An uncountable countable set
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From: Tony Orlow on 25 Aug 2006 08:50 Virgil wrote: > In article <ecikqd$a1o$1(a)ruby.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > >> imaginatorium(a)despammed.com wrote: > >>>> It's easy to say that without mentioning any specific inconsistencies. >>> Well, here are a few statements from you, culled from nearby posts (at >>> least in the google "chronological order")... >>> >>> (Quoted in post by Moeblee above) >>> Tony Orlow wrote: >>>> I have never said I think there is a largest natural. I have said that >>>> some of your assumptions lead to that conclusion. >>> So you think there is no largest natural? >> Right. There is no end to counting. >> >>> (Quoted in post by you, three down the list) >>> David R Tribble wrote: >>>> Tony Orlow wrote: >>>>> But Monsieur, what about the injection from P(N) into N, via the bit >>>>> strings which denote set membership, each of which also corresponds to a >>>>> binary natural? Tsk, tsk. Mustn't forget that one! Remember, the only >>>>> set which doesn't map is the entire set, and that maps to the largest >>>>> natural, that is, ...1111 with all bits in finite positions. >>> But you think that "...1111" _is_ the largest natural. >> I said, given that all bit positions are finite, this constitutes a >> finite value (previously inductively proven), and is the largest finite >> (given that no other string can be greater in any bit position), as well >> as the mapping to the entire set. Since neither finite naturals nor >> countably infinite subsets thereof have an end, and this string doesn't >> either, it represents both equally well. It's not a proper value, >> however, within the T-riffic system. In order to calculate relative size >> of infinite sets, one needs a variable range. > > Thus TO claims that there is no largest and also that there is a largest. > Any system like that is to be avoided at all costs. > >>> I wonder if you see any inconsistency in the Finlayson thingies: where >>> between every pair of rationals is another rational, yet each rational >>> has an adjacent rational to the right of it? >> I'm not sure Ross has ever said any such thing. > > Others are sure. > >> The Finlayson numbers >> are nilpotent infinitesimal "degenerate" intervals, sequential yet >> indistinguishable on the finite scale. > > This garbage about scales again. > > If a < b on any scale then a < b on every scale. A MATHEMATICAL > inequality does not become an equality when you put away your magnifying > glass. > > Nor does a MATHEMATICAL equality become an inequality when you look at > it under higher magnification. > > Does TO also claim that while x is a member of A at one magnification, > it need not be so at another, or that A being or not being a subset of B > depends on the magnification used by the observer? > > Silliness compounded! No, it all rests on the notions of identity and equality. As Leibniz pointed out, when the properties of two objects are all exactly the same, then they are the same object. So, when we say two numbers are equal, that means all properties of the two are equal. However, that assumes that we have looked at all properties. We may consider two objects equal when in fact they are not, due to a property overlooked. For instance, in terms of location alone, using a point set topology, the diagonal line and the staircase in the limit are equal, whereas there is a clear distinction when we take direction into account using a segment sequence topology, as I showed you months ago. Similarly, if a difference less than any finite is always considered 0, then such a difference does not serve as a distinguishing property between values, whereas if it CAN be considered nonzero, then it can be used as a distinguishing property between numbers. :)
From: Tony Orlow on 25 Aug 2006 08:55 Albrecht wrote: > Tony Orlow schrieb: > >> Albrecht wrote: >>> Virgil schrieb: >>> >>>> In article <1156163893.101419.232240(a)p79g2000cwp.googlegroups.com>, >>>> "Albrecht" <albstorz(a)gmx.de> wrote: >>>> >>>>> Infinite sets are self contradicting. >>>> Not in ZF or NBG. What are the axioms of Storz's system? >>> There is no relevance in which system the axiom is found. >>> E.g. the Axiom A: "Axiom A is wrong", is self contradicting, regardless >>> of which other axioms are used, I think. The same holds for the axiom >>> of infinity. >>> >>> Best regards >>> Albrecht S. Storz >>> >> Hi Abrecht - >> >> From that statement I would guess you ascribe to the notion of >> universal consistency within mathematics, that not only must each theory >> be internally consistent, but the axioms of each theory must not >> contradict those of any other, since math describes one universe. How >> far off am I from understanding your position? >> > > > Yes, I think you meet my intention nearly. > But it's not because math is an universe, it's because our world is an > universe. You may study the structure of an strange universe; but the > heart of math should be dedicated to our real universe and the > structures of it. Yes, and the universe is consistent by definition, so math should be consistently overall as well. > And set theorists like to claim their field as to be math itself. Yes, "pure" math supposedly has nothing to do with reality. That's just an excuse for maintaining the belief in transfinitology. > > With concepts like "infinity" we are near to the boundaries of our > understanding. So we have to reason very carefully. Yes, personally I want concepts of infinity to be compatible with the rest of math, as a logical extension. There is something amiss in set theory in this respect. > > I'm just going on holiday for some weeks. > Have a good time and > So long! O! You mean you're not going to continue the argument while on vacation? Understandable. Well, enjoy! Bring me back a souvenir. ;) > > Albrecht > :) Tony
From: Tony Orlow on 25 Aug 2006 09:01 Dik T. Winter wrote: > In article <44ed99b7(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > ... > > > > > Well, you and he are not. The logic is flawed. > > > > > > > > For 1, 2, and 3, order and value are identical, even according to your > > > > logic, I suppose. Where does the first deviation happen? Which one does > > > > deviate first, i.e., which one does first be larger than the other? And > > > > why? > > > > > > It is the case when the set size is not a natural number. > > > > And, with the addition of which natural number to the set does the set > > achieve this unnatural size? > > There is no such specific natural number. It is when we have them all, > but as there is no largest number, this can not be achieved by taking > them one by one. The set of all naturals numbers consists of only natural numbers. There is NO natural number where the count becomes infinite. So there is no point in the set, even if you COULD get to the "last" one, where any infinite set has been achieved. > > > > > > And his proof is not a proof. > > > > > > > > O course not, because every proof which has an unpleasant result is not > > > > a proof. > > > > > > Then show that the set of all natural numbers does not have cardinal number > > > aleph-0 and ordinal number w. A proof please. > > > > I have already shown how the set of bit positions in the binary naturals > > has no cardinal or ordinal that can be assigned to it. > > Yes, but your showing was not a proof. Yes, it is. A finite number of positions is too few, so it must be infinite, but the smallest infinite is too large. So, you have a contradiction when trying to assign a cardinality to this sequence of bits. There is no cardinality for K which does not produce a contradiction. > > > > > > > For my part, I agree that the set of finite > > > > > > naturals is finite, though unbounded, > > > > > > > > > > In that case you are not using standard mathematical terminology. I > > > > > have no idea what a finite but unbounded set is. > > > > > > > > That's why you cannot understand mathematics. You fall back behind > > > > Cantor. He knew it. > > > > > > Oh, perhaps *you* do not understand mathematics? Earlier I have already > > > written that a few things that Cantor has written do not conform with > > > current thinking. But see <http://mathworld.wolfram.com/FiniteSet.html>: > > > A set X whose elements can be numbered from 1 to n, for some positive > > > integer n. > > > > Yes, given the current understanding, any unbounded set is infinite. > > Sorry, that is not current understanding, that is using the definitions > and terminology. Understanding may change, but within the definitions > and the terminology any unbounded set will remain infinite. FIne, the current "system" then. > > > > A set is infinite if it is not finite. A set is Dedekind infinite if > > > it can be mapped to a proper subset of itself, and it is Dedekind > > > finite if it is not Dedekind infinite. When we assume the axiom of > > > choice, the two notions are identical. Without that axiom there > > > can be infinite sets that are Dedekind finite. Do you want to know > > > more about set theory? > > > > > > Now, using, this terminology (pretty standard), what is a "finite but > > > unbounded" set? > > > > Using that system, the phrase is senseless, but that system is not the > > real universe. It's a concoction. > > Well, that is just a statement that (in my opinion) is senseless. Only in the context of the Dedekind definition of infinity, which one is not obligated to consume wholesale. Until one can prove that transfinite set theory is "correct", no one is obligated to accept the theory at all. TOny
From: Tony Orlow on 25 Aug 2006 09:12 Dik T. Winter wrote: > In article <44ed9fcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > > > In article <ecig8i$ta$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > > Dik T. Winter wrote: > > > ... > > > > > The first part of his first proof shows that a complete ordered field > > > > > has cardinality larger than the natural numbers. In his proof he did > > > > > not rely an any properties of the reals other than that they form a > > > > > complete ordered field (he uses reals to exemplify). > > > > > > > > Specifically, he demonstrates that there will always exist an > > > > unaccessible real in any finite interval, given that we are only allowed > > > > a finite (natural, that is) number of iterations. > > > > > > Sorry, you have got that wrong. We are allowed an infinite number of > > > iterations indexed with finite numbers (whatever that may mean). > > > > Thagt means that at no time have we ever completed an infinite number of > > iterations. That's why we cannot arrive at that intermediate value. > > Indeed. Not step by step. And so? I have said this before. I am quite > sure others have said that before. You can not get the set of naturals > by adding the one by one. It is the axiom of infinity that asserts that > that set does exist. And that it's infinite. Why do you need an axiom for that? WHy is it not derivable logically? > > > > > Therefore, indeed, the > > > > number of reals in a finite interval is greater than any finite natural. > > > > That proof is essentially valid, but it's not a proof that the reals > > > > cannot be linearly ordered in a discrete manner. > > > > > > This is approximately right. The number is larger than the number of > > > finite naturals. Whether the reals can be linearly ordered in a > > > discrete manner (eh? quite a few buzzwords here without definition) > > > depends on the axiom of choice, and a few more things. > > > > Using the axiom of choice, it is provable that every set can be well > > ordered, which entails a linear order, but this proof gives no clue as > > to how this can be accomplished. But, well-ordering aside, the H-riffic > > numbers give a sequential list of all reals though exponentiation. The > > other approach is Ross, which relies on the notion of infinitesimally > > different reals, sequential in their natural order. > > But the last is irrelevant in the context of the paragraph above. There > is no proof whether the reals can be linearly ordered or not. Unless > you assume the axiom of choice. And under other discussions you can > prove that the reals can not be linearly ordered (I think). So *of > course* Cantors proof is not a proof that the reals cannot be linearly > ordered. If Ross and I each have our linearizations of the reals, then they exist, independent of the axiom of choice (which might as well be called the axiom of free will). > > > > > > This proof comes > > > > > close to the proof provided by Hessenberg, but is, in my opinion, > > > > > a bit less strict.) > > > > > > > > It is related to the powerset through |P(S)|=2^|S|, but he did it in > > > > decimal, no? > > > > > > No. Nowhere in his papers did Cantor use decimals. > > > > Perhaps it is just the explanations of his second proof of > > uncountability which I have seen. He did use SOME digital representation > > of his list of reals. Was it binary? > > None at all. The second proof (the diagonal proof) is *not* about reals. The second proof of the uncountability of the reals is not about reals? That's news. What was it about, then, in your opinion? > Zermelo has indicated how it *can* be transformed to a proof for the reals > using binary notation and noted the probles in that case with dual > representations and how to solve it. Later somebody has transformed it > to a proof using decimals, but I do not know who has done that. > > > > > In any case, the powerset relation boils down to the > > > > symbolc equation N=S^L, where S is the number of logical states allowed > > > > (there can be more than two in various systems) and where L ultimately > > > > corresponds to the size of the root set. So, the two are related. > > > > > > You make here as much sense as in much earlier articles. > > > > What part of does not make sense? I suppose it's hard to put that > > concept into a sentence or two and be readily understood. > > It works for finite sequences, but your conclusions do *not* work in the > infinite, because transfinite induction fails. I'm not using "transfinite" induction, but straight-up infinite variables and the notion that any infinite is larger than any finite. Quite simple stuff. > > > > > I don't disagree that the first is infinite while the second is > > > > unbounded bt finite, and therefore smaller. > > > > > > Pray, for once, provide a definition. A set is either finite or infinite. > > > And if a set is finite, by the definitions there is a largest element. > > > So, how do you define "unbounded but finite"? > > > > Not with the Dedekind definition. A truly infinite set must have > > elements infinitely many position beyond other elements, the way I see it. > > Yes, but in that case you are not using standard mathematical terminology. > And I have yet to see a *definition* of you about *infinite*. Larger than any finite. The set of naturals is as large as, but no larger than, every natural. > > > > > Yes, the second is really a proof about power set and/or symbolic > > > > representations of quantites. > > > > > > Not at all. There is no question about "representation of quantities". > > > > In the second proof of uncountability of the reals, there are no digital > > strings? What list is he deriving the antidiagonal from, a shopping list? > > Pray read. His second proof is *not* about the uncountability of the > reals. There are no reals involved at all. It is about (infinite) sequences > of symbols (he uses 'w' and 'm'). Then I don't know what proof you are talking about. When people say "Cantor's second", they are generally referring to his second p
From: Tony Orlow on 25 Aug 2006 09:15
MoeBlee wrote: > Tony Orlow wrote: >> MoeBlee wrote: >>> Please say what sentence and its negation you believe are both theorems >>> of set theory. >> I'm not sure how this would be proven in set theory (I don't think it >> is), > > So you don't know that set theory is inconsistent (by 'inconsistent' I > mean the usual definition). > >> but it appears to be a belief, anyway, that all sets can be >> classified through cardinality. > > With suitable axioms, we can define a cardinality operation 'card' so > that we get a theorem: card(x)=card(y) <-> x equinumerous with y. Given a set x, can we always determine card(x)? > >> However, the set of bit positions >> required to list the naturals in binary defies classification in this >> system. > > Whatever you mean by that, based on remark that you know of no proof of > a contradiction in set theory, it is a problem of your own > misconception and not a contradiction in set theory. Fortunately, we > don't obligate ourselves to the burdens of your own misconceptions. No, but you are obligated to define a cardinality for this set which is consistent, if you claim the theory is consistent. You can't. <snip> > > MoeBlee > |