
Re: The uncountability infinite binary tree.
Posted:
Dec 16, 2012 5:29 PM


On Dec 16, 12:39 pm, Rupert <rupertmccal...@yahoo.com> wrote: > On Dec 16, 7:47 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: > > > > > > > > > > > AlJofar writes: > > > 1: "This is a simple Corollary of Cantor's diagonal argument actually. > > Proof: Let G be any countable subtree of the infinite binary tree that > > is 0 rooted and such that all paths ending by 000... or by > > 111...are among the paths of G. " > > > 'Let G be a subtree of the IBBT including all expansions representing > > fractions by powers of two (there are two of those for each of those > > values).' > > > 2: "Let f be a bijective function from the domain N of all > > naturals(except 0) to the set of all infinite paths of G. " > > > For the countable subtree only containing fractions by powers of two > > and their dual representations, that's countable (and not obviously > > all the paths). > > > 3: "Construct the diagonal path d_f in the following manner: The root > > (i.e. the 1st) node of d_f is 0 labeled. Now for n=1,2,3,.. ; The n > > +1_th node of d_f is labeled by a label that is opposite to the label > > of the n+1_th node of the path of G that f sends n to. " > > > Let G be all the paths. In lexicographic order, here as described as > > BT _oo (binary tree's breadthfirst traversal at infinity), augmented > > with .0111... being first to reflect the modification of "the" anti > > diagonal, the generated antidiagonal path is .0111..., that is the > > least element of G in the ordering. > > No, it's not. > > > With an unmodified construction of the antidiagonal, the generated > > antidiagonal path is .111... and at the end of any courseofpassage > > through the paths in their natural order, i.e., never before the end. > > That is, there's a simpler corollary that is the same as the binary > > antidiagonal argument. > > > 4: "Now clearly the diagonal d_f (actually the antidiagonal but it > > shall be called the diagonal for short) is a path and clearly it is > > labeled in a way that is different from labeling of all paths of G. So > > d_f is missing from G. " > > > In the course of passage with the natural order of the paths, d_f = > > f(0) or here f(1) and is not missing from G. > > Nonsense. > > > 5: "So any countable subtree of the infinite binary tree, that is 0 > > rooted and that has all paths ending with 000.. or with 111.. > > among its paths; would be missing a path of the infinite binary tree. > > " > > > That doesn't follow for the natural ordering of the paths by their > > content as expansions. Then though while the paths are totally > > ordered, wellordering them would be as wellordering the reals. > > That's irrelevant. > > > > > > > > > Then I'll happily accept that the antidiagonal argument is much the > > same for the list of expansions or the tree of expansions, then that > > EF's antidiagonal is .111... and at the end of the list and BT's > > antidiagonal is .111... and rightmost of the tree. >
In the breadthfirst ordering, after 0 = .000..., we don't (standardly) have a proof that there's a least element in the normal ordering (or they'd be wellordered by their normal ordering), but we do have a proof that there's a wellordering for each level of the tree. So, we can select from G\0 (here being all the paths setminus zero's) all those less than .1 then all those less than .01 and so on, where then obviously, from the finite to the infinite, there are never none left, but, as well, there isn't a level or row such that there isn't an element starting with that many zeros that would be lesser in the normal or usual or natural breadthfirst ordering. So, as we know that all the elements of the breadfirst traversal would be starting with zero at each relevant place, the only way for it to be different at that place and be part of the antidiagonal, is for the value to be one.
Paths:
.000... = BT(0) .000... = BT(1) .000... = BT(2) ... .111... = BT(2^w)
Antidiagonal, from the beginning:
.111... = BT(2^w)
Then it's quite relevant in as to wellordering the reals, well ordering the paths.
Regards,
Ross Finlayson

