Uncountability of the Rationals?
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From: Virgil on 16 Jun 2010 23:50 In article <ed9633cf-44d1-43c0-b59a-9b81ac39a951(a)19g2000vbi.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 3:26�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <3d816782-ee7e-4974-9202-399716bac...(a)18g2000vbi.googlegroups.com>, > > �Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > > > > > If you prefer, for set H: > > > > > > > 0eH > > > > > xeH -> -(2^x)eH ^ (2^x)eH > > > > > > > For all xeR except for x=0 exists parent element yeR:y=log2(abs(x)). > > > > Such a claim requires proof, and even if true, I doubt that TO could > > prove it in his own. > > > > > > > > > > So, 1/3 is in R because log_2(1/3) is in R, and that's in R because > > > > log_2(log_2(1/3)) is in R and so on? > > > > > > Somehow, I don't get it. > > > > > It sounds like you get it. R is closed under this pair of operations, > > > except that 0 has no parent. > > > > That is a claim that requires proof, as it is no at all obvious, for > > example, that any finite number of iterations of � > > � �{ 0eH \/ (xeH -> -(2^x)eH /\ (2^x)eH)} > > will produce 3 or 1/3 > > No finite number of iteratons using H-riffic base 2 will produce 3. I > don't dispute that. Then your numbers set is incomplete. > This is an uncountably deep structure generating > all the reals. But by your own admission, it does not generate all the reals,. > My original question, years ago, was whether it could > constitute a well order. Using what ordering on what set? It certainly cannot be a well ordering of the reals as it omits all but countably many reals.
From: Virgil on 16 Jun 2010 23:55 In article <28b55d2c-0dcb-470b-be35-d2732420fb8b(a)c15g2000vbl.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 5:22�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? > > > > 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. > > 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. This confusion is the reason that the von Nuemann naturals are to be preferred. > > > > > At the very least, you're contradicting your previous statement > > that Aleph_0 cannot be a member of N+. More egregiously, > > though, is your unjustified "implies" leap. > > OF COURSE I am pointig out a contradiction between two statements > about aleph_0! That's how I am proving, by contradiction, that the > existence of aleph_0 is false. I have no problem with a set not being a member of itself but still existing. t > > > > > So the question becomes, what axiom justifies your logical > > leap, applying a property of finite sets to infinite sets without > > largest members? > > There is no leap. What axiom justifies the existence of aleph_0? The axiom of infinity! Though that axiom does not insist that it be called aleph_0.
From: Virgil on 16 Jun 2010 23:57 In article <a879caac-8fe4-42fd-8f85-03eda87157c0(a)d4g2000vbl.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 5:26�pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > > In standard N ordinals and cardinals are the same thing, really. I > > > have never seen the need for any distinction. As a computer scientist, > > > a number is a number, whether used as a data element (cardinal) or > > > memory address (ordinal). > > > > Does this include the IEEE floating-point numbers +INF and -INF? > > You will recall that arithmetic operations on these values don't > > produce the same results as those on regular numbers. > > Floating point is mostly schlock. Less so than bigulosity.
From: Virgil on 17 Jun 2010 00:13 In article <a26d2f3b-c0e6-41e1-8d0a-2aa370ebbaf4(a)a30g2000yqn.googlegroups.com>, Transfer Principle <lwalke3(a)lausd.net> wrote: > So we drop the Axiom of Infinity from ZF. Why? It's > because TO wants all infinite sets in his theory to > adhere to his rules, yet Infinity proves the existence > of a set, namely standard omega, that doesn't adhere > to TO's rules. So we must drop Infinity. > > It is often pointed out in this type of thread that > Infinity proves the existence of a (nonempty) successor > inductive set, which is strictly stronger than mere > existence of an infinite set. It might be possible to > take advantage of this fact, and see whether it's > possible for tav to be an infinite set whose existence > does _not_ imply (in ZF-Infinity) the existence of a > successor inductive set. > > TO points out that the cornerstone to his theory is > something called ICI, Infinite Case Induction. Since > "induction" sounds like a schema, we can add to our > theory a schema for ICI. I was under the impression that TO's ICI required standard induction as a foundation, but standard induction requires that same alpha_0 that you want to dump.
From: David R Tribble on 17 Jun 2010 01:05
K_h wrote: > [...] > Note, this definition exploits the definition of an ordered > pair which itself is defined in terms of the axioms. > Informally, the bottom line is this: the definitions are > just shortcuts for things that groups of the axioms provide. > If you didn't want to use any definitions then you could > still prove every theorem but it would be a lot more busy > work -- kind of like writing a computer program in binary > instead of a programming language. Perhaps a slightly better example would be writing a program without the benefit of subroutines. Definitions are then like subroutines, i.e., shorthand versions of much longer statements based on the axioms directly. Imagine writing a fairly complex program and having to expand every subroutine call inline Sure it would work, and be just as correct, but what is the advantage of retaining the "purity" of low-level notation? A more mathematical example is having to write "1+1+1" (or even the more primitive "S(S(S(0)))") instead of the more convenient "3" every time we wanted to denote the third successor ordinal of 0. |