Uncountability of the Rationals?
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From: Tony Orlow on 18 Jun 2010 10:18 On Jun 17, 10:17 am, Brian Chandler <imaginator...(a)despammed.com> wrote: > Tony Orlow wrote: > > On Jun 16, 7:45 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. > > How can that be? If the "logical argument" is actually invalid or > nonsensical it wouldn't appeal to me, but then I wouldn't call it a > logical argument. If the logical argument is valid why should anyone > care whether you like it? Does "the moon is made of cheese" imply that "pigs can fly like birds"? What is the value of 0^0? That's really the only logical loophole, in both Boolean and quantitative expressions. > > > > >> 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? > > Invalid? Nonsense? Hard to decide. Anyway this is not what you have > been told, it is your mangled misunderstanding of what some misguided > soul whiled away a few hours trying to drum into your head. So, no axiom violated. Good. > > > If the set is an initial segment of N+ with no largest member it has > > no size. I have said this about a zillion times (or at least tav). > > I think (hope) "N+" means something recognisable like the naturals. The positive naturals, as opposed to my misuse at one point when I meant *N. That's been established. > These go on and on ("to the right") without ever ending. To me > "initial segment" means a proper subset of the naturals that "stops > somewhere", which is equivalent to saying "has a largest element". Can > you explain what an initial segment with no largest element would look > like? Like N+. Say, the initial segment of size aleph_0, or omega, or tav, or whatever you want to call it, as if it exists more than virtually. > > Hereabouts, you seem to have been saying that if a set doesn't have a > last member to "be its size" this means the set "has no size"? Please > confirm: is this the same as "has no bigulosity"? Do you have a > definition of "has a size" for a set? I mean no absolute Bigulosity, but only one in terms of tav, or whatever. > > Brian Chandler Tony Orlow
From: Tony Orlow on 18 Jun 2010 10:26 On Jun 17, 11:42 am, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> [...] the tree of values, the foundation. Additionally, if one wants to > >> include all reals positive and negative then the H-riffics may be > >> redefined as: > >> E(0) > >> E(x) -> E(-2^x) ^ E(2^x) > > David R Tribble wrote: > >> This still omits vast (uncountable) subsets of of the reals. > >> 3, for example, is not a node in the binary tree that is E. > > Tony Orlow wrote: > > It is an uncountable number of iterations down the tree. > > No, it is an countably infinite path of the tree. However, you > don't define the tree as having infinite paths, but as each > H-riffic being a node in the tree, the end of some finite path, > each being at the end of a countable number of left/right > bifurcations. > > You appear to have a basic misunderstanding of "countable" > and "uncountable". I was going to gloss over your drivel, because I don't want to get back into your standard misconceptions of the infinite tree again. Anti-Cantorian or no, Meuckenheim is correct in his arguments about the infinite tree, even if his conclusions are not what I would choose. However, you say things below I should respond to. > > Tony Orlow wrote: > > Unless, of > > course, you think there are only countably infinite segments oof this > > tree, in which case it would apprear to be some kind of well order, > > not havbing any infinite descending sequences. > > Yes, as long as the H-riffics are nodes along finite-length > paths (which is what your definition above produces), then > the set of H-riffics is countable and well orderable. > > But it is composed of only a countable subset of the reals, > and omits a rather large uncountable subset of the reals. > So I wouldn't get too excited about it if I were you. > Only when restricting yourself to the countable tree, or at least that's what I think. I'll play with the H-riffics again hopefully soon, but they aren't very pleasant. :) > David R Tribble wrote: > >> As I pointed out to you several times, any real of the form > >> r = f*k^m, for integers f, k, and m, and k /= 2, is not a node > >> in the E tree. You've never addressed this. > > Tony Orlow wrote: > > I have as I just did. Can you think of any sequence or finitely- > > branched structure that could include all the reals and yet not have > > two that were infinitely distant from each other within that > > structure? I can't. > > I can't either. If I could, I could probably imagine a structure > that could well order the reals (or some other uncountable set). If somebaody could, that would go a long way toward justifying the Axiom of Choice (or its