Uncountability of the Rationals?
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From: Tony Orlow on 26 Jun 2010 20:42 On Jun 26, 8:24 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Transfer Principle <lwal...(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- Hide quoted text - > > - Show quoted text - That's an interesting proof, but you forgot to define the primitive expression for "zize". > ,---- > | (1) N+ = Tav. > | (2) Tav e N+. > `---- Rather, |N+|=tav Thus taveN+ tav=|N+| and |N+|=tav. True. Tony
From: Jesse F. Hughes on 26 Jun 2010 20:53 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 26, 8:24 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Transfer Principle <lwal...(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+. > That's an interesting proof, but you forgot to define the primitive > expression for "zize". No idea what you're talking about. >> ,---- >> | (1) N+ = Tav. >> | (2) Tav e N+. >> `---- > > Rather, > |N+|=tav > Thus taveN+ > tav=|N+| and |N+|=tav. > True. That's not what I proved and you didn't point out any mistake. Nonetheless, I'm fascinated to learn that tav is in N+. N+ contains nothing but finite numbers, as you've said previously. Thus, tav is a finite number. Agreed? -- Jesse F. Hughes "But regardless of my goofs, my reality of journals is different from ANY of yours, as they just treat me in a special way." -- James S. Harris
From: Tony Orlow on 26 Jun 2010 21:20 On Jun 26, 8:47 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > 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 > > [...] > > >> 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. > > Yes, that's what you wanted to say, but what you said is that the > definition is: > > 1 e N+ > n e N+ -> n+1 e N+ > > That does *not* say that *only* such elements are in N+. In order to > say that, one wants to say that N+ is the *smallest* set containing 1 > and closed under successor. Shizzles! I only gave those rules, so what makes you think anything besides all that they have to do with is pertinent? > Hence, we define > > ind_1(x) <-> 1 in x & (Ay)(y in x -> y+1 in x) > > and let N+ be the unique set satisfying > > ind_1(N+) & (Ax)(ind_1(x) -> N+ c x). N+ is not a subset of all inductive sets. Start with 1 and double it every iteration. Just shoft the bit to the left. > > >> >> >> 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? "Nuttin' honey" > > >> >> >> 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+? > > This can be taken as the definition of N+, though one should prove that > there is a unique set satisfying the above condition. I will do so on > request. Perfectly standard. Ummmm, you said, "Er, that was the full definition of ind_1. The definition of N+ is just below." Sorry, the only thing just below was a conclusion supposedly based on some full definition of N+. Incorrect. > > >> >> 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. > > You are confusing my use of the predicate "inductive_1" with the notion > of inductive definition. I've defined inductive_1 explicitly. Yes, using the unit increment generator for a single next element in the sequence. There is nothing that says that is the "smallest" inductive set. It's not. > > >> > 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+. > > Then it is not an ind_1 set, so it's irrelevant. N+ is the smallest > ind_1 set. > Believe that as long as you must. > > >> >> >> 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's no such thing as satisfying *only* the conditions specified. > This is why logicians specify N as the *smallest* inductive set. Piffle. No other conditions are specified, therefore the axiom of separation is satisfied, and the subset is defined. > > >> 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. > > Yes, Z contains 1 and is closed under "unit increment" (which others > call successor). That is, 1 e Z and if n e Z, then n+1 e Z. Oh, but it's not so well founded, at least not in quantitative order, which repeated increments would produce. > > This is getting tedious. If you still don't see *why* we define N+ as > the smallest inductive set, then there's no reason to continue going > point by point here. Probably not. > > One last correction however: > > >> >> 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. > > No. I've told you what ind_1 *means*. A set is ind_1 iff it contains 1 > and is closed under +1. I will not use ind_1 in any other way in this > conversation. That is the sole and complete definition of an inductive set? I beg to differ. > > In this conversation, the meaning of inductive is fixed. It is *not* > something that can be generated from the last. It means it contains 1 > and if n is in it, then so is n+1. Okay, then that is N+, one example of an inductive set. > > That is what inductive means here. > > Note: I've always disliked calling this property "inductive", because > I've found that the term is misleading. Nonetheless, the terminology is > standard. "Inductive" means well founded and transitive, basically. > > -- > "This sucks," said a Pennsylvania State University student [...] " Why > can't the college let me do what I want to do with my computer? The > school computer security guys are being way more annoying than the > spyware was." -- A student pines for his disabled spyware- Hide quoted text - Townee
From: Tony Orlow on 26 Jun 2010 21:21 On Jun 26, 8:57 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > 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. > > Well, *that*'s encouraging! > > Your system is purposely, not accidentally, inconsistent. > > -- > Jesse F. Hughes > > "I think the Iraqi people owe the American people a huge debt of > gratitude." -- G.W. Bush in January, 2007 Disproof by contradiction is a well-established method in logic, dingaling. Tony
From: Tony Orlow on 26 Jun 2010 21:29
On Jun 26, 3:49 pm, David R Tribble <da...(a)tribble.com> wrote: > David R Tribble wrote: > >> R+ is the uncountable set of positive reals. We'll order it > >> "linearly" with the usual ordering ('<' over the reals). So it > >> meets the two requirements you state above. > > >> So then, according to you, R+ "must contain an element > >> which is infinitely distant from the foundation of the set". > >> So: > >> 1. Which element of R+ is that? > >> 2. What is the foundation of set R+? > > Tony Orlow wrote: > > In the H-riffics the foundation is 0, and in base 2, 3 is infinitely > > distant from the foundation (and vice versa). You may require proof of > > this fact, and I'll work on that, because it's getting pretty > > interesting again, but for now, I'm just pretty convinced, personally. > > If you're not, I understand. > > You said nothing about your H-riffics in the requirements for an > uncountable "linearly ordered" set above. No kidding. That's one example, not the "smallest set" or anything dumb like that. > > Are you saying that in addition to the set being a) uncountable and > b) linearly ordered, it must also be c) derived from your H-riffics, > in order to answer the question, which member is infinitely distant > from the foundation of the set?- Hide quoted text - > One of probably a countably infinite number of examples. Tony |