In article <
fc06e396-63dd-4bbe...@googlegroups.com>,
WM <
muec...@rz.fh-augsburg.de> wrote:
> On Friday, 6 September 2013 02:09:46 UTC+2, Ben Bacarisse wrote:
> > WM <
muec...@rz.fh-augsburg.de> writes:
> >
> > > On Thursday, 5 September 2013 15:45:52 UTC+2, Ben Bacarisse wrote:
> >
> > >> WM <
muec...@rz.fh-augsburg.de> writes:
> >
> > >> > On Tuesday, 3 September 2013 23:54:25 UTC+2, Peter Percival wrote:
> >
> > >> >> Ross A. Finlayson wrote:
> >
> > >> >> > I understand that there are only countably many elements of the
> > >> >> > reals in their normal ordering, in a well-ordering, in ZF(C).
There are uncountably many reals in any ordering everywhere, at least
outside of outre places like WM's wild weird world of WMytheology.
> >
> > >> >> How can order affect cardinality?
It can only effect the ease by which one determines cardinality, at
least for infinite sets.
For finite sets all orderings are essentially the same as far as
determining cardinality is conserned.
> > >> > The rational numbers are countable when well-ordered.
Changing ordering of a set does not change its cardinality.
> > >> > In the binary tree however, which is constructed from all
> > >> > rational numbers only, we get an uncountable number of paths.
A complete binary trees has a node for each binary rational, but has no
nodes for non-binary rationals lie 1/3 or 2/3.
> > >> > How can order influence the presence or absence of the
> > >> > irrationals (and with that cardinality)?
It doesn't
> > >>
> >
> > >> You meant to say "how can taking the limit influence the presence or
> >
> > >> absence of the irrationals". Then you'd see the answer for yourself.
> >
> > >
> >
> > > The set of all rationals does not contain irrationals. If reordered
> >
> > > (or simply left as it is, the set of all rationals has no canonical
> >
> > > ordering), so if reordered in form of a tree, the irrationals enter
> >
> > > the scene unavoidably and without further ado? That sounds like magic
> >
> > > or matheology. Who will believe that? I certainly not. You? Really?
The difference is between nodes and paths. One can identify the nodes of
a complete infinite binary tree with proper binary fractions, but we
cannot do that with the paths, because, among other reasons, there are
far more paths than nodes.
> >
> >
> >
> > All of 3, 3.1, 3.14, 3.141, 3.1415... are rational. Put just them,
> >
> > nothing else, into a binary tree using the 0/1 encoding suggested and
> >
> > what happens? Either you stop at some point and leave the tree as
> >
> > finite (containing only rational numbers) or you "complete" just this
> >
> > one path and, bang!, an irrational is represented in it! Where did it
> >
> > come from???
It is the entire set of those lower decimal approximations, or, if you
like, the LUB of them.
>
> Why make it so difficult, using pi?
>
> Use the following sequence:
>
> 0.1
> 0.11
> 0.111
> ...
>
> According to your view I get the limit 0.111... - unless I stop somewhere.
According to our view we get the convergent sequence
{ 0,1. 0,11, 0.111, ...}
which in a Cauchy model of the reals represents its limit, 1/(b-1) where
b is the base of the number system in which { 0,1. 0,11, 0.111, ...} is
being written.
> That is not a mathematical view.
It may not be a WMytheological view, but it is totally a mathemtical
view.
> And it is not substantiated by any axiom. In
> analysis the sequence does *not* contain its limit.
In the Cauchy model the sequence IS the limit, or at least one
representation of it.
> Same is true when I write
>
> Why is the limit not contained in the sequence and in the tree? Simple
> answer: Because it does not exist other than by a finite definition
The Cauchy definition is a finite definition.
>
> But I understand
NO! You don not!
> >
> >
> >
>
> The tree does not contain ist limits.
The Complete Infinite Binary Tree does contain its paths as subsets of
the set of its set of nodes, and for every real in [0,1] there is at
last one such path, and for some of them two paths.
>
> > You think
> >
> > of it as always unfinished. That's fine -- there are many unfinished
> >
> > things in your world.
>
> Yes, all infinite things are infinite or, in English, unfinished.
In English, "infinite" and "unfinished" are not at all the same thing.
WM's mathematics is in many ways unfinished, rough, and
self-contradictory without being in any way infinite.
The paths in a Complete Infinite Binary Tree which become eventually
constant (all 0 branchings from some point onwards or all 1 branchings
from some point onwards) all represent binary rationals, each
expressible as m/2^n with m <= 2^n, m+1 and n in |N ( m can be 0).
The paths whose branching never become constant in one direction, but do
become periodic, produce the non-binary rationals like 1/3 and 2/5.
"Eventually periodic" means that from some node onwards a fixed finite
sequence of two or more left AND right branchings repeats forever.
In a binary tree, irrationals appear only as those convergent (weakly
monotone increasing) infinite sequences of binary rationals called paths
which never become either constant or periodic in the sense defined
above.
To the extent to which WM's notion of a COMPLETE Infinite Binary Tree
differs from the above, WM's notion is wrong!
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