Cantor Confusion
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From: Andy Smith on 10 Jan 2007 23:32 OK, thanks for that, content appreciated. I know I said I was calling it quits, but, once more into the fray... What about the ordered set of the zero crossings of the function: sin(exp(1/(x**2)) between say x = -1.0 and 1.0? a) is that set infinite (yes, for sure?) b) does it have a well defined first and last member? So you can have an infinite set with a first and last member?
From: mueckenh on 11 Jan 2007 11:19 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > ,----[ <45a3ae6b$0$97254$892e7fe2(a)authen.yellow.readfreenews.net> ] > | No. I do refer to "complete" in the sense of "finished" in contrast to > | "potential" inifite. Since neither the list and hence nor the diagonal > | is finite there are no largest elements. OTOH since we do not assume > | every member of the list or the diagonal to provably exist, the list > | and the diagonal are "potential" infinite. > | > | This said William Hughes has shown that the assumption of the > | exist[e]nce of a "potentially infinite" "last line" L_D leads to a > | contradiction. Hence "L_D exists" is wrong. > `---- > > So you agree to this part you have cut? Fine! There exist only some members "since we do not assume every member of the list or the diagonal to provably exist". Or what does it mean that we do not assume every member to provably exist (actually)? Only every second member exists (actually)? > > >> Since there is no largest element in "potentially" infinite sets (in > >> "actual/complete/finished", too) this sentence makes no sence at all. > > > > A potentially infinite quantity (set or not) is always finite. > > There is no time in maths. But maths is in time. God, if existing, may be out of time. But even worshipping sets does not make them divine. > >> > If it turns out, that the diaogonal has a larger element, then it > >> > turns out that a line containing this (and all smaller ones) does > >> > also exist. > >> > >> > In any case, a line containig all elements of the diagonal does > >> > exist. > >> > >> Proof? It is the set of all existing numbers. > > > > The existence of the diagonal (if existing) and the fact that a given > > set of natural numbers is always a subset of a natural number. > > Why not say "the pope said so"? Because I can show of every set of natural numbers *given to me* that it belongs to one line (= finite set). > Regards, WM
From: mueckenh on 11 Jan 2007 11:23 William Hughes schrieb: > > > > No. But the property of being the greatest line can and does change. > > > > > > Irrelevant. The question is whether L_D exists. > > > > It does. Does the tallest man exist? When did it start, when did it > > cease? > > Recall: > > L_D is a line that contains every element that can > be shown to be in the diagonal. > > > The "tallest man" is something that can change. L_D is a line. > A line cannot change. The analogy is not valid. L_D is the name of a line, like a championship title. > > > It is possible to find L_D > > > > It is not assumed, but it is obvious that for every given set of > > natural numbers there is one line containing it. > > But since the set of natural numbers can change > this "one line" can change. It is not L_D. > > > > > > > If you assume actual infinity then L_D > > > does not exist. > > > > It does exist, in potential infinity. But it is not fixed. > > L_D is a line. A line is fixed. L_D does not > exist. It is the line containing the whole set. > > > > > > > Therefore you cannot assume actual infinity. > > > > > > You now admit that it is not possible to find L_D, > > > > In actual infinity (everything including L_D being fixed) it is not > > possile to find L_D. > > And it is also not possible to find L_D in potential infinity. Let L_D go from 1 to oo. > Regards, WM
