```Date: Jan 12, 2013 11:33 AM
Author: ross.finlayson@gmail.com
Subject: Re: FAILURE OF THE DISTINGUISHABILITY ARGUMENT. THE TRIUMPH OF<br> CANTOR: THE REALS ARE UNCOUNTABLE!

On Jan 11, 5:57 pm, Virgil <vir...@ligriv.com> wrote:> In article> <6c4f8c18-2d3d-43ad-b680-8e972d357...@eo2g2000vbb.googlegroups.com>,>>  WM <mueck...@rz.fh-augsburg.de> wrote:> > On 11 Jan., 01:35, Virgil <vir...@ligriv.com> wrote:>> > Do you think that the union of the set of finite initial segments of> > paths> > 0.1> > 0.11> > 0.111> > ...> > contains the path 0.111...> > or not?>> Since I have no idea what the "union" of even "0.1" and 0.11" would be,> since neither of them appears to refer to any set, I cannot answer until> WM explains what sets he is referring to.>> > If not, what node is missing? None.>> What nodes are present, unless you first tell us that you are working> within a tree?>> Rephrase you  questions less ambiguously!> --What he asks is what is the union of the finite sets.Does induction precede the axiomatization of infinity?  Because, itcan be built from unbounded finite cases.Then, does the unbounded imply the existence of the infinite?  (Yes,else it would be bounded.)Not saying there's enough paper to write it out, saying paper isn'tneeded to read it out.For the path defined by a set of naturals, with the presence of anatural in the set indicating a 1 else 0 for branch, that an n-setdescribe .111 ... {n-many 1's} , what is the union of the n-sets'paths?Then, where the union of the n-sets is N, does induction not requireaxiomatization of infinity?  Because, a union of m<n-sets would defineinduction through n, and there exists n > m for each m.Then, wouldn't the set of finite sets, be Russell's that all don'tcontain themselves?  Now, that's a result.Then, casually about one or zero being the first, ordering the n-setswith all the sets containing one following all those not, and allthose containing two, and one, after all those not containing two, andcontaining one, and all those containing two, and not one, after allthose containing neither two nor one, etcetera ad infinitum, theordering of the breadth-first traversal so defined sees theantidiagonal result not follow.1^1 + 2^2 + 3^3 + 4^4 ... = sum_i=1^n n^n rules define breadth firstordering of n-sets as pathsThe breadth-first traversal of the C.I.B.T. sees the antidiagonalresult not follow.  And:  rays through countably many ordinal pointsare dense in the paths.Regards,Ross Finlayson
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