Cantor Confusion
in [Math]
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From: David Marcus on 18 Feb 2007 12:07 mueckenh(a)rz.fh-augsburg.de wrote: > 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). For a second there I thought WM had seen the light. But, although his logic contains an error, all of his conclusions still (miraculously) hold. -- David Marcus
From: David Marcus on 18 Feb 2007 12:11 mueckenh(a)rz.fh-augsburg.de wrote: > On 18 Feb., 02:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > If my house contains no dogs, in what way does the set of dogs in my house > > not exist in reality? > > There is no dog and no set of dogs, because it would be the same set > of cats. Your dog would be your cat. That is impossible. They would > not live together, let alone form the same (identical) set. So, I can't have no dogs and no cats because my dogs and cats won't get along? -- David Marcus
From: Carsten Schultz on 18 Feb 2007 12:33 mueckenh(a)rz.fh-augsburg.de schrieb: > The limit {1,2,3,...} = N does exist according to set theory. Why > should the limit {1}, {1,2}, {1,2,3}, ... = N not exist? > > There is no arguing in set theory that the infinite binary tree would > not contain all the real numbers of the unit interval. You seem to me to be very, err, confused. Best, Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: David Marcus on 18 Feb 2007 14:04 mueckenh(a)rz.fh-augsburg.de wrote: > That is the problem of set theory. If you let n --> oo then n > nevertheless remains finite. Set theory needs an nfinite set which s > larger than every set you get from letting n --> oo. The set of all > fractions of the form 1/n, for instance, can be divided in subsets, as > we do when proving the divergence of the harmonic series. The n-th set > has 2^n elements. If aleph_0 was reached by the number of subsets, > then 2^aleph_0 was reached by the fractions 1/n and, therefore, there > were 2^aleph_0 natural numbers. This has to be one of WM's better results: There are 2^aleph_0 natural numbers. And, the proof is so simple: We let A_n be a set of natural numbers of cardinality 2^n. Then, we let n go to aleph_0. But, why stop there? Why not let A_n be a set of cardinality 2^{2^n}? The possibilities are limitless! -- David Marcus
From: Dik T. Winter on 18 Feb 2007 18:57
In article <1171789678.561730.262020(a)t69g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 16 Feb., 16:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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. > > ??? What do you not understand about my remarks? *if* p(1) = {0.0}, p(1) has only one element, which is (if I understand you correctly) not a node. Or is "0.0" a node? > 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? Yes, as i wrote, your initial notation was misleading. > > > 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. Wrong. The union of sets of paths is *not* the tree. According to your notation above, the tree is the union of paths. There is a huge difference between the union of sets of paths (which is a set of paths) and the union of paths (which is a set of nodes). And the union of sets of things that do not have property A can not be a set that contains an element that *has* property A. This is fundamental. > 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). It need not be and is not in that union. That is your funamental misunderstanding. It *is* that union. Moreover, the union of all finite numbers (using von Neumann representation) *is* N, it does not contain N. > "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). What is the illogic? > 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." There is not infinite path as an element *in* that union, that union *is* the infinite path. Pray look at the difference. > > > 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. But that one is correct. But in P(0) U P(1) U P(2) U ... you are *not* uniting path, you are uniting sets of paths. > > > 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 Yes, so what? In P(0) U P(1) U P(2) paths are not united. P(0) U P(1) U P(2) is a set of paths. T(2) is a set of nodes, so the two can not be equal (unless both are empty). To take this case (with your node numbering): P(0) = {{0}} P(1) = {{0, 1}, {0, 2}} P(2) = {{0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}} and so P(0) U P(1) U P(2) = {{0}, {0, 1}, {0, 2}, {0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}} and T(2) = {0, 1, 2, 3, 4, 5, 6} so T(2) != P(0) U P(1) U P(2). > > > >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. You did argue it. Reread 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)? It is. Why do you think I am arguing it is not? I am not arguing that way, I am only stating that p(oo) is *not* an element of P(0) U P(1) U P(2) U ... because in that union you are not uniting paths but sets of paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |