Virgil
Posts:
4,491
Registered:
1/6/11
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Re: Endorsement of Wolfgang Mueckenheim from a serious self-confessed non-mathematician?
Posted:
Jan 31, 2013 6:02 PM
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In article
<9920e842-55f7-49b1-abbe-2a826fd56ba0@l9g2000yqp.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 31 Jan., 15:27, forbisga...@gmail.com wrote:
> > On Wednesday, January 30, 2013 1:03:49 PM UTC-8, WM wrote:
> > > Now I
> > > construct the complete infinite Binary Tree by means of countably many
> > > paths,
> >
> > You have never done such a thing.
>
> Of course I have. I construct one infinite path through each node.
> That means there are uncountably many paths and every combination of
> nodes residing on levels L_n with finite index n is constructed.
That may give all nodes, but omits anything like an explicit
construction of most paths.
It is no more complete than constructing a non-empty set involves a
complete and explicit construction of all of its subsets.
> >
> > > such that every node and every possible combination of nodes is
> > > covered by at least one path (in fact by infinitely many).
> >
> > Yes, infinitely many paths. Uncountably infinite many paths.
>
> That is wrong. I need only countably many paths.
You need a to more to get all paths.
>
> > You have never given any means to count them all. You just
> > assert you have done so.
>
> Here is a means that everybody with a minimum of intelligence can
> understand.
>
> I construct a finite path to every node and then append that path by
> an infinite tail, for instance, 010101...
>
> But I don't tell you what tail I used, because this tail is completely
> meaningless. It will only prevent that you can argue the path of 1/3
> or the path of 1/7 is not in the tree.
But for every tree built with only one such infinite tail, there are
uncountably many differing tails that do not appear in your tree, unless
you let all tails appear in it.
In any Complete Infinite Binary Tree, for every suser of |N there is a
path that branches left at just the levels enumerated in that set and
right at every other level.
So everyone but WM has as many such paths as one has subsets of |N.
And until WM can prove a surjection from |N to its power set, we will
continue to hold that that number of subsets of |N is greater than the
number of members of |N.
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