Cantor Confusion
in [Math]
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From: Virgil on 8 May 2007 14:11 In article <1178636442.004124.84910(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > Could you explain, please, how you define a set of at least two > different paths, p' and p'' (or even more) which are required to > accompany p for all n? It is WMs delusion that anyone other than himself has claimed anything like that. What has been claimed, and what is quite true, is that for any n in N, there are uncountably many other paths that accompany any given path up through level n. But unless WM is claiming that there is a last member to N, which is false in ZF and NBG anyway, that does not mean that any two paths agree for all n in N.
From: Virgil on 8 May 2007 13:56 In article <1178629393.151679.20430(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 7 Mai, 23:13, Virgil <vir...(a)comcast.net> wrote: > > In article <1178566976.513253.63...(a)p77g2000hsh.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 7 Mai, 18:50, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1178536779.509069.124...(a)l77g2000hsb.googlegroups.com>, > > > > > > There are all sorts of things in naive set theory texts that formal set > > > > theories do not allow. > > > > > I am not interested in formal set theory but only in the question: Are > > > there uncountably many real numbers? > > > > There are uncountably many binary strings, and only countably many of > > them map to duplicates of other strings when interpreted as reals in > > [0,1], so, in ZF or NBG, yes! > > > > > > > > > > > Fn = (2-1-1) > > > > > It is the number of separated path coming out of the n-th element of > > > > > the tree minus the number of ingoing separated paths minus the number > > > > > of nodes of this element. > > > > > > In that case, it is irrelevant. > > > > > Why? > > > > Because you never are counting anything but "bunches" each of which > > represents a set of uncountably many paths, so the number of bunches is > > not relevant to the number of paths. > > Why are you never countaing paths? Perhaps because there are no paths? > I there were paths, then they could be counted. Since in a complete infinite binary tree one different path for each different subset of N, one can no more "count" all the paths that all the subsets of N. I.e., there cannot be any surjection form N to P(N), at least in ZF or NBG. What can or cannot happen in WM's system only WM can say, as he keeps its properties secret from us..
From: WM on 8 May 2007 17:26 On 7 Mai, 21:32, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 7, 3:19 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > That is not the point here. The point is that there are subsets of > > countable sets which do not allow for a bijection with N. > > No. Why "no"??? > There are subsets of a finitely definable set which are not > finitely definitable. And they do allow for a bijection with N? > The difference is... they do not allow for a bijection with N. (That's the same with the paths in the tree.) Regards, WM
From: WM on 8 May 2007 17:29 On 8 Mai, 17:11, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 8, 11:00 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 8 Mai, 14:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 8, 8:20 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 7 Mai, 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > Every node, n, of path p is shared with another path > > > > > > > p'(n). Note that p'(n) can be different for every n. > > > > > > > Note that p'(n) need not diffrent for every n. > > > > > > However, it cannot be the same for all n. > > > > > For which n must it change? > > > > > Could you explain, please, how you define a set of at least two > > > > different paths, p' and p'' (or even more) which are required to > > > > accompany p for all n? > > > > Here is a very simple, countable example. > > > > Let p be the path 000... > > > > Let p'(1) be the path 0111... > > > Let p'(2) be the path 00111... > > > Let p'(3) be the path 000111... > > > ... > > > > let P' be the set of all the p'(n) > > > > In general p'(n) has a n 0 nodes, followed by only 1 nodes. > > > (Note that there is no last p'(n)) > > > > A: for any node k, of p, there are an infinite number of paths > > > in P' which contain node k > > > But why do you write down these paths: > > > Let p'(1) be the path 0111... > > Let p'(2) be the path 00111... > > Let p'(3) be the path 000111... > > > ? > > Any set p(1) up to p(n) can be substitued by a single path. In fact? Why then did you write down so many useless paths? > > Look! Over there! A pink elephant! > > The set P' can be substitued by a single path. Oh, does the set P' consist of any paths which are not paths p(n) with a finite argument n? Regrads, WM
From: WM on 8 May 2007 17:32
On 8 Mai, 17:28, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 8, 11:09 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 8 Mai, 15:09, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 8, 8:27 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 7 Mai, 21:49, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > Since there is no last path which branches off, this list never ends. > > > > > So every node of p is shared by other paths. > > > > > However, every path branches of somewhere, so there > > > > > is no path p' which shares every node of p. > > > > > You state: > > > > (1) Every node of p is shared by other paths. > > > > (2) Every path branches off somewhere. > > > > > (1) ==> Not every path has banched off somewhere (because there must > > > > remain at least one "other path" and "is shared" is the opposite of > > > > has branched off). > > > > (2) ==> Every path has branched off somewhere. > > > > > So your claim can be summarized: > > > > Not every path has branched off somewhere. > > > > Every path has branched off somewhere. > > > > The statement > > > > "Every path branches off somewhere" > > > is true. > > > > However. this implies "Every path has branched > > > off somewhere" only after the last path has branched off. > > > If "*there is* a set such that every path branches off somewhere", > > then every path *has branched off* somewhere. > > No. 'every path branches off somewhere' needs only existence, no > order (temporal or otherwise) is needed To branch off is a temporal act. > > 'every path *has branched off* somewhere' needs some sort of > order (temporal or otherwise) Yes, factual existence and completion of the generaton process. > > There are ordered sets that do not have a last element. Why do you stress this? Pinky Billy? Regards, WM |