Uncountability of the Rationals?
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From: Tony Orlow on 16 Jun 2010 17:54 On Jun 16, 2:52 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <f6d161b0-150a-4c47-bea4-bc898e5a0...(a)i31g2000yqm.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> 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? > > Using the von Neumann naturals, or any other set of naturals which > starts with 0, as is becoming fairly standard in mathematics, no natural > need be anything like a member of itself, which is a consummation > devoutly to be wished. Did you just state an axiom? Why don't you answer this simple question? You know why, don't you? Tony
From: Tony Orlow on 16 Jun 2010 17:58 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. This is an uncountably deep structure generating all the reals. My original question, years ago, was whether it could constitute a well order. Tony
From: Virgil on 16 Jun 2010 18:01 In article <c6f1e881-44d9-45ca-9ae1-4715d3571d26(a)32g2000vbi.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > So it appears to me that I don't know 1/3 is in R at all. �This chain of > > reasoning does not end in a statement that I know. �Can you give me a > > finite proof (or, let's be generous, anything that looks like a proof) > > that 1/3 is a real number. > > > > You already know that fact. 1/3 is a real number, rest assured. There are two distinct definitions of the reals here. The standard set of reals and the set, R` generated by: 0 e R` and [ x e R` ==> ( 2^x e R` and -(2^x) e R` ) ]. While it is fairly obvious that R` is a subset of R, there has not yet been any proof that R` = R. And I do not believe it.
From: Virgil on 16 Jun 2010 18:03 In article <1f0f83e5-6123-417f-84ff-701f82219790(a)37g2000vbj.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 16, 1:09�pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > "Jesse F. Hughes" <je...(a)phiwumbda.org> writes: > > > > > So it appears to me that I don't know 1/3 is in R at all. �This chain of > > > reasoning does not end in a statement that I know. �Can you give me a > > > finite proof (or, let's be generous, anything that looks like a proof) > > > that 1/3 is a real number. > > > > In fact, I'd be interested also in a proof that 8 is a real number. > > That, of course, would follow from the existence of 3. But in TO's system of "reals" we do not know that 3 is a "real".
From: Tony Orlow on 16 Jun 2010 18:04
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. > > 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. > > 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? TOny |