applications). > > David R Tribble wrote: > >> Likewise, your definition is inherently described by a countable > >> binary tree, where each node is an H-riffic. Therefore, E can't > >> possibly contain all the uncountable reals, nor any uncountable > >> subset of them. > > Tony Orlow wrote: > > Not by a countable tree, no, but by an uncountable one, yes. > > Right. As longs as the each H-riffic is a node along a finite > path in the tree, the H-riffics are only a countable set. Your > definition above is exactly that. Not exactly, but whatever. > > David R Tribble wrote: > >> Now (as Walker pointed out previously) if you want to amend the > >> H-riffics to include all infinite paths in the tree, that's fine, as > >> there are an uncountable number of such paths. However, once > >> you do that, you lose the well-ordering of the tree. The countable > >> nodes are still well-orderable, of course, but the infinite-length > >> paths are not. (Choose any infinite path in the tree, then ask > >> what the "next" path is; there is none.) > > Tony Orlow wrote: > > If there is an infinite path from 2 to 3, then there is a next one > > after that, as Transfer pointed out. > > Not within any tree within infinite paths. The infinite path that > is (the Cauchy limit to) 3 does not have a sibling path. Each > finite path (H-riffic node) has left and right sibling nodes, sure, > but no infinite path does. Sure, the sibling path is -3, or 1/3, depending on whether you are using the 2nd or 1st form I offered for the set. > > It's the same problem with any infinite binary tree containing > infinite paths. Choose any infinite path (which has no end node) > and you can't get to its "next" right or left neighboring path. Different problem. This tree's uncountable. > > That's essentially the continuum problem, and the whole problem > with trying to well-order the reals (or any other uncountable set) > using countable constructions. Yep, there's a bunch of "problems" to address. > > David R Tribble wrote: > >> So you can't have both an uncountable number of elements > >> and a well-ordering on them. > > Tony Orlow wrote: > > Then you believe there is no well ordering on the reals, and you must > > therefore reject the Axiom of Choice? > > What on god's green earth gave you that idea? Because you said you can't have a well ordering on an uncountable set. :o > > When I say "you can't have" I mean "you", Tony Orlow. Specifically, > "your" H-riffics can't be both an uncountable set and a well-ordered > set. Perhaps I should have said: > | So you can't have both an uncountable number of elements > | and a well-ordering on them with a structure like yours. Oh, perhaps. So, "you" see my confusion. Tony
From: Tony Orlow on 18 Jun 2010 10:45 On Jun 16, 11:31 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 14, 10:50 pm, Brian Chandler <imaginator...(a)despammed.com> > wrote: > > > > Don't you make any connection between a set of pairs and, say, spatial > > > coordinates? When you talk about points in n dimensional space, do you > > > not define them as unique n-tuples in the spatial set of points? > > Look, let's not use "omega" to refer to any of your stuff. It's an > > abuse of respected terminology. As I understand it, there's a stage in > > which we "declare a unit infinity", so I declare mine to be Tav. OK? > > If we are specifically trying to make TO's theory more > rigorous, than I agree with Chandler, to some extent. Hi Transfer - I appreciate all the thought and input you're providing. Thanks. > > So let's start with ZF and add a new primitive symbol > to denote this "unit infinity." We could call it "tav" > as per Chandler. > > 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. Well, wait a minute. I don't object to the set N or N+ or omega, as a construction. So, as far as I am concerned, the axiom of infinity can stay. I just wouldn't pretend that the set transitively closed under membership is necessarily the greatest model of the natural numbers. My rules don't say such a set doesn't exist - all countably infinite sets are measured with respect to N+ in my theory. What I object to is assigning this set some absolute size. What we need, really, is a primitive pertaining to |S|, the size of a set. Defining axioms would dictate how such a size works, and would differ between cardinality and Bigulosity. In Bigulosity, tav=|N+| is only a "virtual" or "relative" number, and not an absolute quantity in any sense, because of the following logic, which doesn't appear to violate any of ZFC that I have heard. If we consider the set N+ as the standard countable infinity, and note that Ax (element[x]=x), and that having a set of x elements from the set, with size x, therefore means that one of them is at least as large as x, then we see that having an infinite number of them entails that one be infinite. However, since the set is defined such that it can have only finite values, such a set cannot contain its size, though it cannot lie beyond the set. Therefore, such a set size is self-contradictory, and cannot exist. I can use the term "countably infinite" so you know which kind of sets I refer to, but I really think of them as "unboundedly finite" and only "potentially infinite". > > 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. N+ is successor inductive. No problem. It just doesn't have an absolute size. > > 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. > > But how can we state the ICI Schema in a manner that's > rigorous enough for Chandler and others? We notice > that the schemata labeled TA1 and TA2 are stated in > terms of real numbers, leading posters like MoeBlee > to ask for the definition of "<" and other symbols. For > after all, in standard theory, we start with omega and > define (rational numbers, then) real numbers, but here > there is no omega, so there aren't any reals yet. The definition of '<' is not at all difficult. It is an operation that follows the following rules: (a<b ^ b<c) -> a<c ~(a<b ^ b<a) ...or... (a<b) -> ~(b<a) ~(a<b v b<a) <-> (a=b) Likewise we can define the field axioms for the behavior of +, -, *, /, ^, and log to determine the behavior of such operators when applied in formulas. We can define ICI as: (Ea>0 : AneN+ f(n)-g(n)>a) -> ((AneN+ tav>n) -> (f(tav)>g(tav))) That is, "If there exists an 'a' greater than 0 such that for all positive naturals n, f(n)-g(n)>a, then if tav is greater than any finite, f(tav)>g(tav)". Here, really "tav" can be any ifinite set, countable or uncountable. IFR can be stated, in terms of well formed formula g: |{x: x<=a ^ x>=b ^ g(x)eN+ ^ g(x+1)>g(x)}| = floor(g(b))-ceiling(g(a)) +1 (monotonically increasing) and |{x: x<=a ^ x>=b ^ g(x)eN+ ^ g(x+1)<g(x)}| = floor(g(a))-ceiling(g(b)) +1 (monotonically decreasing) Then, we can define IFR *with* ICI, and state that: |{x: g(x)eN+ ^ g(x+1)>g(x)}| = floor(g(tav))-ceiling(g(1))+1 (the floor() here is superfluous) and |{x: g(x)eN+ ^ g(x+1)<g(x)}| = floor(g(1))-ceiling(g(tav))+1 (now ceiling() is superfluous) Doesn't tht sound like a relatively valid outline? If we take the field axioms to be inviolable, then we have to discard certain conclusions of transfinite set theory as a result of ensuing contradictions, no? > > Since our language contains the primitives "e" and "tav" > it would be preferable to state ICI in terms of these > primitives, not reals or the "<" relation. In another > thread, I once posted the following schema, which is a > rewriting of TA2 using these primitives: > > If phi doesn't contain "tav" then all closures of: > > (phi(0) & Ax (phi(x) -> phi(xu{x})) -> phi(tav) > > are axioms. By "xu{x}" you mean, successor(x) or x++ or x+1, yes? Or does the set- theoretic expression actually serve more purpose than to denote a next step? The problem with "phi" is it's too general. restrictions must exist on the nature of "phi". > > But then Hughes let phi be the statement "x is finite" > (using a standard definition of finite) to prove that tav is > finite, yet TO wants tav to be infinite. Furthermore, if > we let phi be the statement "0ex or 0=x," we can prove > that either 0etav or 0=tav. At this point, I suspect that > tav is actually the empty set -- which is definitely not > what TO intends tav to be. No, that wouldn't do. Of course "x is finite" has always been the big objection to ICI. I don't "want tav to be infinite". It's neither finite nor infinite, and so, only virtually exists. It's the Bigulosity of N+, with which all other countably infinite sets must be compared for Bigulosity to be determined. > > So let's change the schema to the following: If phi doesn't > contain "tav" then all closures of: > > (En (n fin. nat. & phi(n)) & Ax (phi(x) -> phi(xu{x})) -> phi(tav) > > are axioms. Here "fin. nat." refers to the (standard) definition > of finite natural. > > This new schema reflects TO's comments about how he > can prove 2^tav > tav^2. We can let phi be the statement > "2^x > x^2" and n be 5. > > But we still have problems. Notice that by letting phi be > the statement "x is finite," we still have tav finite -- yet > by letting phi be the statement "mex" for each natural > number m in turn (and letting n = m+1, of course), we > can prove "me(tav)" for _every_ natural number m. Yes, that's why my restriction is on the nature of phi itself. > > So somehow tav is finite, yet for every natural number m > we can prove that tav has m as an element? At first this > sounds like a contradiction, but it was mentioned in earlier > threads that: > > For every natural number n, it is provable that tav contains n. > > and > > It is provable that for every natural number