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Topic: Matheology � 233
Replies: 36   Last Post: Apr 2, 2013 5:56 PM

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Posts: 908
Registered: 2/15/09
Re: Matheology § 233
Posted: Mar 30, 2013 5:19 PM
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On Mar 30, 1:17 pm, fom <fomJ...@nyms.net> wrote:
> On 3/30/2013 8:38 AM, WM wrote:

> > On 29 Mrz., 19:34, Virgil <vir...@ligriv.com> wrote:
> >>> So we have established the fact that an irrational number has no node
> >>> of its own.

> >> No number in any infinite binary tree has any node "of its own", as
> >> every node has two child nodes belonging to necessarily different
> >> numbers.

> > That is correct, but only establishes the fact that no actually
> > infinite path can be distinguished from all rational paths as should
> > be possible in a Cantor-list - but is not.

> WM failed the science lesson again today.
> The Cantor argument is an argument scheme.
> It presupposes a standard, classical use of
> of the quantifier "all".
> WM has never defined his non-standard uses
> for the word "all".  It has no agreed upon
> usage.  It is meaningless by WM's own standards
> of meaning through pragmatic agreements between
> language users.
> By definition, all paths in the complete infinite
> binary tree are infinite whether or not they
> become eventually constant.
> Any purported countable listing of all the paths
> of that tree will result in a successful
> defeat of the claim by a Cantor argument.

A breadth-first traversal of the paths, as the level of the tree goes
to infinity, is the same as depth-first of the nodes, for the infinite
tree, for each path, and each node, in reading out that paths as
nodes, as paths. The depth first traversal goes through nodes in a
transfinite sequence, as it were, that is the same as the readout of
the nodes of the paths of the breadth-first traversal of paths, as
modeled from the finite, in exhaustion, in the infinite.

And, it is very similar to n/d, with n, d e N and d -> oo, n->d (or,
"EF"), in this manner: unfilled nested intervals and the antidiagonal
result don't follow from the premises.

And besides that rays through countable ordinal points are dense in
the paths, of the infinite balanced binary tree (and ordinals are well-
ordered). This is similar to the notion of that the rationals are
dense in the reals: that for other sets dense in the reals (countable
or not): the rationals are dense in those, and stronger: for the
ordinal points as well-ordered, each is identified with a distinct
path, here regardless of what its elements as nodes are, except:
first, and last.

Then transitively that 2^w <-> P(N) and P(N) <-/-> N that N <-/-> 2^w,
due the set-theoretic Cantor's theorem of set and powerset (where even
infinite sets are _axiomatized_ to be well-founded and the universe
doesn't exist), well, with ubiquitous ordinals that S of Cantor's
theorem is {} (not containing any elements of the set) for n -> n+ 1 ,
again there are systems with mapping the continuum of the naturals
through the sweep of the continuum of the segment, that it is 1-1,
onto, and furthermore constant monotone: and where it is not onto if
not constant monotone or sweep.

So: draw a line. Cantor's theorem for reals is simply read as that
the drawing of the point is through the stroke, not each mark, for
each mark besides beginning or end as of the others: of the drawing,
of the points, of the line: all of them.

The drawing, the stroke, the sweep of the line, is through all, of
infinitely many: only at once.

There are proofs that irrationals exist without the uncountable. It
would be of tremendous interest to many to find application solely due
transfinite cardinals. To accommodate the results of real analysis
matching geometry: countably additivity of no greater than
infinitesimal reals, to accommodate the existence of a universe: non-
sets in set theory, it's easier to find non-application, and eschewal,
of transfinite cardinals: for results.

Then, with regards to the universal quantifier: sometimes: transfer
carries. That is where: "for each" _is_ "for all". This with
regards to "the" universal quantifier, here over sets.

So: make a science lesson today.


Ross Finlayson

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