An uncountable countable set
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From: Dik T. Winter on 20 Aug 2006 21:46 In article <ec9qh6$762$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: > > In article <ec8h1q$prm$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > ... > > > Yes, that's a shame, isn't it? It would seem to me that any good theory > > > of infinite sets could be applied to infinite sets of points, such as > > > the reals in (0,1] or those in (0,2], and be able to draw conclusions > > > such as that there is twice as many points and twice as much space in > > > the second as in the first, rather than coming to to useless conclusion > > > that the infinite sets are equal in size. > > > > You are looking for measure theory. > > I looked at measure theory a little. Lebesgue measure doesn't come close > to what I'm concocting. I am trying to integrate set theory with > combinatorics, set density, Lebesgue measure, topology, etc, and all it > takes is a housecleaning and a few new simple rules. O. > > > Transfinite cardinalities > > > don't really measure anything, though, do they? > > > > They measure something, but not what you want them to measure. > > As far as I can see, they "measure" with a fluid yardstick, based on > non-logical arbitrary axioms. If cardinalities measured set size at all > accurately, then they would always reflect the addition or removal of > elements with a change in value. It's actually really not hard to > achieve that. Arbitrary axioms? They are soundly based on equivalence relations and bijections. What are the non-logical axioms used in that? > > > Indeed, they don't form > > > a field. Tsk tsk. > > > > There is not something inherently wrong with a collection of objects > > not forming a field or a division ring, or whatever. However, if that > > is the case you must be prepared to find that division is not generally > > possible. Take for instance the Sedenions, an extension of the Cayley > > numbers (or Octonians). In the Sedenians general division is not > > possible because there are zero divisors. > > That is an area I don't know too much about except that it's an > extension of the complex numbers. That shows. If you do not know what I am talking about, do not demean it with your "Tsk tsk.". > Complex numbers are really not single > quantities, at least in terms of linear order, and so I wouldn't expect > them to form a ring, but perhaps more of a 2-D ring, or toroidal > surface, in that sense. They not only form a ring; they form a field. The Quaternions form a division ring. The Cayley numbers do not form a ring (associativity is not guaranteed), but division can be defined. In the Sedenions division can not be defined. > However, when we are talking about raw counts of elements in a set, that > lies somewhere on the infinite real number line, and it is quite > possible to formulaically compare and order infinite values along this > line. Don't you think? I was not talking about comparisons, I was talking about division. > The projectively extended reals, in the end, form a circle and a ring. Not a ring in the mathematical sense. > 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. -- 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 20 Aug 2006 21:58 In article <ec9r9u$7tb$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: > > In article <ec8hlf$qq8$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > If the formula applies to an infinite number of finites, then does the > > > sum of the aleph_0 finite naturals equal (aleph_0^2+aleph_0)/2? > > > > I think there is a misunderstanding. The formula > > sum{i = 1 .. n} i = n * (n + 1) / 2 > > holds for every finite number n, so it holds for infinitely many finite > > numers n (as there are infinitely many finite numbers n). But we can > > not switch to an infinite sum (that is something different). > > Well, we can. If we turn to what we know about infinite series, we can > apply notions such as, if every term in series A is greater than its > corresponding term in series B, then the sum is obviously greater. Yes, you can claim that, but you might get in trouble. We may similarly state that if every term in a sequence A is greater than its corresponding term in a sequence B, then the limit is obviously greater. Now apply that to: lim{n -> oo} 1/n > lim{n -> oo} 1/(2n). You have to formalise what you mean with an infinite sum of a diverging series before you can state things like that. > claim there are infinitely many n e N, aleph_0 of them. So, if that > formula represents the sum of the first x terms, and you plug in aleph_0 > for x to include all of them, then you get that result. But that is no proof. > > > In > > > standard theory, would this not equal aleph_0, and if so, does it make > > > sense that sum(n=1->aleph_0: 1) = sum(n=1->aleph_0: n), when n>1 for > > > all n>1? The scond would appear to be clearly a larger sum. > > > > 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. > > All that needs doing is declaring a unit oo and allowing it to be used > formulaically. sum{n = 1 .. x} 1=x, so sum{n = 1 .. oo} 1=oo. sum{n = 1 > .. x} n=(x^2-x)/2, so sum{n = 1 .. oo} n=(oo^2-oo)/2. It was perhaps a > year ago that three of us independently said the sum is (|N|^2-|N|)/2. > It's not a problem dealing with infinite values, once you declare an > infinite unit. "Diverges" doesn't specifically describe the value of the > sum. :) Are you sure there will no contradictions come up? How do I operate with oo? > > He may be trying to point that out, but not in a way that I do understand. > > Moreover, I have stated over and over again that aleph_0 does not behave > > like a normal number. It is just people like WM and you that wish that > > if behaves like a normal number, but it does not do so. I see no problem > > with that. You can not find a midpoint in the ordered set of natural > > numbers using a measure derived from the standard measure of the reals. > > True. WM and I and others want numbers to behave like numbers, and > cardinalities and ordinals simply do not. Yes. The only problem you and WM have is with the terminology. But that is just labels. > We find them useless and a > digression from the study of quantity and representation which we see as > being the foundations of math. Yes, you