Date: Mar 30, 2013 9:35 PM Author: ross.finlayson@gmail.com Subject: Re: Matheology § 224 On Mar 30, 6:09 pm, Virgil <vir...@ligriv.com> wrote:

> In article

> <4f5abe9e-4a74-4ada-ab2c-3f6cab383...@ia3g2000vbb.googlegroups.com>,

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> WM <mueck...@rz.fh-augsburg.de> wrote:

> > On 30 Mrz., 19:15, Virgil <vir...@ligriv.com> wrote:

> > > In article

> > > <ab85409a-eabf-4b68-b505-d194ed33a...@c15g2000vbl.googlegroups.com>,

>

> > > WM <mueck...@rz.fh-augsburg.de> wrote:

> > > > On 30 Mrz., 10:17, William Hughes <wpihug...@gmail.com> wrote:

> > > > > On 24 Mrz., 18:09, WM <mueck...@rz.fh-augsburg.de> wrote:

> > > > > <snip>

>

> > > > > > > The only difference is that in the second case you consider

> > > > > > > some subsets of the nodes to be paths, that are not considered

> > > > > > > to be paths in the first case.

>

> > > > > > Well, that is a correct description. It implies that these additional

> > > > > > subsets cannot be distinguished by nodes from the finite subsets

>

> > > > > Piffle. It is trivial to distinguish a subset that has a node

> > > > > at a last level from a subset that does not have a node

> > > > > at a last level.

>

> > > > No, that is impossible if an infinite path consists of infinitely many

> > > > finite subsets.

>

> > > All infinities consist of infinitely many finite parts.

> > > But the infinite set of all naturals is distinguishable be from the

> > > infinite set of all FISONs,

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> > And so is the path of 1/pi distinguishable from all its finite initial

> > segments which are in the tree. But as you said, 1/pi is not distinct

> > from them.

>

> WM conflates the set of all FISONs with the union of that set, |N, so,

> as he does far too often, fails to distinguish between the subsets of a

> set and the members of a set.

>

> Until he has learned ro distinguish between the members of a set and

> the subset of a set reliably, he should avoid anything to do with sets.

>

> It comes into tze construction automatically. The limit is

>

> > in any case a member of the sequence. That is unmathematical.

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> > > > It is impossible to distinguish the actually infinite path of 1/pi

> > > > from a path that only is built of all finite initial segments of the

> > > > path of 1/pi.

>

> > > It may be so in Wolkenmuekenheim, but a set of only finite

> > > approximations to an irrational number can elsewhere be distinguished

> > > from the number itself.

>

> > Then explain why this is not possible in the Binary Tree. You said

> > that the irrationals come into the tree automatically, impossible to

> > distinguish by nodes.

>

> I never said that they were impossible to distinguish by node, because

> they are.

>

> In a Complete Infinite Binary Tree, every binary rational path has only

> finitely many left-child nodes or only finitely many right-child nodes,

> whereas every other path has infinitely many of each.

>

> Something that everyone who understands anything about Complete Infinite

> Binary Trees should know but WM apparently does not.

> --

Those aren't all the rational sequences, only integral products of

negative powers of two. Each infinite path that has as a subpath

(subsequence of nodes, in sequence) any ((0|1)*)\infty is a rational

sequence. Of course, (0|1)* isn't bounded, where to be a rational

sequence it's finite. Here to write * for the Kleene star, \infty is

for an infinite number in sequence, in the language of the paths of

the tree. Here of course then the rational expansions end ((0|1)+)

\infty, where they don't simply end, and the Kleen closure includes

the empty set. Every other path than a binary-rational has infinitely

many 0- and infinitely many 1-branches, its children in the tree.

There are only (as it were) countably many nodes to distinguish the

paths.

Ordinals conflate the set of all finite initial ordinals to the union

of that set, at least as so axiomatized in ZF.

Pot: kettle black.

Regards,

Ross Finlayson