Uncountability of the Rationals?
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From: Tony Orlow on 16 Jun 2010 09:15 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... > > 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. 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? > > (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. Okay, perhaps I am "wrong" about this. I am going over the axioms of ZFC, and I simply don't see any reference to any primitive referring to ordinality or cardinality, or any statement that the infinite set declared in the Axiom of Infinity has any particular "size". While the von Neumann ordinals are consistent with ZFC, are they really logically entailed by the axioms? I must be missing something. > > 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. In standard N ordinals and cardinals are the same thing, really. I have never seen the need for any distinction. As a computer scientist, a number is a number, whether used as a data element (cardinal) or memory address (ordinal). > > 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. That would seem to be a valid conclusion.
From: Tony Orlow on 16 Jun 2010 09:24 On Jun 15, 8:37 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 15, 4:44 pm, Tony Orlow <t...(a)lightlink.com> wrote: > > > On Jun 15, 5:40 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > I believe that among the sci.math posters who argue against > > > ZFC, the most common axioms to reject are Infinity and > > > Choice, followed by Powerset (if they accept infinite but > > > not uncountable sets). > > That's generally the Anti-Cantorians, as opposed to us here Post- > > Cantorians. See? > > Another interesting label, "Post-Cantorian." So here TO > calls himself a "Post-Cantorian." I wonder what some of > the key differences between these two labels are. I explained to you that your "substandard" and "superstandard" theorists already have the labels "Anti-Cantorian" and "Post- Cantorian". So, you already know what the differences are. :) > > > > We notice that in this theory, ~Infinity would be a theorem > > > (proved via Deduction Theorem/contrapositives). > > Okay. Do you agree wth that axiom, or are you just trying it on for > > Halloween? ;) > > The latter, I guess. I was trying to see what sort of > results it leads to, and whether it leads to any > interesting results. Okay. Not sure. > > > > It's doubtful that any textbook discusses sets that have no > > > transitive closure, since most textbooks are grounded in ZF, > > > which proves that every set does have one. A good starting > > > point might be old zuhair threads, since zuhair mentioned > > > transitive closures in his theories all the time. > > The H-riffics have transitive closure, after some uncountable number > > of iterations. Does that count? > > In that case, the poster whose ideas are represented by > my theory isn't TO, at least if we're talking about his > set of H-riffics. But there might be another sci.math > poster whose ideas might be formulated in this theory. The H-riffics are a bit of a sidebar. Besides, it occurs to me that within R there exists one element which cannot be a child node of an element in R, namely, 0. So, 0 can be taken to be the root case for the tree of values, the foundation. Additionally, if one wants to include all reals positive and negative then the H-riffics may be redefined as: E(0) E(x) -> E(-2^x) ^ E(2^x) Just thoughts, really unrelated to Bigulosity, and apparently unqualified to be an explicit well order of the reals. TOny
From: Tony Orlow on 16 Jun 2010 09:34 On Jun 15, 9:36 pm, David R Tribble <da...(a)tribble.com> wrote: > David R Tribble wrote: > >> I'm still confused, because it looks for all the world as if the sets > >> N = { 1, 2, 3, ... } > >> and > >> R = { sqrt(1), sqrt(2), sqrt(3), ... } > >> have exactly the same number of members. For a given k in N+, > >> there is a sqrt(k) in R at exactly the same position. (Forget > >> bijections, I'm applying your BO set ordering and member indexes.) > > Tony Orlow wrote: > > You're still thinking within the bax, David. You know AneN (n>1 -> > > E(reR: r=sqrt(n)), and that it's a countable set that occurs more > > often than the naturals, as you move at some measurable speed upward > > along the number line from that 1. > > > So, you must admit, there is some kind of "moreness" that occurs with > > the square roots of naturals, and the same measure of "lessness" that > > occurs with the squares of naturals (of course with a natural density > > of 0). > > I know you think so, but I don't see how. If every natural k in N+ > has a square root sqrt(k) in S, how are there more roots than > naturals? Does each natural have only a single (positive) square > root, or are there more roots than that? > > I'll ask the question again, since you didn't bother answering > it before: Are there square roots r in S such that r*r is not a > natural in N+? If so, can you provide an example r? If not, does > this mean that there are more members in S than in N+?- Hide quoted text - > Imanswered this already. I do not disagree that there exists a bijection, but within any segment of R greater than measure 2 exist more square roots of naturals than naturals. Sure, you can find a member in each set corresponding to a unique member of the other. They are equicardinal. They are not equibigulous. Perhaps you should ask this question about natural density? How can one possibly assert that there are half as many evens as naturals, or 1/3 as many multiples of 3, etc? How can we say, in natural desnity that the measure of the set of squares is 0 when the measure of N is 1? Obviously, there is a bijection between N and the multiples of any finite neN. Why do they not have the same natural density? Obviously, natural density is wrong, since it contradict cardinality. Which axioms must be abandoned in order to entertain natural density? Tony
