Uncountability of the Rationals?
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From: Tony Orlow on 26 Jun 2010 03:34 On Jun 26, 12:01 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 24, 10:59 pm, Tim Little <t...(a)little-possums.net> wrote: > > > On 2010-06-24, Transfer Principle <lwal...(a)lausd.net> wrote: > > > So tav appears to contain all the pofnats. But it also contains > > > elements which aren't pofnats. > > And so we have a contradiction. Gee, that didn't take long. > > But now TO is posting again, so we can fix this error based > on TO's latest comments. There is no error. The contradiction concerning omega or tav is deliberate. It doesn't exist as a number, as evidenced by its own self- contradiction. > > Let's summarize what TO has stated. > > 1. The set omega is exactly what we expect it to be. The set N+ (the positive naturals) is as normally defined. Omega, or tav, is not any kind of exact number of elements. > > 2. N+ is the set of all pofnats, i.e., omega\{0}. It is a > countably infinite set. Yes, the standard countably infinite set. > > 3. *N contains the nonstandard naturals (which Chandler calls > "tnats," but TO calls T-riffics). They include both nonstandard naturals and reals, both infinite and initesimal, given digital poisitions infinitely distant from the 0- power digit. > > Notice that sometimes the notation *N is used for Robinson's > hyperreals as well, with *R being the set of hyperreals and > *N the set of hypernaturals. Although TO has stated that his > T-riffics are _not_ the hypernaturals, the use of similar > notation implies that the concepts are analogous. Yes, similar, though perhaps not identical. Or, perhaps ultimately so. > > How can we fix the axioms? At this point, it appears that we > can just go back to ZF and assign Bigulosities to the sets > of ZF, as I originally intended to do. This means that we can > go back to determining which bijections are "strong." That's not a bad idea, but I forsee some possible objections. > > TO states that "tav" isn't a set, at least in the way that I > have been using it. But in ZF, _everything_ is a set (i.e., > there are no urelements), so tav must be a set. N+ is the standard countably infinite set of Bigulosity "tav". All other countably infinite sets are measured therewith. > > A similar dilemma occurs when we wish to extend the set R of > standard reals to the extended reals. We wish to append two > new elements, oo and -oo, to R, but what sort of objects are > oo and -oo, anyway? They must be sets since every object in > ZF is a set. If one were to use Dedekind cuts to construct > the reals, then one might define oo to be Q and -oo to be the > empty set, but what if we wanted to find a definition that > wasn't depended on the D-cut construction of R? It doesn't > matter what sort of objects we use for oo and -oo as long as > they differ from each other and from every element of R. (For > example, the Metamath website defines oo to be P(U(C)) and > -oo to be P(P(U(C))).) P(x) being the powerset and C being the continuum? Er, interesting, at least. > > And likewise with tav. All we demand of tav is it to be > different from any other object assigned to be the Bigulosity > of some set. It doesn't matter whether tav is a finite or an > infinite set -- what matters is that the object assigned the > Bigulosity tav (such as N+) be infinite. This is trickier > than the extended reals since we only needed two new objects, > oo and -oo, while we need infinitely many new objects to be > Bigulosities (since tav-n is a separate Bigulosity for each > natural number n). Yes, among other examples. IFR and ICI combined give us the ability to order infinite quantities. > > After doing so, then we simply treat tav and the other > Bigulosities as symbols, just as TO demands. Well, tav and zillion are required as primitive units in addition to 1. They must be declared. Tony
