Cantor Confusion
in [Math]
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From: WM on 12 May 2007 16:28 On 12 Mai, 20:46, Virgil <vir...(a)comcast.net> wrote: > In article <1178959303.279713.119...(a)h2g2000hsg.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 12 Mai, 00:20, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On May 11, 4:41 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > It need not change. Take just the path constructed in my last posting. > > > > This path does not exists. As usual, your construction made use > > > of an unjustified assumption that a set of paths could be chosen > > > in such a way that *all* paths were the same. > > > It can. See below. Or say why it cannot, according to your opinion. > > > > > My question concerned your idea always to use that path p* which takes > > > > the next exit > > > > In a counterexample I gave I used a path p* that takes the next exit. > > > However > > > in general I do not claim that this must be true. I do claim that you > > > always > > > have to choose a path with will eventually exit. > > > > >(irony warning). Why did you make this poor choice? > > > > Because you need a path that is not p, and any path that is not p will > > > eventually exit. > > > That is contradicted by the fact that there is always another path p* > > with p. > > And always a node at which p* leaves p, provided p* != p. > > > Hence not every path can have exited at a node k(p, n) with a > > finite index n. > > Each path other than p exits p at some node, but in an infinite tree in > which each p has infinitely many nodes, there is no node of p by which > all other paths have exited. May be the reason is that the CIBT tree does not exist? You should know: The union (as I defined it for trees) of all finite trees is an infinite tree with countably many nodes and countably many finite paths. (If you like, you can union the nodes and the paths separately.) If we switch to the CIBT, then there is not a single node added. Only the number of paths and their lengths changes from countable to uncountable and from finite to infinite, respectively. Can a mathematician, who advocates this, be in complete possession of his mental health? Regards, WM
From: WM on 12 May 2007 16:35 On 12 Mai, 21:11, Virgil <vir...(a)comcast.net> wrote: > In article <1178979741.537850.290...(a)w5g2000hsg.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 11 Mai, 21:45, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178866486.658963.253...(a)y5g2000hsa.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > My personal opinion is not the starting point here. Since mathematical > > > > entities exist only in the mind, such entities, which cannot exist in > > > > the mind, cannot exist at all. > > > > Since your own mind is the only one for which you can speak, what goes > > > on it other minds is beyond your competence to decide. > > > Do you think mathematical entities exist other than in minds? > > Do you thing that everything that can go on in any mind is limited to > what goes on in your mind? No, most minds are trapped in by far narrower limits. But why don't you answer my question which is independent of the limits of your mind - or does my question already fall outside of your point of view? Regards, WM
From: William Hughes on 12 May 2007 16:54 On May 12, 4:28 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: <snip> > You should know: > The union (as I defined it for trees) of all finite trees is an > infinite tree with countably many nodes and countably many finite > paths. (If you like, you can union the nodes and the paths > separately.) > > If we switch to the CIBT, then there is not a single node added. Only > the number of paths and their lengths changes from countable to > uncountable and from finite to infinite, respectively. > This is nothing more than the statement Let R be a countably infinite set. the set of finite subsets of R is countable the set of infinite subsets of R is uncountable. Note the number of elements in R does not change. When you change the size of the sets (the lengths of the paths) you change the number of sets (the number of paths) > Can a mathematician, who advocates this, be in complete possession > of his mental health? Yes. - William Hughes
From: Virgil on 12 May 2007 18:29 In article <1179001706.180042.42660(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 12 Mai, 20:46, Virgil <vir...(a)comcast.net> wrote: > > In article <1178959303.279713.119...(a)h2g2000hsg.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 12 Mai, 00:20, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 11, 4:41 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > It need not change. Take just the path constructed in my last posting. > > > > > > This path does not exists. As usual, your construction made use > > > > of an unjustified assumption that a set of paths could be chosen > > > > in such a way that *all* paths were the same. > > > > > It can. See below. Or say why it cannot, according to your opinion. > > > > > > > My question concerned your idea always to use that path p* which takes > > > > > the next exit > > > > > > In a counterexample I gave I used a path p* that takes the next exit. > > > > However > > > > in general I do not claim that this must be true. I do claim that you > > > > always > > > > have to choose a path with will eventually exit. > > > > > > >(irony warning). Why did you make this poor choice? > > > > > > Because you need a path that is not p, and any path that is not p will > > > > eventually exit. > > > > > That is contradicted by the fact that there is always another path p* > > > with p. > > > > And always a node at which p* leaves p, provided p* != p. > > > > > Hence not every path can have exited at a node k(p, n) with a > > > finite index n. > > > > Each path other than p exits p at some node, but in an infinite tree in > > which each p has infinitely many nodes, there is no node of p by which > > all other paths have exited. > > May be the reason is that the CIBT tree does not exist? Then the power set of N cannot exist either, as it provides a concrete model of a CIBT. But in ZF and NBG, for every set there is a power set, so that in ZF and NBG, CIBTs exist! > > You should know: > The union (as I defined it for trees) of all finite trees is an > infinite tree with countably many nodes and countably many finite > paths. (If you like, you can union the nodes and the paths > separately.) Then it is incomplete, as in a complete infinite binary tree there must be a separate path for every separate subset of N. > > If we switch to the CIBT, then there is not a single node added. Only > the number of paths and their lengths changes from countable to > uncountable and from finite to infinite, respectively. So that WM's tree must have omitted uncountably many paths. > > Can a mathematician, who advocates this, be in complete possession > of his mental health? Yup!
From: Virgil on 12 May 2007 18:32
In article <1179002100.262022.320270(a)e65g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 12 Mai, 21:11, Virgil <vir...(a)comcast.net> wrote: > > In article <1178979741.537850.290...(a)w5g2000hsg.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 11 Mai, 21:45, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1178866486.658963.253...(a)y5g2000hsa.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > My personal opinion is not the starting point here. Since mathematical > > > > > entities exist only in the mind, such entities, which cannot exist in > > > > > the mind, cannot exist at all. > > > > > > Since your own mind is the only one for which you can speak, what goes > > > > on it other minds is beyond your competence to decide. > > > > > Do you think mathematical entities exist other than in minds? > > > > Do you thing that everything that can go on in any mind is limited to > > what goes on in your mind? > > No, most minds are trapped in by far narrower limits. Mathematician's minds, however, are less limited, in that they can and do imagine things that WM says he cannot. > But why don't > you answer my question which is independent of the limits of your mind > - or does my question already fall outside of your point of view? There are things that can 'go on' in mathematicians minds that do not go on in WMs. |