find them useless. Others do not find them useless. Just opinion. > It irks people like us when set theory > claims to be the foundation of math, and yet makes all sorts of > exceptions and new rules for infinite values. That is not the case. The rules come from the way these things are *defined*, not from anything else. Consider Conways surreal numbers, they act much more in the way of numbers (they even form a field). > It's some kind of logical > construction, but for folks like us, it's really not related to > mathematics. It is related to mathematics. Perhaps it is not related to physics, but it is related to computer science. Read about the computable numbers as defined by Turing. -- 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 20 Aug 2006 22:18 In article <ec9rfa$7tb$2(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: > > In article <ec8hsd$r5t$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > Dik T. Winter wrote: > > > > In article <ec7gek$frg$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > > > Inductively provable: Any set of consecutive naturals starting > > > > > at 1 always has its largest element equal to size of the set. > > > > > > > > What is the size of the empty set? What is the largest element of the > > > > empty set? > > > > > > The empty set is not a set of consecutive naturals starting at 1, > > > because it does not include 1 as an element, and therefore it does not > > > stand as a counterexample. However, even in your von Neumann ordinals, > > > the empty set is 0, and contains zero elements, and could be considered > > > the 0th in this sequence of sets. > > > > And the next in the sequence is {0} = 1, followed by {0, 1} = 2... > > That's the von Neumann ordinals. I am talking about {}, {1}, {1,2}, ... > If by convention one says the empty set has max=0, then the empty set > can be included this way. Otherwise, just leave it out and start with {1}. But you make conclusions about an ordinal number. I question your "inductively provable". When you consider the von Neumann ordinals I can state that the value of an ordinal is larger than each of its elements. I can use quite the same induction here as you are using and come to the conclusion that the ordinal of the set of all naturals is larger than each natural. You assume that what holds for natural numbers also should hold for either ordinal or cardinal numbers. Finite cardinal numbers can be equated in some sense to the non-negative integers, but they are not identical. The same holds for finite ordinals. Consider the following. The ordinal number of each initial segment of ordinal numbers is larger than each of the ordinal numbers in that segment. That works for finite and for infinite ordinals. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 Aug 2006 22:24 In article <eca31m$itn$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > David R Tribble wrote: .... > > Not without proof that the formula works for infinite n, anyway. > > But we're still waiting for Tony to provide that proof. > > David, this is not a matter of proof given standard definitions, but a > redefinition of the principle of induction such that it applies to the > infinite case. When an inductive proof demonstrates that f(x)>g(x) for > all x greater than some y, all infinite values falls into that category. > So, it simply amounts to removing the restriction that induction only > applies to finite n. Once this restriction is lifted, then we can > "prove" that, for instance, for all n>2, n^2>2n, and therefore this also > holds for all infinite n. And as 1/n > 1/2n this holds also for infinite n, so 0 > 0? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: imaginatorium on 20 Aug 2006 22:30
Tony Orlow wrote: > imaginatorium(a)despammed.com wrote: > > Tony Orlow wrote: <snip> > >> PS - you still owe me an actual answer to my question, or at least an > >> explanation as to why you don't accept my answer to yours. I thought we > >> had a deal, that didn't include pink elephants. > > > > I'm sorry, too many posts - you'll have to remind me exactly which > > question I'm supposed not to have answered. Is it isomorphic to the one > > about drive shafts in a dog? > > You've obviously been drive-shafting your dog's chocolate a little too > much lately. It's not healthy for the concentration. > > It would help for a start if you check for > > the presence of i-words and f-words, because if you can't rewrite it > > without using them, I can guarantee I will not be able to attach any > > clear meaning to your question. > > And yet, I was able to attach meaning to your question about the angle > of a circle's vertex. You tried to make me look foolish, and I answered > you in a way that you can't even respond to, much less offer the > response you promise when you proposed our deal. Your sense of humor is > a tad tiresome, Brian. A circle has no vertices in mathematics (nor does it in normal parlance, actually). I was not in fact trying to make you look foolish, I was vainly hoping you might one day see that arguments depending on supposed properties of nonexistent objects are invalid. Never mind. <snip...> > >>>> That is an area I don't know too much about except that it's an > >>>> extension of the complex numbers. Complex numbers are really not single > >>>> quantities, at least in terms of linear order, and so I wouldn't expect > >>>> them to form a ring, but perhaps more of a 2-D ring, or toroidal > >>>> surface, in that sense. > > > > Well, contrary to your expectations, the complex numbers do form a > > ring. I can't make any sense of your comment - what is a "2-D ring" for > > a start? Actually, I have never understood why the name "ring" is given > > to what mathematicians mean by the word. Can you explain? > I think you know. I told you I do not know. Why do you try to argue with my statement of what I know or think? (A hallmark of religionists, btw.) Perhaps someone else can answer: why is a (mathematical) ring called a ring? If the term 'ring' were used for a cyclic group I could understand, but I don't see any obvious way in which a ring is characterised by being "circular". Am I missing something? Wouldn't it have reduced Tony's confusion (at least infinitesimally!) if a different name had been used? Brian Chandler http://imaginatorium.org |