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Topic: § 424 Actual Infinity: We never get it - but
we get it!

Replies: 4   Last Post: Feb 4, 2014 8:21 PM

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Ben Bacarisse

Posts: 1,387
Registered: 7/4/07
Re: § 424 Actual Infinity: We never get it - but
we get it!

Posted: Feb 4, 2014 8:53 AM
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WM <wolfgang.mueckenheim@hs-augsburg.de> writes:

> Am Dienstag, 4. Februar 2014 01:07:19 UTC+1 schrieb Ben Bacarisse:
>

>> > The set of all finite sequences is countable.
>>
>> That's the set of paths? You just come right out and say that they are
>> all finite?

>
> Not the paths are finite but their definitions like "0.111..." with
> its eight symbols.

>>
>> Please define the set of paths. The set that you claim is countable.
>>

>
> So it is.


Unless you define it to be the only paths in your WMtree it is simply a
subset of the paths in the real tree. The WMtree is inconsistent, but
you can't even start to discuss it unless you make the stipulation that
it does not contain every path, just those with "finite definitions".

In graph theory, the set of paths in a graph are those sequences of
nodes whose consecutive pairs are in the edge set of the graph. That
set is uncountable in the case of the infinite binary tree.

>> Go on, you know you want to. It will all be over just as soon as you
>> define it to be some countable subset of the paths in the tree everyone
>> else it talking about.


> Everyone else is talking about? All your talk consists of finite
> words. Paths defined by finite words are coutable or sub-countable
> (but the latter does not exist in ZFC).


And finite definitions can result in uncountable sets. This is one such
example.

I have defined the tree (and its paths) that everyone else it talking
about several times. Every time you cut the definition without
significant comment. You can't explicitly accept it, since it has a
different path set to the one you want, and you can't correct the
definition because that would make it clear you are not talking about
the same tree that everyone else is. Your only option is to cut it
without agreement or disagreement.

Here it is again... The infinite binary tree, B = (N, T) is a graph
whose noes are the natural numbers (1, 2, 3...) and whose edges are the
pairs T = { (i, j) | j = 2i or j = 2i+1 }. The set if infinite rooted
paths in B are the set of sequences p(n) with p(1) = 1 and p(n+1) =
2p(n) or p(n+1) 2p(n)+1. Is this set of paths countable? No.

--
Ben.



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