From: Tony Orlow on 26 Jun 2010 03:42 On Jun 26, 12:46 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 25, 11:30 am, Brian Chandler <imaginator...(a)despammed.com> > wrote: > > > > When it comes to set C, it *looks* like set A, however, if there are > > > tav+1 naturals (starting at 0) in decimal, there are log_10(tav+1) bit > > > positions. Given N=S^L, with S=2 and L=log_10(tav+1), we have > > > 2^log_10(tav+1)=log_5(tav+1) elements. > > This doesn't seem to be right. Just suppose tav=999. Then 2^(log10(tav > > +1)) = 2^3 = 8. And 5^8 does not equal 1000. Not that details like > > this seem very important, somehow. > > Let's see whether we can simplify this correctly. Yes, I messed up... If > t+1 is a finite number, then: > > 2^log_10(t+1) = (t+1)^log_10(2) > > Checking this with t = 999 gives: > > (999+1)^log(2) = 1000^log(2) = 8 > > as expected. > > So via ICI, we conclude that the Bigulosity of set C > must be (tav+1)^log(2). Via IFR, without the use of ICI to extend to the infinite case. But, yes. :) Tony
From: Tony Orlow on 26 Jun 2010 03:53 On Jun 26, 12:46 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 25, 11:30 am, Brian Chandler <imaginator...(a)despammed.com> > wrote: > > > > When it comes to set C, it *looks* like set A, however, if there are > > > tav+1 naturals (starting at 0) in decimal, there are log_10(tav+1) bit > > > positions. Given N=S^L, with S=2 and L=log_10(tav+1), we have > > > 2^log_10(tav+1)=log_5(tav+1) elements. > > This doesn't seem to be right. Just suppose tav=999. Then 2^(log10(tav > > +1)) = 2^3 = 8. And 5^8 does not equal 1000. Not that details like > > this seem very important, somehow. > > Let's see whether we can simplify this correctly. If > t+1 is a finite number, then: > > 2^log_10(t+1) = (t+1)^log_10(2) > > Checking this with t = 999 gives: > > (999+1)^log(2) = 1000^log(2) = 8 > > as expected. > > So via ICI, we conclude that the Bigulosity of set C > must be (tav+1)^log(2). Oops again. I misspoke. I'll look again in the (later) morning. Sleepy, TOny
From: Tony Orlow on 26 Jun 2010 03:53 On Jun 25, 7:37 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > On Jun 25, 12:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Tony Orlow <t...(a)lightlink.com> writes: > >> > On Jun 25, 10:48 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> >> Tony Orlow <t...(a)lightlink.com> writes: > >> >> > For the purposes of IFR, N+, the positive naturals, is the standard > >> >> > countably infinite set. > > >> >> Can you tell us exactly how you define N+? This question may seem > >> >> silly, but it really is important. > > >> > 1 e N+ > >> > x e N+ -> n+1 e N+ > >> ^n > > >> That "definition" does not specify a single set. There are many sets > >> that satisfy that definition. The set R of real numbers, the set Q of > >> rationals, the set Z of integers *all* satisfy the definition you've given. > > > Uh, what? > > I don't know where the "^n" came from, but it wasn't mine. Please > > don't insert deliberate obfuscation. 1 is an element of N+ and if x is > > an element of N+ then so is x+1. > > It wasn't an obfuscation. I was trying to point out a typo in your > formula. You wrote "x e N+ -> n+1 e N+". It is obvious that you meant > to use the same variable in both places. Oh, sorry. My bad. But.... > > And, of course, that typo is irrelevant to my comment. You didn't pick > out a defining characteristic of N+, since the sets Z, Q, R, w+w, and so > on, all satisfy the condition you wrote. N+ contains all and only the elements described by the rules mentioned, plus the term "standard", or "finite", if you so desire. > > >> >> For comparison, I would define N+ to be the least inductive_1 set, where > > >> >> ind_1(x) <-> 1 in x & (Ay)(y in x -> y+1 in x). > > >> > I like mine better, but same difference. > > >> You're confused. I haven't yet defined N+, so it does not compare to > >> your "definition". > > > If that wasn't the full definition then you have little business > > drawing conclusions prematurely. > > Er, that was the full definition of ind_1. The definition of N+ is just > below. That was equivalent to my full definition. What else is there? > > >> >> Thus, N+ has the property > > >> >> ind_1(N+) & (Ax)(ind_1(x) -> N+ c x). > > >> > How do you draw that conclusion? Okay, "thus" you drew this conclusion, without having defined N+? > > >> That's what I mean when I say that N+ is the *least* inductive_1 set[1]. (I > >> haven't proved there *is* a least inductive set here, but I will if you > >> wish. It's perfectly standard.) > > > Mmmmmm....standard. Nummies! > > > So, your proof is that N+ is a member of the set of inductively > > defined and well-founded sets, and supposedly the "least" among them. > > "Inductive", not "inductively defined". And while it happens to be > well-founded, that's not something that I've