Cantor Confusion
in [Math]
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From: Virgil on 17 Feb 2007 22:59 In article <JDMyHH.HDp(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <1171702351.590890.177460(a)h3g2000cwc.googlegroups.com> > mueckenh(a)rz.fh-augsburg.de writes: > > On 16 Feb., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > This is not a representation of 3 other than in a perverted system, > > > > which calls 0 the first number, 1 the second and so on. Of course > > > > {{{{}}}}, or better and easier {{{{, denotes the fourth number which > > > > is 4 and not 3. > > > > > > Can you tell me a form of set theory where 0 is *not* the first ordinal > > > or cardinal number? If so, how many elements does the empty set have > > > in such a system? > > > > 0 may be the first (or better the zeroest) ordinal or cardinal number > > (if you wish to have the empty set in the theory). Nevertheless it is > > not the first natural number and not a natural number at all. > > But you said "perverted system which calls 0 the first number". The only > place where it is called the first *natural* number is in Bourbaki and its > followers, but that is only a simple renaming of the term "natural number". > But following your reasoning, {{{}}} is the third number, which is 3. > BTW, I can quote you as saying: > > > > > {{{}}} > > > > It was page 93 of my book. > > > I have seen it earlier than that. > > By the way, above is only number 2 given. > So earlier you said it is 2. What is it? > > > > So how many natural numbers precede the first natural numbers? We are > > > counting natural numbers, so it should be a natural number? > > > > Natural numbers are counting the elements of natural sets, i.e., of > > sets which exist in reality (in nature, as Cantor woud have said). > > If my house contains no dogs, in what way does the set of dogs in my house > not exist in reality? > > > > > |{1,2,3,...}| = ... i.e. potentially infinite, not fixed, capable of > > > > growing without bound, denoted by oo but not by a fixed number omega. > > > > > > And |{}| = ? > > > > A set which "also streng genommen als solche gar nicht vorhanden > > ist" (Cantor) > > A set which "ist also verm�ge der Definition von S. 4 gar keine > > Menge" (Fraenkel) > > An unnatural set cannot have a natural number of elements. > > If you think {} to be an unnatural set, so be it (that is not mathematics, > because there is no mathematical definition of natural set). But if > somebody asks how many coins I have in my purse, he is asking for the > cardinality of the number of coins in my purse. And I can correctly > answer 0 at some times. In that case the set is the set of coins in > my purse, and that can be empty (and is quite often in reality). > According to current definitions and axioms in set theory, {} *is* a > set. The distinction "natural" vs. "unnatural" set is not known. How about the unnaturalness of WM's Mueckematics?
From: mueckenh on 18 Feb 2007 04:01 On 16 Feb., 15:19, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 16, 6:58 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 16 Feb., 00:32, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > True and absolutely irrelevent. Which of the following do > > > you disagree with? > > > > All initial segments of E have a fixed maximum > > > yes. > > > > E does not have a fixed maximum > > > yes. > > > > If every initial segment of E has a fixed maximum > > > then E has a fixed maximum > > > no. > > Your claim is > > M: If all initial segments of E have [a] property and if > M: no element of E is outside of every initial segment, then E has > that > M: property. > > So if > i: All initial segments of E have a fixed maximum > and > ii: no element of E is outside of every initial segment > then > > iii: E has a fixed maximum > > You have claimed that i is true, and iii is false. > Given the ii is clearly true, we have i [true] plus ii [true] implies > [Claim above] iii. How do you explain this contradiction? > You are right. The claim in its generality is clearly wrong, as one can see already by the fact that every finite set is even or odd and bounded and the infinite set is neither even nor odd and is unbounded from above. The claim holds only for certain properties (like the existence of elements larger than the cardinal number in any set of even numbers which has a cardinal number). Regards, WM
From: mueckenh on 18 Feb 2007 04:07 On 16 Feb., 16:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171626163.912450.149...(a)m58g2000cwm.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 16 Feb., 02:38, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1171575875.054198.56...(a)q2g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 15 Feb., 14:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > I think we should first clarify the notation before continuing: > > > > P(n) is the set of all paths of the tree T(n) with n levels. > > P(0) is a set with one path, namely p(0) = {0.