Uncountability of the Rationals?
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From: Aatu Koskensilta on 19 Jun 2010 11:37 Transfer Principle <lwalke3(a)lausd.net> writes: > Whatever action AP was going to use to stop Hughes from using his > quotes in the .sig randomizer, it obviously hasn't worked yet, since > Hughes is still quoting AP in the .sig. A penetrating observation. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Tony Orlow on 20 Jun 2010 06:40 On Jun 17, 4:44 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. > > 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?- Hide quoted text - > First of all, I never said anything about *finite* initial segments of N+. You cannot insert that condition into my argument, then claim that the inconsistency it introduces into my argument is my fault. I stated that every initial segment of N+ with size x has an xth element whose value is x, without reference to finiteness. You agree with the statement in the finite case, but object when I apply it to the infinite case. Why? You say it is "unjustified", but your objection is far more so. If it is true that there exists an xth element in any sequence of size x, and that when the sequence is an initial segment of N+ that element is always equal to x, then how do you justify discounting this contstant equality between element count and value in the infinite case, for N+ in general? Unless you can come up with some kind of contradiction *within* my argument, saying "I just don't like it" is an unjustiified objection. Tony
From: Tony Orlow on 20 Jun 2010 06:47 On Jun 17, 5:04 pm, David R Tribble <da...(a)tribble.com> wrote: > 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.] Sure. Given the definition of cardinality you can involve at least one of the axioms in proving that there is a cardinal that applies to all countably infinite sets. In other words, the definition of ordinal and cardinal numbers use terms to which the axioms can be applied in order to deduce new statements. That is a very different matter from saying that the definition of the ordinal is a necessary consequence of the axioms. When you say the existence of omega "follows from the axioms," that sounds to me like it can be deduced solely from the axioms. It really follows from the definitions of the ordinal and the cardinal, when the axioms and basic logic are applied to them to deduce new statements. Tony
From: Tony Orlow on 20 Jun 2010 06:52 On Jun 17, 5:28 pm, David R Tribble <da...(a)tribble.com> wrote: > 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. Hmmm... I've always held AC at arm's length withh some suspicion. It's always seems to be very strangely applied, for what it appears to say on its face. So, I'm not surprised if someone considers its use to possibly imply cardinality as some sort of theorem. I suppose if there is an axiom I don't like, it's Choice. I don't think I have any basic problems with ZF, anyway. Tony
From: Tony Orlow on 20 Jun 2010 07:07
On Jun 18, 1:10 am, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-17, Jesse F. Hughes <je...(a)phiwumbda.org> wrote: > > > I don't really see that we can guarantee to get as near to any > > arbitrary real as we want by traversing this tree. > > Shouldn't it really have a +/- at the front as well, or are we talking > only of nonnegative reals? With either adjustment, it is true. Well, the positive H-riffics would be defined the way I originally did, more or less: 0eH xeH -> 2^xeH xeH -> 2^-xeH Or all H-rrifics can be defined as: 0eH xeH -> (2^x)eH xeH -> -(2^x)eH > > Such a sequence can be obtained by taking successive logarithms of > absolute values, noting the signs at each step. For example, 3 is > approximated to eight decimal digits by > > 2^2^2^-2^-2^-2^-2^2^-2^2^-2^2^-2^-2^2^2^-2^2^-2^2^2^-2^2^-2^2^-2^2^-2^-2^2^0 Okay, I'm excited to hear there is perhaps a way to do this but I am not quite following what you're doing. How do you necessarily guarantee you are the path approaching a given value? > > It is fairly easy to prove that a slightly modified lexicographic > order on distinct sign sequences agrees with the usual ordering on the > reals. Proving that distinct reals yield distinct sign sequences > requires somewhat more, but can be done by analysing stability under > iteration of log_2|x|. > > - Tim Thanks, Tim. I'd love to hear more. Have a nice day. Tony |