Uncountability of the Rationals?
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From: Tony Orlow on 15 Jun 2010 19:33 On Jun 15, 2:17 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > On Jun 14, 4:49 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> David R Tribble <da...(a)tribble.com> writes: > > >> > Serves me right for trying to think like Tony. > > >> > Oh well, this is easily solved! The problem is that TA1 is too > >> > general, so we need to limit its scope. Instead of any property > >> > P, let's just consider arithmetic properties and operators. > >> > So, perhaps something like this: > > >> > Tony's Axiom [TA2] > >> > Given arithmetic function f(x) = y, where x and y are reals, > >> > then f(w) = u, where w and u are infinite ordinals. > > >> > Yeah, I know this really isn't any better when taken any further, > >> > but Tony really needs to put his money where his mouth is and > >> > just go ahead and write his assumptions down already. He can't > >> > proceed with any of his other statements about bigulosity, > >> > ICI, IFR, or whatever until he starts with an axiom or two. > >> > And he knows it. > > >> It seems to me that he's stated his assumption, but he doesn't get quite > >> how much it assumes. > > >> Given any real valued functions f and g, if lim (f - g) > 0, then f(x) > >> > g(x) where x is any infinite number. > > > That's actually a nice succinct way to state it, however I generally > > include the condition that finite-case induction applies to f-g as > > well, to make clear that it is an extension of the finite form of > > proof to the infinite case. > > >> As a consequence of this, I guess, it follows that f and g are defined > >> on infinite numbers, though we don't know anything about their values > >> aside from the fact that f(x) > g(x). > > > No, that's correct. At best you can say one set is greater then the > > other, and you can quantify that difference as the ratio of the sizes > > functionally defined over [1,w]. There is some math that can be done, > > but it's nothing all that super-special. The special part is that it > > provides for a countably large spectrum of countably infinite sets > > which satisfy normal intuitions, such as the proper subset always > > being smaller, as well as more general notions. That seems like a > > little bit of an advance to me. > > That little bit of math is utterly unclear. > > For instance, you claim that sqrt(w) * sqrt(w) = w, but that does *not* > follow from the above principle, *even if* we assume that the above > principle entails that sqrt(w) is defined. > > The above principle just allows us to conclude that sqrt(w) < w. > -- > "You are beneath contempt because you betray mathematics itself, and > spit upon the truth, spit upon decency, and spit upon the intelligence > of the world. You betrayed the world, and now it's time for the world > to notice." -- James S. Harris awaits Justice for crimes against Math.- Hide quoted text - > > - Show quoted text - True. So? It doesn't mean there isn't an easy way to envision omega^2 or sqrt(omega). TOny
From: David R Tribble on 15 Jun 2010 19:39 Tony Orlow wrote: >> Number 3. The set of square roots of naturals has a Bigulosity of omega^2. > Tony Orlow wrote: > First of all, by IFR, in harmony with the fact that square roots of > naturals occur more often as we move along the real line than do > naturals, we can conclude that over N+ there exist more naturals' > square roots than naturals, omega^2 square roots of naturals. So you're saying that the set of square roots of naturals, which I'll call R for brevity, has more members than N+. N+ = { 1, 2, 3, ... } R = { sqrt(1), sqrt(2), sqrt(3), ... } size(R) > size(N+) Since you say that R has more members that N+, this in turn means that there exist members k in R such that k*k = m is a natural but m is not a member of N+. Tell us, then, how N+ can contain all of the naturals yet not contain all of the k*k squares? Are these extra k*k values larger than every natural in N+? Can you provide an example of a natural square root whose square is not a natural in N+?
From: Tony Orlow on 15 Jun 2010 19:44 On Jun 15, 5:40 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 14, 8:43 pm, David R Tribble <da...(a)tribble.com> wrote: > > > Jesse F. Hughes wrote: > > > It seems to me that he's stated his assumption, but he doesn't get quite > > > how much it assumes. > > > Given any real valued functions f and g, if lim (f - g) > 0, then f(x) > > > > g(x) where x is any infinite number. > > > As a consequence of this, I guess, it follows that f and g are defined > > > on infinite numbers, though we don't know anything about their values > > > aside from the fact that f(x) > g(x). > > It's obvious that Tony can't and won't answer any of these > > questions, and no one here (except Walker) really sees him > > as capable of going any further in any meaningful sense. > > So except for deriving some entertainment value and learning > > a few new things, I don't see the point in indulging him any more. > > Teaching pigs to sing, and all that. > > I've stood back and watch this thread grow, allowing TO to > discuss his ideas in more detail before posting again. I > can't respond to every post to which I want to respond, but > let me start here since my name is mentioned here. I was wondering where you were. ;) > > As expected, I don't believe that working with theories > other than ZFC or set sizes other than standard cardinality > is analogous to "teaching pigs to sing." I definitely > prefer to believe that there is a theory in which infinite > sets work differently from how they work under ZFC, and > perhaps working as TO or another poster would like them to. > > 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. I rather think not. The Axiom of Infinity declares something like N a set, when it's really a sequence, but asserting the existence of this object doesn't lead to all the confusion. The Axiom of Choice, on