Uncountability of the Rationals?
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From: Brian Chandler on 20 Jun 2010 12:29 Jesse F. Hughes wrote: > Tony Orlow <tony(a)lightlink.com> writes: > > > Yes, the size of P is not derivable given the methods I have > > suggested. However, I would maintain that if lim(x->oo, y->x: P(y)/N+ > > (y)) = y/log y, then in this sense ICI applies, even if IFR is > > irrelevant. Do you see the distinction? > > Well, I first notice that P(y) and N+(y) are so far utterly undefined > and thus you haven't expressed yourself clearly. As far as ICI, I'm > afraid I don't recall what it is, ... Come on. It's "Infinite Case Induction". > ...but I'm pretty sure that you haven't > said anything that implies the conclusion you've just drawn. Not in what you are talking about Jesse. But the fundamental problem is obvious: Tony is talking about something else entirely. You're not sure -- I'm not sure -- what the "+" is in "N+", but you interpret it as referring to the naturals. Tony *claims* to be talking about something with the same name, but he is talking in another world. To you or me, "N", or the naturals, is a simple set, containing p iff p is either 0 or 1 more than another element of N. By ("non-infinite- case") induction, all of these p are finite, normal, naturals ("Pofnats"). Tony has several(!) totally bogus "proofs" that if there is a set containing all of these Pofnats, some of them must be "infinite" (which is actually undefined). So his "N+" includes normal numbers like 1, 2, 57, 1200000000000000000, and so on, but as the 'so on' progresses, these meld mysteriously into wormy things with 100... at the left end and ....789 at the right end and "infinite sequences" of other digits in between. Further 'so on' yields even more mysterious wormy things, perhaps now with an "uncountable infinite sequence" of dots in the middle. And really this baroque creatibility never ends. Tony declares this and that -- how can you possibly object? Brian Chandler
From: Tony Orlow on 20 Jun 2010 12:32 On Jun 20, 11:53 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > Yes, the size of P is not derivable given the methods I have > > suggested. However, I would maintain that if lim(x->oo, y->x: P(y)/N+ > > (y)) = y/log y, then in this sense ICI applies, even if IFR is > > irrelevant. Do you see the distinction? > > Well, I first notice that P(y) and N+(y) are so far utterly undefined > and thus you haven't expressed yourself clearly. Sorry, I thought maybe you could interpret this and understand the connection to ICI, and relate it to the page I gave the link for regarding the aymptotic nature of P. P(y) is the number of primes less than or equal to y, and N+(y) is the number of positive naturals less than or equal to y (which would be y). As far as ICI, I'm > afraid I don't recall what it is, but I'm pretty sure that you haven't > said anything that implies the conclusion you've just drawn. ICI is infinite case induction, which asserts that if any inequality holds for all neN+ greater than some value a, then it also applies to any positive infinite value, provided the difference establishing the inequality does not have a limit of 0 in the positive infinite case. There are always fewer primes than naturals in any initial segment of N + greater than 3, and if the asymtotic limit is x/log x for the number of primes per x naturals, then the xth tends to fall around x/log x. Then we may map the primes, "on average", from N+ using n=x/log(x), and the primes may be "close" to the inverse of that function. What is the inverse of x/log(x)? In the page I cited it estimates the nth prime by two methods: n log n and the better n(log n_log log n -1). Which is closer to the inverse of n/log n? Hmmmm... > > -- > "Sure, maybe I have a tiresome task that is nearly impossible, but > part of who I am is an endless amount of energy as long as there is > hope. Without hope, I find that I start to lose focus, and feel, just, > well, hopeless." -- James S. Harris I can kind of relate to that, being human and all. TOny
From: Jesse F. Hughes on 20 Jun 2010 13:30 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 20, 11:53 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Tony Orlow <t...(a)lightlink.com> writes: >> > Yes, the size of P is not derivable given the methods I have >> > suggested. However, I would maintain that if lim(x->oo, y->x: P(y)/N+ >> > (y)) = y/log y, then in this sense ICI applies, even if IFR is >> > irrelevant. Do you see the distinction? >> >> Well, I first notice that P(y) and N+(y) are so far utterly undefined >> and thus you haven't expressed yourself clearly. > > Sorry, I thought maybe you could interpret this and understand the > connection to ICI, and relate it to the page I gave the link for > regarding the aymptotic nature of P. > P(y) is the number of primes less than or equal to y, and N+(y) is the > number of positive naturals less than or equal to y (which would be > y). > > As far as ICI, I'm >> afraid I don't recall what it is, but I'm pretty sure that you haven't >> said anything that implies the conclusion you've just drawn. > > ICI is infinite case induction, which asserts that if any inequality > holds for all neN+ greater than some value a, then it also applies to > any positive infinite value, provided the difference establishing the > inequality does not have a limit of 0 in the positive infinite case. > There are always fewer primes than naturals in any initial segment of N > + greater than 3, and if the asymtotic limit is x/log x for the number > of primes per x naturals, then the xth tends to fall around x/log x. > Then we may map the primes, "on average", from N+ using n=x/log(x), > and the primes may be "close" to the inverse of that function. What is > the inverse of x/log(x)? In the page I cited it estimates the nth > prime by two methods: n log n and the better n(log n_log log n -1). > Which is closer to the inverse of n/log n? > Hmmmm... As I recall it now, ICI simply said: if lim f(x) - g(x) > 0, then f(N+) > g(N+). All you can say here, as far as I can tell, is: There are more naturals than primes. This stuff about mapping primes "on average" is not a consequence of ICI. You're simply grasping at straws. -- Jesse F. Hughes "Yesterday was Judgment Day. How'd you do?" -- The Flatlanders