From: mueckenh on 11 Jan 2007 11:29 Virgil schrieb: > > > > Definition: Denote the nodes of the tree by > > > > > > > > (0,0) > > > > (1,0) (1,1) > > > > (2,0a) (2,1a) (2,0b) (2,1b) > > > > ... > > > > (n,0a) (n,1a) ... > > > > > > > > The union of all trees up to the n-levels tree is > > > > > > > > {(0,0)} U {(1,0), (1,1)} U .. U {(n,0a) (n,1a) ...} > > > > which is obviously the same as > > > > {(0,0)} U {(0,0), (1,0), (1,1)} U .. U {(0,0), (1,0), (1,1),..., > > (n,0a) (n,1a) ...} > > > > End of definition. > > This presumes falsely, that every tree has the same nodes and edges and > thus that all trees are subtrees of some ur-tree. > But suppose I have a family of trees in which no two trees share any > nodes or edges, what is the union of such a family? Suppose I have a set of initial segments {o}, {a,b}, {1,2,3}, in wich no two segments share any element. What is the union of such a set? Both, question and answer are completely irrelevant for the present discussion. > > Books on set theory would not require the union of a set of disjoint > trees to be a single tree. But would allow for the union of such trees with identical nodes. > Then there isn't any one L_D, there are endlessly many of them, in fact > an infinite sequence of them, for no sooner is one created than it is > eclipsed by its successor. If you create with staying power. (Numbers are created, Dedekind said.) > But in ZFC and NBG, the completed diagonal exists alas, ZFC and NBG do not exist. Regards, WM
From: mueckenh on 11 Jan 2007 11:39
Dik T. Winter schrieb: > That still does *not* show that the union of the sets of paths in the > finite trees contains infinite paths. You still fail to provide a > proper definition of the union of trees. So I can only guess. Let us > assume two finite trees, where I number the nodes: > 1 1 > / \ / \ > / \ / \ > / \ / \ > 2 3 2 3 > / \ / \ / \ / \ > 4 5 6 7 > / \ / \ / \ / \ > We can easily identify like numbered nodes from both trees with each other. > And so, with respect to sets of nodes, the union of the two is the tree on > the right hand side. With a bit of imagination we can identify the edge > going to the left from 2 in the left-hand tree with the edge going from > 2 to 4 on the right-hand tree, and in that way see that also with respect > to edges, the union of the sets of edges is the set of edges of the tree > on the right hand side. If we accept finite paths (and assume that they > all start at node 1), there are 6 paths in the tree on the left There are 4 paths, because the finite trees have leaves only at the end. > and 14 > paths in the tree on the right. There are 8 paths. > And in this way we accept that the union > of the sets of paths is the set of paths on the right. > > Do you agree with it so far? Yes, except that you mistake edges for paths (or invent leaves). But that is not important. > > If you agree, there is however one problem with the complete infinite tree. > *None* of the constituent threes contains an infinite path, and so the union > of the sets of paths also does not contain an infinite path. None of the finite initial segments of N contains N. Is there a problem with N? > > You are wrong. The union of sets of paths is a set of paths were each of > the paths is element of at least one of the sets that define the union. > If there is a constituent of the union that contains a particular path, > that path is element of the union set. If there is none, that path is > not an element of the union. That's not required. The set of infinite paths is the union. > > The relevance is that the uninon {1,2,3,...} is *not* an infinite set N > > with cardinal number aleph_0, as long as there is not an "infinite > > number" *in* the set. > > But that is just what you can find when you allow the axiom of infinity. Why do you refuse to allow it for the segments of paths? > Suppose the set X consisting of all the finite natural numbers plus a > single infinite number. That set does not satisfy the Peano axioms, > as a set with a largest element *can not* satisfy the Peano axioms. Therefore it cannot actually exist. > Simply because there is no successor of the largest element. Now remove > that largest element. If there is still a largest element, the Peano > axioms can not be satisfied. However, the axiom of infinity states that > there is at least one set that satisfies the Peano axioms, and that N is > the smallest of those sets. You are claiming that there is no such set. You claim it for the set of paths. > > > But you do not see this small difference. You > > claim that the union of all finite numbers or of all finite initial > > segments {1,2,3,...,n} is the infinite set {1,2,3,...}. > > Elaborate on the difference. Why do you see it different in case of the union of all finite trees or paths? > > Therefore you > > must add, in addition to all finite segments of that very infinite > > path, something which you cannot describe. Something, however, which is > > certainly not a finite segment of that path. > > I think I was talking about sets of paths. representing finite and infinite sequences of numerals of real numbers. Regards, WM |