorems at hand), and no logical conclusion can be drawn from it. It's a logic error of omission, or drawing a conclusion from a statement without justification. Tell us, what axiom or theorem justifies that jump? David R Tribble wrote: >> By your logic, you state that all sets of the form {1,2,3,...,k} >> have a largest member, therefore sets of the form {1,2,3,...} >> also have a largest member. I assume you have an axiom >> or theorem in mind to help you reach this conclusion. > Tony Orlow wrote: > No, read carefully. I said that every initial segment of N+ with size > x has an xth element with value x. The contrapositive would be that if > an initial segment of N+ does NOT include an xth element, or the > element x, then it is smaller than x in size. Yes, correct so far. That is true of all sets that are finite initial segments of N+. > If N+ cannot contain aleph_0 then the size cannot be aleph_0. > Thus, the assignment of an absolute size to this set results in contradiction. That is the same fallacious logical jump. N+ is not a finite initial segment of N+, so your requirements for finite initial segments of N+ (to have an xth element, or for the size to be a member of the set) do not apply to N+. So again, the same question is, what axiom or theorem justifies you to state that your property of finite sets also applies to infinite sets?
From: David R Tribble on 17 Jun 2010 17:04 Tony Orlow wrote: >> What axiom justifies the existence of aleph_0? > David R Tribble wrote: >> Axiom of Infinity. Duh. > Tony Orlow wrote: > No, it simply asserts that the countably infinite set "exists". It > says nothing about the size, zize, or cardinality of the set. And yet that is the axiom that is used to prove that there is a cardinal for N. A given cardinality is, as you'll recall, the class of all sets that are equipollent. Equipollence, in turn, is defined in terms of one-to-one functions on sets. So given that a 1-to-1 function exists between the members of set A and set B, A and B are equipollent, and there exists a cardinal for A and B. Substitute N (from the Axiom of Infinity) for B, and there is a cardinal for all sets equipollent to N, a.k.a. Aleph_0. [I'm paraphrasing from "Axiomatic Set Theory" by Suppes, btw.]
From: David R Tribble on 17 Jun 2010 17:28 K_h wrote: >> Definition: >> The ordinals can be defined as follows: A set S is an >> ordinal if it is well-ordered by membership and if a set U >> is a member of S then U is also a subset of S. > Tony Orlow wrote: > Well, that defines the von Neumann ordinals, as described by the Axiom > of Infinity. For N+ it is certainly true that it's well ordered, but > to say that 2 is exactly a subset of 3 doesn't sound particularly > necessary. The subset relation is what guarantees the well-order property of the ordinals. Other kinds of set constructions can be used instead of the von Neumann ordinals, obviously, just as long as they meet the same requirements (and other people have done this). It's just that the v.N. ordinals are so simple and particularly easy to use. K_h wrote: >> Cardinals can then be defined as follows: >> Definition: >> For an ordinal S, the cardinal of S, notated |S|, is the >> least ALEPH_K equinumerous to S (if S is infinite) or is S >> itself (if S is finite) where K is an ordinal. > Tony Orlow wrote: > Right, that definition is based on the previous one, which is what is > in question, not in terms of consistency with ZFC, but in terms of > whether the axioms imply the definition. Suppes actually gives an axiom for cardinals, as: K(A) = K(B) <-> A ~ B That is, card(A) = card(B) iff A and B are equipollent. He later says that this axiom can be downgraded to a theorem if we add the Axiom of Choice as an axiom.
From: K_h on 17 Jun 2010 20:51
"MoeBlee" <jazzmobe(a)hotmail.com> wrote in message news:2085fa31-97e6-4c6f-af42-7937491aa470(a)x21g2000yqa.googlegroups.com... On Jun 17, 8:51 am, Tony Orlow <t...(a)lightlink.com> wrote: > On Jun 16, 7:59 pm, "K_h" <KHol...(a)SX729.com> wrote: > > Cardinals can then be defined as follows: > > > Definition: > > For an ordinal S, the cardinal of S, notated |S|, is the > > least ALEPH_K equinumerous to S (if S is infinite) or is > > S > > itself (if S is finite) where K is an ordinal. > > Personally, I wouldn't define cardinals in terms > of alephs, but rather vice-versa. I like defining cardinals in terms of alephs because alephs can be easily defined by using a class function, ALEPH, on all ordinals using transfinite induction. For the initial, successor, and limit cases we have: ALEPH_0 = w ALEPH_K+ = smallest initial ordinal T satisfying ALEPH_K < T ALEPH_L = Union{ALEPH_K : K is in limit ordinal L} _ |