From: Virgil on 20 Jun 2010 15:15 In article <42564f99-5e6a-48af-85de-a73c363483c6(a)z10g2000yqb.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 17, 4:44�pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > >> Sometimes there are logical arguments themselves which don't appeal to > > >> everyone equally. What I have been told is that the size of omega must > > >> be larger than every natural since no natural is large enough to > > >> express it, and so it is some infinite number, aleph_0. However, that > > >> logical argument, as I pointed out, really just proves that aleph_0 > > >> cannot be finite. Along with the argument that any initial segment of N+ > > >> of size x contains an xth element whose value is x, which would > > >> imply that aleph_0 or omega is a member of N+, we arrive at a > > >> contradiction implying that aleph_0 cannot actually exist. Which axiom > > >> contradicts this logic? > > > > David R Tribble wrote: > > >> The problem is that there is no logic there. > > >> You consider finite sets of the form {1, 2, 3, ..., k}, all of which > > >> have a least and a greatest member, to have size k and to also > > >> have k as a member. Then you take an unjustified leap and > > >> claim that this implies that sets of the form {1, 2, 3, ...}, > > >> which do not have a largest member, to also have a size equal > > >> to one of their members. > > > > Tony Orlow wrote: > > >> No, they all have a size, and if that size is n, then there exists an > > >> nth. If there does not exist an nth, then there are not n elements in > > >> the sequence, which this set also is. > > > > David R Tribble wrote: > > >> But what if a set has an nth member, for every n in N+, but has > > >> no largest member? Does the set still have a size? Is that size > > >> a member of the set? If so, is it the largest member or some > > >> other member? > > > > Tony Orlow wrote: > > > Is that supposed to answer my question? Is that the Axiom of > > > Avoidance? Which axiom did I violate? > > > > No, it points out the error in logic that you made. > > > > You didn't violate any specific axiom. There is no "axiom that > > contradicts your logic". However, you didn't use any of the > > existing axioms when you made the logical jump from stating that > > a property of sets of the form {1,2,3,...,k} also applies to sets of > > the form {1,2,3,...}. Therefore the second half of your argument > > is unfounded (not based on the axioms and theorems at hand), > > and no logical conclusion can be drawn from it. > > > > It's a logic error of omission, or drawing a conclusion from a > > statement without justification. Tell us, what axiom or theorem > > justifies that jump? > > > > David R Tribble wrote: > > >> By your logic, you state that all sets of the form {1,2,3,...,k} > > >> have a largest member, therefore sets of the form {1,2,3,...} > > >> also have a largest member. I assume you have an axiom > > >> or theorem in mind to help you reach this conclusion. > > > > Tony Orlow wrote: > > > No, read carefully. I said that every initial segment of N+ with size > > > x has an xth element with value x. The contrapositive would be that if > > > an initial segment of N+ does NOT include an xth element, or the > > > element x, then it is smaller than x in size. > > > > Yes, correct so far. That is true of all sets that are finite initial > > segments of N+. > > > > > If N+ cannot contain aleph_0 then the size cannot be aleph_0. > > > Thus, the assignment of an absolute size to this set results in > > > contradiction. > > > > That is the same fallacious logical jump. N+ is not a finite initial > > segment of N+, so your requirements for finite initial segments > > of N+ (to have an xth element, or for the size to be a member of > > the set) do not apply to N+. > > > > So again, the same question is, what axiom or theorem justifies > > you to state that your property of finite sets also applies to > > infinite > > sets?- Hide quoted text - > > > > First of all, I never said anything about *finite* initial segments of > N+. You cannot insert that condition into my argument, then claim that > the inconsistency it introduces into my argument is my fault. I stated > that every initial segment of N+ with size x has an xth element whose > value is x, without reference to finiteness. Then your N+ is incompatible with the von Neumann naturals. You agree with the > statement in the finite case, but object when I apply it to the > infinite case. Why? Because it isn't true for any standard set of naturals, so your definition of the naturals must have these two unnatural properties" 1. It does not contain 0, and 2. It contains a largest member. > You say it is "unjustified", but your objection is > far more so. If it is true that there exists an xth element in any > sequence of size x Since "it" doesn't, the issue is moot.
From: Virgil on 20 Jun 2010 15:21
In article <98e5dd81-5220-4c92-bd4d-f995ea093830(a)d37g2000yqm.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > When you say the existence of omega "follows from the axioms," > that sounds to me like it can be deduced solely from the axioms. Since omega in not even mentioned in the axioms, it does not sound to me like it can be deduced solely from those axioms, at least not without at least a definition for it over and above those axioms. But given the appropriate definitions, which really add nothing but abbreviations to the system, then the existence of omega does "follow from the axioms". |