the face of it, doesn't seem exactly wrong, but it seems to, perhaps in combination with assumptions of truth about the ordinals, lead to bizarre conclusions, being interpreted strangely. That there exist countably infinite sets seems obvious to me. > > 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? > > And thus, a good starting point is to start with ZF and > replace Infinity with a new axiom. This new axiom won't > merely be ~Infinity, since we aren't just trying to get rid > of infinite sets but replace them with new objects that > work differently from the infinite sets under ZF. My problem's with von Neumann. He's gonna pay, I tell ya... > > In another thread, I mentioned how Infinity is used to > prove that every set has a transitive closure. Therefore, > we can take ZF, drop Infinity, and add a new axiom: > > There exists a set with no transitive closure. > > 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? ;) > > But the problem here is that it isn't obvious how this > theory matches the intuition of any sci.math poster. Also, > it's not evident how this theory is related to math for the > sciences, either. Until those objections are addressed, no > one is going to accept this theory. Nah, it's less the axioms than the models, I believe. > > 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? :) TOny
From: Tony Orlow on 15 Jun 2010 19:46 On Jun 15, 6:07 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <0cef0e10-6e81-4031-9673-2df1cad85...(a)k39g2000yqb.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > On Jun 14, 7:55 pm, Virgil <Vir...(a)home.esc> wrote: > > > In article > > > <b95c48f1-4244-4d6b-b218-5dcebc166...(a)30g2000vbf.googlegroups.com>, > > > Tony Orlow <t...(a)lightlink.com> wrote: > > > > > On Jun 14, 3:08 pm, David R Tribble <da...(a)tribble.com> wrote: > > > > > Tony Orlow wrote: > > > > > >> Actually, in Bigulosity, there are much larger countably infinite > > > > > >> sets. The square roots of the naturals have a higher "density" than > > > > > >> the naturals. > > > > > > David R Tribble wrote: > > > > > >> How is that? The two sets should be exactly the same size > > > > > >> (your size(), not just cardinality). Every natural has a square > > > > > >> root, > > > > > >> and every natural square root has a corresponding natural. > > > > > >> (I assume you're only dealing with the positive roots.) > > > > > > Tony Orlow wrote: > > > > > > Holy cannoli! There exists a bijection, f:N->S and g:S->N, so indeed > > > > > > they have the same cardinality. They do not share the same > > > > > > bigulosity, > > > > > > since f and g are not the identity function. The number of square > > > > > > roots over omega is omega^2. > > > > > > Sorry, but it's not clear what you mean by "the number of square > > > > > roots over omega is omega^2". Do you mean: > > > > > 1. the number of naturals in omega that are square roots is omega^2? > > > > > 2. the number of naturals in omega that have square roots is > > > > > omega^2? > > > > > 3. the set of the squares roots of all the naturals in omega has > > > > > omega^2 members? > > > > > 4. or something else? > > > > > Number 3. The set of square roots of naturals has a Bigulosity of > > > > omega^2. > > > > Since every natural has 2 square roots, why wouldn't it be 2*aleph_0?- Hide > > > quoted text - > > > > - Show quoted text - > > > First of all, by IFR, in harmony with the fact that square roots of > > naturals occur more often as we move along the real line than do > > naturals, we can conclude that over N+ there exist more naturals' > > square roots than naturals, omega^2 square roots of naturals. Now, if > > you want to include the negative square roots of naturals, then the > > inverse is not a function, since it includes two values for each neN. > > In such cases, which we haven't gotten to yet, one must divide the > > domain of the mapping from N to S at their extrema (mins and maxs) to > > get monotonically increasing or decreasing sequence segments, then add > > the results for each to get the total membership. Thus, the number of > > squeare roots should be roughly double, perhaps plus the 0th one. > > as in 2*N^2 + 1? > But is such an N an ordinal (omega) or a cardinal (aleph_0)? > > It makes considerable difference! > > And why (ordinal vs cardinal)?- Hide quoted text - > > - Show quoted text - Those terms are meaningless in Bigulosity. It's a "count". Tony
From: Tony Orlow on 15 Jun 2010 20:06
On Jun 15, 7:39 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> Number 3. The set of square roots of naturals has a Bigulosity of omega^2. > > Tony Orlow wrote: > > First of all, by IFR, in harmony with the fact that square roots of > > naturals occur more often as we move along the real line than do > > naturals, we can conclude that over N+ there exist more naturals' > > square roots than naturals, omega^2 square roots of naturals. > > So you're saying that the set of square roots of naturals, > which I'll call R for brevity, has more members than N+. "more", yes. greater Bigulosity. > > N+ = { 1, 2, 3, ... } > R = { sqrt(1), sqrt(2), sqrt(3), ... } > > size(R) > size(N+) Yes. > > Since you say that R has more members that N+, this in > turn means that there exist members k in R such that > k*k = m is a natural but m is not a member of N+. No, there exists a bijection, but equibigulosity is not indicated by the existence of a bijection alone, but by a bijection which boils down using IFR and ICI to an identity relation, which almost never happens. > > Tell us, then, how N+ can contain all of the naturals yet > not contain all of the k*k squares? Are these extra k*k > values larger than every natural in N+? You were talking about the set of square roots of naturals and now you switch to squares? Nice try. > > Can you provide an example of a natural square root > whose square is not a natural in N+? Wow, gee, no. I guess you got me, exacept that your point is completely irrelevant. Tony |