From: Tony Orlow on 16 Jun 2010 09:51 On Jun 16, 2:12 am, Brian Chandler <imaginator...(a)despammed.com> wrote: > Tony Orlow wrote: > > On Jun 15, 1:50 am, Brian Chandler <imaginator...(a)despammed.com> > > wrote: > > > Tony Orlow wrote: > > > > On Jun 14, 12:15 pm, Brian Chandler <imaginator...(a)despammed.com> > > > > wrote: > > > > > Tony Orlow wrote: > > > > > > On Jun 14, 1:21 am, Brian Chandler <imaginator...(a)despammed.com> > > > > > > wrote: > > Excuse me taking your points largely in reverse order... > > > > 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+. 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. > > Right, so you have now managed to read "set of pairs {p,p}" (call it > pp) correctly. That's progress. And you agree that this is the > "diagonal" of the "unit infinite square", so B(pp) = Tav. > > > Tav us a bad name, but whether you are talking about omega or a > > zillion, the equivalence relation holds. > > Why is "Tav" a bad name? In any conversation with you, I am certainly > not talking about the "real" omega, because you don't understand what > it is. If there is "declaring of unit infinities" to be done, I shall > do it in the Hebrew alphabet. (Irrelevant anecdote: I remember Conway > using 'Beth', and commenting that the Hebrew alphabet is underused...) > > > Not exactly. 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. > > I'm sure Lebesque is (posthumously) delighted to have your agreement. > Personally I have no idea what you are talking about. But working back > towards my original question: > > > > > Don't you make any connection between a set of pairs and, say, spatial > > > > coordinates? When you talk about points in n dimensional space, do you > > > > not define them as unique n-tuples in the spatial set of points? > > > > I can make all sorts of connections, but if I want to persuade someone > > > else of something, I make those connections explicit. > > > Yes, okay. If you have a matrix x wide and x tall, then you have an > > x*x grid of values, perhaps with some values repeating, but with > > expressions unique. > > Sure, a matrix is a matrix. But consider (for example) a two- > dimensional vector space over a finite field: say the set of ordered > pairs of elements of GF17 (pairs {a, b} where a, b are integers modulo > 17). > > This is a vector space; we could write all 17^2 elements in a grid, > but this is not a matrix -- there is no "perhaps with some values > repeating". FWIW, I don't know of a definition of "matrix" which > forces "expressions" to be unique. But I think this is just another > example where you write (much) faster than you think. But anyway: > finally my original question... > > [to David Tribble, I think] > > > > > > > > Suppose, for example we thought about the bigulosity > > > > > > > of the set NQ > > > > > > > > NQ = { {p,q} | p^2 = q e N} (set of pairs of naturals with their > > > > > > > roots) > > > > > > > Done. You have an omega X omega^2 matix. You omega^3 elements. > > Right, well let's avoid the "mat[r]ix" word, and call our "unit > infinity" Tav, as usual. Fine, if you want, we can call Bigulosity(N+) "tav". "Zillion" is uncountable. > > Yes, you tell us there are Tav^2 natural roots in the vertical > direction, and Tav naturals in the horizontal direction. I'm asking > about the set NQ containing values {p,q} such that p^2=q. I think you > have probably withdrawn your original claim that there are Tav^3 of > these points. But how many are there? > > I envisage there is some sort of trail of these points, each one above > and to the right of the preceding one. If I count them vertically I > seem to get Tav^2, horizontally I get Tav. This is what I am asking > you to resolve. > > Brian Chandler Oh, I see your question. It's not a bad one, though it is outside of the scope of IFR abd ICI, ultimately, since it is not a mapping between N+ and another set of reals, but an expression of the relation mapping N+ to the squares. Your point is that, according to Bigulosity, there should be tav elements since there are tav naturals, but there should be sqrt(tav) elements if there are sqrt(tav) squares of naturals. That would appear to be a contradiction, granted. However, Bigulosity does not claim to assign a size to such a set. It is something worth considering though. TOny
From: Jesse F. Hughes on 16 Jun 2010 09:53
Tony Orlow <tony(a)lightlink.com> writes: > The H-riffics are a bit of a sidebar. Besides, it occurs to me that > within R there exists one element which cannot be a child node of an > element in R, namely, 0. So, 0 can be taken to be the root case for > the tree of values, the foundation. Additionally, if one wants to > include all reals positive and negative then the H-riffics may be > redefined as: > > E(0) > E(x) -> E(-2^x) ^ E(2^x) I suppose this means 0 exists if x exists, then so does -2^x and 2^x. But, it does not follow from this that[1], for instance, 3/4 exists. I suppose that you want to include at least finite sums of things that exist, so then 3/4 would indeed exist. But then you still wouldn't have 1/3, unless you include some infinite sums. Of course, you don't want to include all infinite sums, but only those that converge. I don't know whether it's easy to specify this condition (convergence of an infinite sum) in the context you have in mind. Also, you claimed that you'd get all of the reals positive and *negative*, but you'll need another assertion to get the negatives. But obviously, you do *not* get all of the reals by only the two axioms you wrote above. Again, you presume too much. > Just thoughts, really unrelated to Bigulosity, and apparently > unqualified to be an explicit well order of the reals. Footnotes: [1] I'm ignoring the fact that statements such as "0 exists" are not really formalizable in classical FOL. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce |