An uncountable countable set
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From: imaginatorium on 22 Aug 2006 04:45 Tony Orlow wrote: > Dik T. Winter wrote: > > In article <ecb84s$b1q$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > Dik T. Winter wrote: > > ... > > > > > With the addition of infinitesimals and specific infinite quantities, > > > > > even division by 0 can be handled, probably even for complex numbers, > > > > > > > > Division by 0 can not be handled in a ring. > > > > > > According to MathWorld, indeed, there is an exception for 0 in the > > > optional Multiplicative Inverse condition. Only required due to the > > > aversion to infinitesimals. Alas! Why cannot we tie this all together? > > > > No. Required because if 0 has also to have an inverse in a ring, trivially > > there is only one ring. The ring consisting of the single element 0 with > > the standard operations. Because, suppose we have a ring. It can be > > proven that in a ring 0*a = 0 for every a in that ring. But if 0 has > > an inverse, say b, we also have 0*b = 1, and so 0 = 1. > > > > On the other hand, do not confuse 0 with the infinitesimals. And there > > is no aversion to the infinitesimals. Look at the surreals, look at > > non-standard analysis. Infinitesimals abound. Still, in order to be > > logically consistent, 0 has a special role. > > Yes, true. Absolute 0, the origin, is a single point, and > infinitesimals, neighboring points, are something ever so slightly > different. So should we understand that in Tonyspeak "absolute 0" just means what mathematicians call zero. What is the point of using different words? > But then again, oo as a concept and a limit is something > absolute and different from, say, the number of points per unit of > space, or some other infinite unit. To have infinitesimal numbers, you > need to have non-absolute infinities also. The ring is broken by the > reluctance to allow absolute oo to balance 0 in this respect. What do you mean by a "ring" being "broken"? What would it mean for "absolute oo" (whatever that is, exactly) to "balance" 0? Who is reluctant? Suppose I agree not to be reluctant: what happens then? Brian Chandler
From: Dik T. Winter on 22 Aug 2006 10:29 In article <ecdu6d$r81$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: > > In article <ecb927$ce6$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > Dik T. Winter wrote: > > > > In article <ec8if0$rtd$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > ... > > > > > Indeed, I agree with WM's logic concerning the identity relationship > > > > > between element count and value in the naturals. He's quite correct in > > > > > that regard. > > > > > > > > Well, you and he are not. The logic is flawed. > > > > > > How so, precisely? > > > > Because you by the same logic the ordinal number of the set of naturals > > is *larger* than the largest natural, or *smaller*. > > No, I say the ordinal is larger than the greatest finite, since that's Sorry, the discussion was about something else. The logic concering the identity relationship between element count and value in the naturals. I presume you mean that the number of naturals in an initial segment of naturals is the largest of those naturals. And using that that so the size of the set of all naturals should be in the set of all naturals. But using that logic you can also prove that that number should *not* be in the set of all naturals. > > You can not prove it inductively based on properties for finite x. > > The induction axiom only is about finite x. > > That is what I am suggesting changing. When starting with the assumption > that oo>n for n e N, any proof of f(x)=g(x) for x>n should also apply to oo. You can suggest changing it, changing it does not make it true. If you change it you get inconsistencies. > > > > In that case you are not using standard mathematical terminology. I > > > > have no idea what a finite but unbounded set is. > > > > > > Yes, that's contradiction in terms given the Dedekind definition of an > > > infinite set. WM's point is very subtle. > > > > A *subtle* contradiction in terms? > > Yes. Given the Dedekind set-theoretic definition, the set is infinite > because it is bounded only by finiteness, which offers no clear > boundary. For sets truly infinite in membership, it applies, but it > misses the subtlety of the Twilight Zone, the lack of boundary between > finite and infinite. This is Zen mathematics I think. > > > I know that because I have > > > raised it a bunch of times over the last year+, as I think has he. But, > > > where there is a constant finite distance on a line between points, and > > > no point is infinitely distant from any other, there is a finite range, > > > and only a finite number of such disjoint intervals can occupy that > > > space. > > > > Yes, you are arguing that, and WM is arguing that, but offering no proof. > > Not according to set-theoretic principles, no, but according to > numerical and quantitative principles, yes. Still not a proof. > > > Dik - have you given much thought to infinite values? > > > > I have studied the surreals quite thoroughly. > > And does that lead you to any thoughts as to how one might improve the > precision of comparisons between the sizes of infinite sets? In my opinion bijections are doing very well. Thank you. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 22 Aug 2006 10:22 In article <ecdtie$q90$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: .... > > No. Required because if 0 has also to have an inverse in a ring, trivially > > there is only one ring. The ring consisting of the single element 0 with > > the standard operations. Because, suppose we have a ring. It can be > > proven that in a ring 0*a = 0 for every a in that ring. But if 0 has > > an inverse, say b, we also have 0*b = 1, and so 0 = 1. .... > The ring is broken by the > reluctance to allow absolute oo to balance 0 in this respect. Not reluctance. If we add oo to a ring we have to define the operations on it (addition and multiplication). Moreover, oo should have an additive inverse (otherwise it is not a ring). Now suppose we define oo to be its own additive inverse: oo + oo = 0. Further suppose we define n + oo as oo for all n. We now have: 3 = 1 + 0 + 2 = (use oo + oo = 0) 1 + oo + oo + 2 = (use n + oo = oo + n = oo) oo + oo = 0. So we can not define n + oo as oo. So we *must* have n + oo is some other number, say m. But now we have oo = m - n... Conclusion, it is not possible to add a single oo without breaking the ring structure. No reluctance present. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Tony Orlow on 22 Aug 2006 11:57 David R Tribble wrote: > Dik T. Winter wrote: >>> No that does not make sense. sum{n = 1 .. oo} 1 is not defined, the same >>> holds for sum{n = 1 .. oo} n. If you want to use them you have to >>> provide a definition for them. > > David R Tribble wrote: >>> But Tony thinks he has provided a definition, based on his "Big'un" >>> number. The problem, of course, is that he simply assumes that >>> arithmetic operations on Big'un work "es expected" without providing >>> any proof of that whatsoever. >>> >>> It's one thing to provide a definition, it's quite another to prove >>> that it is a consistent definition. > > Tony Orlow wrote: >> I think the responsibility lies with you to point out an inconsistency >> that arises from my assumptions. > > You're kidding, right? > >> Set theory is not "proven true". It >> cannot prove itself consistent. It took years of trying, with some >> succcess in detecting failures, to refine set theory so that it was >> somehow actually consistent. But, it cannot "prove" itself so. > > It's true that in any sufficiently powerful axiomatic system a theorem > cannot be formed that proves the consistency of the that system. > > However, within the framework of the theory (the axioms and theorems > that it comprises), you can prove theorems true and prove false > statements to be false. In that sense, the theorem that, for example, > the set of all naturals is infinite is a provably true theorem within > standard set theory. > No argument there. > >> Please >> show where I am being inconsistent, not with set theory, but within my >> own assumptions. Remember, I don't claim to believe in transfinite set >> theory, and don't intend to be consistent with it. > > The problem is that your own assumptions are not consistent with > each other - they don't form a coherent theory. Many of these > inconsistencies have been pointed out to you before, but you choose > to not believe them or simply to ignore them. > No, the objections to my ideas are based on their inconsistency with set theory. Otherwise, which two of my ideas produce a contradiction?
From: MoeBlee on 22 Aug 2006 14:00
Albrecht wrote: > 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. You miss the point. Since you've not shown any contradiction in set theory, whatever contradiction you claim to have found must be a contradiction between set theory and something else outside of set theory. But if you can't articulate that something else as a mathematical formula, then no one much cares that set theory conflicts with your not mathematically articulated principles. So I take it from your response that you don't have a set of axioms for your mathematics. So I wonder how you expect people to evaluate whether something is or is not a theorem of your mathematics. MoeBlee |