} > > P(1) is a set with two paths, namely p(1) = {0.0} and q(1) = {0.1} > > P(2) is a set with four paths, namely p(2) = {0.00}, and some others. > > Using that representation for paths is misleading, As you state that > each path is a set of nodes, and if that is the case, with that notation > 0.00 is a node, and each path is a set containing a single node. ??? Here is the Tree T(2) which can be represented by a simple chain (as the structure of the tree is unique) 0 1,2 6,5,4,3 The tree is a set of nodes which can be represented by the set of paths P(2) = {p(2), q(2), (2), s(2)}, namely T(2) = p(2) U q(2) U r(2) U s(2) = {0,1,6} U {0,1,5} U {0,2,4} U {0,2,3} = {0,1,2,3,4,5,6} It has the subtrees: The subtree T(0) = p(0) = {0} The subtree T(1) = p(1) U q(1) = {0,1} U {0,2} = {0,1,2} We see that T(2) is also the union of the paths belonging to the sets of paths P(0) and P(1) and P(2). T(2) = p(0) U p(1) U r(1) U p(2) U q(2) U r(2) U s(2) = {0} U {0,1} U {0,2} U {0,1,6} U {0,1,5} U {0,2,4} U {0,2,3} = {0,1,2,3,4,5,6} Can we use this convention? Are your objections met? > > > The union of all sets of finite paths is P(0) U P(1) U ... is the set > > of all paths. > > Why? It is the set of all *finite* paths, because there can be no > infinite path in that union because none of the constituent sets > contains an infinite path. The union of sets of alle finite paths is the union of all finite trees, and this union is the complete tree T(oo) if the union of all initial segmens of N is N. This is the fundamental problem of set theory. The union of all finite segments of N is and must be the same as the union of all natural numbers. But while the latter union is infinite by defintion and by axiom, the union of all finite segments cannot be the infinite segment N (because N is not contained in this union). "The infinite union of finite numbers" is illogic but it has been swallowed by most of us (yes, I did also swallow it until deeper thinking lead me the correct way). But the infinite union of finite segments is obviously incapable of yielding an infinite path, as you say above: "there can be no infinite path in that union because none of the constituent sets contains an infinite path." > > > In this union P(0) U P(1) U ... there is the path p(0) as an element > > and there is the path p(1) as an element and so on. Therefor the set > > of all the paths p(0), p(1), p(2), ... is a subset of P(0) U P(1) > > U ... > > The set of all the *finite* paths. Correct. Even more precisely: The infinite set of finite paths p(0), p(1), p(2), ... is a subset of the set of all finite paths p(i), q(j), r(k), .... The union of the infinite set of finite paths p(0), p(1), p(2), ... is the infinite path p(oo), if the union of all finite segments of N is N. > > > The tree forms unions from sets of paths. > > The tree T(2) is the union of the sets of paths P(0) U P(1) U P(2). > > But the tree T(2) does not contain the paths of P(0) and P(1), namely > > p(0) and p(1) and q(1). > > So it is *not* that union. If T(2) = P(0) U P(1) U P(2) it should > contain all paths of P(0), P(1) and P(2). T(2) does not contain the paths of P(1) but it is the union of the elements p(j) of the sets P(k) , j,k < 3.> > > > It is. p(oo) = p(0) U p(1) U p(2) U ... > > The finite paths are elements of the union of sets of paths P(n). The > > tree unites some of these elements which fit together to form an > > infinite path. > > In that case, yes. > > > >But anyway, > > > yes, P(oo) contains infinite paths, and so can *not* be a subset of > > > U P(i), as that one contains (as you note) only finite paths. > > > > It need not be a subset of U P(i) in order to be in the tree. > > That is what I am arguing all the time, but you did maintain that it > *is* a subset of that union, No. The set of the finite paths p(n) is a subset of the union. p(oo) exists in the tree because the set of paths p(n) exists in the tree. and need that in your proof that P(oo) > is countable. Mynheer Winter, all arguing converges to the decisive point: The union of all finite segments of natural numbers {1}, {1,2}, {1,2,3},... is the infinite set N {1,2,3,...}. Why is the union of all finite paths {0.}, {0.0}, {0.00}, ... not the infinite path {0.000...} = p(oo)? Regards, WM
From: mueckenh on 18 Feb 2007 04:09 On 16 Feb., 16:36, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171626693.139060.112...(a)v45g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 16 Feb., 02:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > But it is > > > > correct that there are more obvious and less obvious numbers: > > > > ... > > > > > > An ellipsis. > > > > > > > ||| > > > > > > A sequence of bars. > > > > > > > {{{ > > > > > > A sequence of braces. > > > > > > > abc > > > > > > The first letters in some of the alphabets that use the standard order. > > > > > > In none of them I do immediately see the number three. What I see is that > > > there are three of something, not the number three. > > > > What is common to all? > > I stated that already just above. Yes. You said "three of something". That is exactly the meaning of the number three. Regards, WM
From: mueckenh on 18 Feb 2007 04:14
On 16 Feb., 17:30, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > As I understand it : > > P(1) is the set of all numbers of 1 bit length - 2 elements (0.0 & 0.1) > P(2) is the set of all numbers of 2 bits length - 4 elements (0.00, > 0.01, 0.10, 0.11) > P(n) is the set of all numbers of n bits length - with 2^n elements. > > The set P(n) includes all P(m) for m<n, if a right fill of 0 for the > bits is inferred (i.e. 0.0 is the same as 0.00 is the same as 0.00..) The set P(n) contains only the n-bits numbers. In the tree with n levels as I defined it, there are no shorter paths. (We could change that definition without impairing the proof, but it would confuse the readers.) > For any bit-length (path-length) n, P(n) is countable, so for any finite > n, P(n) is countable. Yes. > > But, real numbers require an infinite number of bits to represent them > (or, if you prefer, we can define the set of reals as the set of numbers > with an infinite random binary expansion). But if the number of bits is > not finite, neither is P(oo), and the reals are uncountable. The question is simply as follows: The union of all finite segments of natural numbers {1}, {1,2}, {1,2,3},... is the infinite set N {1,2,3,...}. Why is the union of all finite paths {0.}, {0.0}, {0.00}, ... not the infinite path {0.000...} = p(oo)? If it were the union, then the union of all finite paths would contain all subsets required to "be", when united, all infinite paths like p(oo) above. Then the reals were uncountable and countable. Regards, WM |