mentioned. It is a > consequence of the definition, not a requirement. Yes, it has a first element. And a set can be defined inductively, if you don't mind, so that becomes a vacuous correction. > > > It's a member, but far from the least. COnsider the size of the set of > > natural tetrations. > > Is that set an inductive set? That is, does it satisfy the property > > 1 in x & (Ay)(y in x -> y+1 in x)? No, that's the definition of N+, not tetration(n) for neN+. > > >> >> Is this also your definition? > > >> > Apparently not. > > >> I guess I wasted my time with this question. You don't really > >> understand the need for specifying N (or N+) as a *least* set satisfying > >> some condition, because you mistakenly think that your two conditions > >> already define N+. I hope that my examples showed you why your > >> so-called definition does not suffice, but I honestly don't expect so. > > > What doesn't suffice for your purposes? What can't you prove? > > That there is a unique set satisfying the so-called definition. There is one set which satisfies all and only the conditions specified. > > There are just oodles of sets containing 1 and closed under successor. > Lots of them. Z contains 1 and is closed under successor. Not unit increment, aka, +1. > > So your definition does not suffice and you still don't understand why. > Maybe now you understand why it does (finally). > > > >> Footnotes: > >> [1] That is, "least" modifies "inductive_1 set", not "inductive_1". N+ > >> is the smallest set which is inductive_1. This is the definition of N+. > > > Consider any subset of N+, with a somewhat naive perspective, if only > > for a moment. > > Any proper subset S of N+ either does not contain 1, or contains an n such > that n+1 is not in S. Thus, no proper subset of N+ is inductive_1. Piffle beside the point. An inductive set is one where each member can be generated from the last, not necessarily by adding 1. "A fart by a waterfall" - Demetri Martin > > -- > Jesse F. Hughes > > "You're terrified of your daughters dreaming about me." > -- James S. Harris, on why mathematicians fear him- Hide quoted text - > I dream sometimes of James' daughters..... TOny
From: Jesse F. Hughes on 26 Jun 2010 08:24
Transfer Principle <lwalke3(a)lausd.net> writes: > On Jun 25, 8:57 am, Tony Orlow <t...(a)lightlink.com> wrote: >> On Jun 24, 10:01 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> > In Walker's defense, he said that the formula phi should not contain the >> > symbol tav. >> Good point. Maybe phi should not include "finite" either, but at >> least, specifying an inequality without a limit of 0 is sufficient to >> avoid supposed contradictions. > > This was something that I was wondering earlier as well. We > could add a new primitive "finite" and declare that phi > cannot contain the "finite" primitive. Then we can add some > axioms stating which objects are finite, such as: > > finite(0) > (Ax (finite(x) -> Ay (finite(xu{y})) > ~finite(tav) I don't see what "good" such a primitive would do. It still wouldn't preclude proving that tav is Dedekind-finite, by using the defining formula for Dedekind-finite. By the way, here's a keen proof. Let N+ be defined as usual, so that ind_1(N+) & (Ax)(ind_1(x) -> N+ c x) where ind_1(x) <-> 1 in x & (Ay)(y in x -> y+1 in x). Using your statement of Tony's induction principle, ((En fin. nat. (phi(n)) & Ax (phi(x) -> phi(x+1))) -> phi(Tav) We will prove two weird things: ,---- | (1) N+ = Tav. | (2) Tav e N+. `---- Theorem: Tav e N+. Proof: Let phi(n) be the formula n e N+. Then, clearly, (En fin. nat. (phi(n)) & Ax (phi(x) -> phi(x+1))) is true, because N+ is ind_1. Hence, phi(Tav) is true by the above inductive principle and so Tav e N+. Theorem: N+ <= Tav. Proof: Tav is ind_1, since it contains 1 and is closed under successor. Thus, N+ is a subset of Tav and hence N+ <= Tav. Theorem: Tav <= N+. Proof: Let phi(n) be the formula n <= N+. Again, clearly, (En fin. nat. (phi(n)) & Ax (phi(x) -> phi(x+1))) is true, because N+ is ind_1. Hence, phi(Tav) is also true and thus Tav <= N+. Corollary: Tav = N+. -- "Yeah, I know, it's quantum [computing], and all kind of interesting physics associated with what is to many a mystical word, but I have a B.Sc. in physics, and I know that you're just talking about specialized mechanical devices when you talk about quantum computing." -- James S. Harris |