n, tav contains n. Is it true that AneN+ neN+? > > are distinct (in first-order logic at least -- I'm not sure about > second-order logic). But even so, this would make tav a > really strange set to contain every natural number as an > element, yet still somehow be "finite" -- and it certainly > doesn't describe tav as intended by TO. It sort of does, actually, but if you could please state your two English sentences there in FOL it might be illuminating. > > Furthermore, since we have all the other axioms of ZF > besides Infinity, we do have the Separation Schema, and by > this schema, the set: > > w = {me(tav)} | m (standard) finite natural} > > must exist. But what set is w? It would appear to equal > omega, yet is still a "finite" set? Problems, problems.... > > What we really need to avoid this contradiction/paradox is to > allow some instances of the ICI schema and not others, but > MoeBlee uses the word "oracle" to describe this. Indeed, back > when TO was describing ICI, I even _warned_ him that his > schema might be described as "oracular," since "oracle" is a > buzzword commonly used by MoeBlee to object to theories > (just as Hughes uses the phrase "ad hoc"). So in order to > find a schema acceptable to MoeBlee, we must avoid these > "oracles" at all costs. They like their dismissive phrases. I have been pretty explicit, and even more so than ever, above. > > Note that, since we are trying to avoid the existence of omega, > we can't allow for a set to which the Separation Schema can > be applied to it to recover omega. This is why, in a post that I > wrote yesterday, I wish to consider sets without a transitive > closure, since one _can't_ apply Separation to such a set to > recover omega. But then again, it's not obvious how such sets > can describe anything that TO, or any other poster (save > perhaps zuhair), is writing. We aren't trying to do any such thing. Maybe Brian is. I dunno. Thanks for your input. Hope I've tipped my hand enough to keep you thinking. TOny Tony
From: Jesse F. Hughes on 18 Jun 2010 10:40 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 17, 9:31 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Tony Orlow <t...(a)lightlink.com> writes: >> >> And, yes! subset is DEFINABLE from 'e'. >> >> >> Just as 'is an ordinal' is DEFINABLE from the mere primitives. >> >> > Is it? Demonstrate as I did for you just now. >> >> x is an ordinal iff (Ay,z in x)(y in z or z in y) & >> (Aw,y,z in x)((w in y & y in z) -> w in z) & >> (A z)( ((Ay)( y in z -> y in x ) & (Ey)( y in z )) -> >> (Ey in z)(Aw in z)( y in w ) ) & >> (A y in x)(A z in y)( z in x ). >> >> Not so hard. The first two clauses assert that elementhood is a >> linear order on x (anti-reflexivity is trivial, assuming Regularity), >> the next clause that it is a well-ordering and the final clause that x >> is transitive. Someone else will surely correct any errors in the >> above. >> >> Tell me, Tony, did you *really* think it was impossible? Did you >> *really* think that "x is an ordinal" is literally undefinable in the >> language of ZFC? And that generations of mathematicians have been >> pulling one over on the public? > > You miss the point. I never said there was no deifnition. I said the > definition does not "floow from the axioms" in the sense that it is > not a theorem of the axioms. You're mighty confused. Definitions are never theorems of the axioms. > It may be a statement "consistent" in that there is verifiably no > contradiction, however, that does not mean that deifnition is entailed > by the axioms in any way. I rather think my definitions for IFR, IC, > and the combination are considerably more succint and therefore > matheatically beautiful. Just my opinion.... Maybe you mean: (Ex)(x is an ordinal) should be a theorem. Well, it is. In fact, so is (Ex)(x is an infinite ordinal), (Ax)(if x is an ordinal, then so is s(x)) and a whole slew of other theorems. But this complaint that the definition is not a theorem of the axioms is quite literally confused. I honestly have no idea what you think you're expressing this way. -- Jesse F. Hughes "There's absolutely no information here." -- Hank Hill, on blogging and information theory.
From: Jesse F. Hughes on 18 Jun 2010 11:27
Tony Orlow <tony(a)lightlink.com> writes: > IFR can be stated, in terms of well formed formula g: > > |{x: x<=a ^ x>=b ^ g(x)eN+ ^ g(x+1)>g(x)}| = floor(g(b))-ceiling(g(a)) > +1 > (monotonically increasing) Tony, you do understand, I'm sure, that a well-formed formula is not a function? I don't think you meant "well-formed formula g". -- "This confused and outraged many Matrix fans, who'd already spent hours on the web explaining that man and computers could never really live in such a state of harmony and mutual benefit." -- http://www.pointlesswasteoftime.com |