Uncountability of the Rationals?
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From: Tony Orlow on 26 Jun 2010 21:47 On Jun 26, 8:53 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(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? Absolutely not!!! As already proven, tav is neither finite nor infinite, and therefore exists only virtually, or "potentially". If we assign this size "tav" to that set, then it is not defined as belonging to either group of quantities. > -- > 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- Hide quoted text - > Journals get me all hot. Tony
From: David R Tribble on 26 Jun 2010 22:19 Jesse F. Hughes wrote: >> Is it your opinion that N+ has a last element? > Tony Orlow wrote: > No, and that's beside the point. If there are an infinite number of > unit increments between any two values then there is an infinite > difference between them. If not, then it's a countable sequence. In R (the reals), there is no infinite "number of unit increments" and no infinite difference" between any two elements in R. Yet it is not a countable set. In contrast, the ordered set: S = N U { w } = { 0, 1, 2, 3, ..., w } has an "infinite distance" between w and all of its other members, yet it is a countable set. > N+ is countable. Some subsets of R *might* be measurable even though not > continuous (perhaps the Cantor set?), Consider the discountinuous subset of R: D = { x | 0 < x < 1 or 2 < x < 3, for x in R }, i.e., the union of the real intervals (0,1) and (2,3). Unless I'm mistaken, it is "measurable".
From: David R Tribble on 26 Jun 2010 22:36 Tony Orlow wrote: >> Where you have a minimum positive difference between successive >> elements (like 1), then any infinite number of them means an infinite >> difference between the first and last. > Jesse F. Hughes wrote: >> Is it your opinion that N+ has a last element? > Tony Orlow wrote: >> No, and that's beside the point. You entirely missed the point of Jesse's question. You claim that for an infinite ordered set with a minimum difference between members, the set must have an infinite difference between the first and last elements. N+ is an infinite set with a natural order, and has a minimum difference between its members. So by your claim, the set must have an infinite difference between its first and last members. So Jesse asked the obvious question: does N+ therefore have a last element? If not, then perhaps you meant something other than what you actually wrote?
From: Jesse F. Hughes on 26 Jun 2010 22:32 Tony Orlow <tony(a)lightlink.com> writes: >> > 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? > > Absolutely not!!! > > As already proven, tav is neither finite nor infinite, and therefore > exists only virtually, or "potentially". If we assign this size "tav" > to that set, then it is not defined as belonging to either group of > quantities. I don't understand any of those words. Let's try again. Do you agree with each of the following? (1) Every element of N+ is a finite natural number. (2) Tav is an element of N+. You said (2) *just now* and, unless I'm much mistaken, you've agreed with (1) quite recently. Now, bear with me here and see if you can't understand how (1) and (2) imply (3) Tav is a finite natural number. -- "So why talk [about my factoring method] out on Usenet? Because it's a highly public place so I'm unlikely to disappear[...] You people are my protection. [...] You may be what's keeping me free and walking out in the open air." -- James S. Harris, theory guy on the edge.
From: Jesse F. Hughes on 26 Jun 2010 22:37
Tony Orlow <tony(a)lightlink.com> writes: > 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. Yes, it is. So? It doesn't mean that people find inconsistent theories in classical logic very interesting. (Though, I really do appreciate your patient tutelage in logic, kind master. Thanks!) -- "Not all features that are found on the Security tab are designed to help make your documents and files more secure." --Microsoft on Office security features (after it was pointed out by a third party that a certain password setting is easily bypassed.) |