Uncountability of the Rationals?
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From: Tony Orlow on 15 Jun 2010 16:43 On Jun 14, 8:08 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <3b89977b-eb55-4134-803c-2d2b32bd7...(a)g19g2000yqc.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > 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? > > Hmmmm..... > > Not necessarily. One can easily have a space without having to have a > coordinatized space. What bits of geometry are so dependent on > coordinates that they cannot be done without it? Point-set topology. How do you define each unique point?
From: Tony Orlow on 15 Jun 2010 16:45 On Jun 14, 11:37 pm, Brian Chandler <imaginator...(a)despammed.com> wrote: > David R Tribble wrote: > > Tony Orlow wrote: > > >> 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". > > OK, let me try to guess. > > let Qn be the set of (real) values p<=n, such that p^2 is a natural. > (the "roots up to n") > > Then for any n, |Qn| (call it q(n)) equals (approximately) n^2. So > > lim (n->oo) q(n) = n^2 > > Therefore, by a general principle with many names (QD basically), > declaring 'omega' to be our unit infinity we have > > q(omega) = omega^2 > > Isn't that it? > Brian Chandler- Hide quoted text - > > - Show quoted text - That's an exatrpolation of the theory, but as you well know, it begins with an inerrant finite model, as does cardinality.
From: Jesse F. Hughes on 15 Jun 2010 17:03 Tony Orlow <tony(a)lightlink.com> writes: >> Then you are claiming that EVERY subset of N+ has a well defined >> bigulosity? > > IF it has an algebraic bijection with N+ it can be compared therewith. What's "algebraic bijection" mean? -- Jesse F. Hughes "I thought it relevant to inform that I notified the FBI a couple of months ago about some of the math issues I've brought up here." -- James S. Harris gives Special Agent Fox a new assignment.
From: Transfer Principle on 15 Jun 2010 17:40 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. 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 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). 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. 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). 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. 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.
From: Virgil on 15 Jun 2010 18:07
In article <0cef0e10-6e81-4031-9673-2df1cad85016(a)k39g2000yqb.googlegroups.com>, Tony Orlow <tony(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)? |