Uncountability of the Rationals?
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From: Tony Orlow on 17 Jun 2010 08:54 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). > > 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. > > 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. 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. 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. > > > What axiom justifies the existence of aleph_0? > > Axiom of Infinity. Duh.- Hide quoted text - No, it simply asserts that the countably infinite set "exists". It says nothing about the size, zize, or cardinality of the set. > > - Show quoted text - So, you refuse to answer my question? What axiom did my argument above violate? Thanks, Tony
From: Jesse F. Hughes on 17 Jun 2010 09:06 Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes: > "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > >> I think 8 is Julius Caesar. > > Ah, so Caesar /isn't/ a prime after all! I didn't say that. Caesar is also 23. Twenty-three is prime. These are things I heartily believe. Now, it is up to Walker to come up with a theory in which they are true. It is an exciting task, to be sure, and I see why he's so fascinated with such random theories that he endlessly talks about his task. -- I thought the wreck was over. I thought the fire was out. I thought the storm had passed and I was safe at last. I thought the wreck was over, but here she comes again. --The Flatlanders
From: Jesse F. Hughes on 17 Jun 2010 09:31 Tony Orlow <tony(a)lightlink.com> writes: >> And, yes! subset is DEFINABLE from 'e'. >> >> Just as 'is an ordinal' is DEFINABLE from the mere primitives. > > Is it? Demonstrate as I did for you just now. x is an ordinal iff (Ay,z in x)(y in z or z in y) & (Aw,y,z in x)((w in y & y in z) -> w in z) & (A z)( ((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> (Ey in z)(Aw in z)( y in w ) ) & (A y in x)(A z in y)( z in x ). Not so hard. The first two clauses assert that elementhood is a linear order on x (anti-reflexivity is trivial, assuming Regularity), the next clause that it is a well-ordering and the final clause that x is transitive. Someone else will surely correct any errors in the above. Tell me, Tony, did you *really* think it was impossible? Did you *really* think that "x is an ordinal" is literally undefinable in the language of ZFC? And that generations of mathematicians have been pulling one over on the public? -- Quincy (age 5): Baba, play some [computer games]. Mama: Quincy, if you want [Baba] to live, don't make those suggestions. Quincy: Make those suggestions. Got it.
From: Tony Orlow on 17 Jun 2010 09:51 On Jun 16, 7:59 pm, "K_h" <KHol...(a)SX729.com> wrote: > "Tony Orlow" <t...(a)lightlink.com> wrote in message > > news:f6d161b0-150a-4c47-bea4-bc898e5a0f87(a)i31g2000yqm.googlegroups.com... > On Jun 15, 8:22 pm, Transfer Principle <lwal...(a)lausd.net> > wrote: > > > > > > > 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 they 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. > > Actually, I don't see any explanation of how ZFC > > proves their existence. I will probably be told to > > just go read a book... > > //section removed > > > While the von Neumann ordinals are consistent > > with ZFC, are they really logically entailed by the > > axioms? I must be missing something. > > You are missing the fact that the definitions are created > using the axioms. For example, using the axiom of > extensionality, ZF1, the empty set axiom, ZF2, and the axiom > of pairs, ZF3, you can then define an ordered pair as > follows: > > Definition: > Given sets A and B, let <A, B> denote the set {{A},{A,B}}and > then <A,B> is the ordered pair with A the first member and B > the second member. > > This is but one example. The theory ZFC, and other theories > like ZF+(V=L), allow you to define the ordinals and the > cardinals, for example, totally in terms of the axioms of > the theory. Note, it is bad to use a definition that is not > ultimately expressed from the axioms. For example, one can > define a measureable cardinal but that definition doesn't > follow from the axioms of the theories mentioned above. If > you wanted a theory with measurable cardinals you would have > to first define an axiom like MC="measurable cardinals > exist" and then add it to get a new theory ZFC+MC. You > can't add it to ZF+(V=L) because ZF+(V=L)+MC is an > inconsistent theory. > > Another example, in addition to ZF1, ZF2, and ZF3 mentioned > above, if you also add ZF4, axiom of separation, ZF5, > powerset axiom, and ZF6, union axiom, then you can define > the Cartesian product. Let P(S) denote the powerset of S. > > Definition: > The Cartesian product of sets A and B, written C=AxB, is the > set: > > {c in P(P(AUB)):EaEb((a in A ^ b in B) ^ c=<a,b>)} > > Note, this definition exploits the definition of an ordered > pair which itself is defined in terms of the axioms. > Informally, the bottom line is this: the definitions are > just shortcuts for things that groups of the axioms provide. Hi K_H - The definitions of basic non-logical operators can be defined by a set of axioms concerning the behavior of those operators, and other non- logical operators may be built logically from the basic ones, as union and intersection are defined using logical operators and 'e'. Sure. I have no problem with defining Cartesian product this way. > If you didn't want to use any definitions then you could > still prove every theorem but it would be a lot more busy > work -- kind of like writing a computer program in binary > instead of a programming language. As for your question > about ordinals and cardinals, here they are: > > 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. 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. I understand that that provides the set-theoretic basis for N, but to view quantities as sets might not capture all the qualities of quantities. That may sound like gibberish, but my point is this. While one MAY define something this way and call it an ordinal, it doesn't mean this definition is ENTAILED by the axioms, it simply means it does not CONTRDICT the axioms, and so is CONSISTENT with the axioms. Otherwise, the definition of the ordinal could be deduce directly from the axioms, no? > > 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. 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. > > Equinumerosity can be defined in terms of bijections and > bijections are defined in terms of functions and functions > are defined by the axioms. Again, it can all be reduced > back to the axioms. The cardinality of an arbitrary set can > then be defined in terms of bijections, and so on. One can define operations consistent with the axioms which the axioms don't directly imply. Thanks for your explanation. That's exactly what I was looking for. Tony
From: Brian Chandler on 17 Jun 2010 10:08
Tony Orlow wrote: > On Jun 16, 5:51 pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > > The H-riffics are a bit of a sidebar. ... > > > ... the H-riffics may be redefined as: > > > E(0) > > > E(x) -> E(-2^x) ^ E(2^x) > > > > This still omits vast (uncountable) subsets of of the reals. > > 3, for example, is not a node in the binary tree that is E. > > It is an uncountable number of iterations down the tree. What does it mean to say "an uncountable number of iterations"? If one iteration is followed by another iteration, how does the "magic leap" occur? Brian Chandler |