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Topic: Another AC anomaly?
Replies: 280   Last Post: Dec 27, 2009 3:53 PM

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Virgil

Posts: 870
Registered: 7/27/09
Re: Another AC anomaly?
Posted: Dec 16, 2009 12:07 AM
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In article
<9904ae9d-de2c-4d5f-86da-be165e8e9d7e@p30g2000vbt.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 15 Dez., 13:32, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
>

> >  > It is nonsense to define a pink unicorn.
> >
> > That statement is nonsense.  Let pu be a pink unicorn is a proper
> > definition.
> > However there is nothing that satisfies that definition.

>
> That is a matter of taste.


So WM is declaring that it is a matter of taste whether there is
something that satisfies a definition of a pink unicorn?


> >
> >  >                                          The set N does not exist as
> >  > the union of its finite initial segments. This is shown by the (not
> >  > existing) path 0.000... in the binary tree.
> >
> > As you have defined your tree as only containing finite paths it is trivial
> > to conclude that the infinite path 0.000... does not exist in your tree.
> > However, this does *not* show that N does not exist.

>
> If N does exist, then the tree containing all finite paths contains
> also infinite paths.


In that case, you have proved that N "must" exist.
> >
> >  > Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}.
> >  > What then is
> >  > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} ?
> >
> > Ambiguous as stated.  What is "..." standing for?

>
> Look two lines above:
>
> Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}
> hence
> {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...} .


An ellipsis with last term, as WM has indicated above, can only
represent a finite sequence in HONEST mathematics.

What WM may have meant, but was too dishonest or too dumb, or both, to
express honestly is
( {1} U {1, 2} U {1, 2, 3} U ... ) U {1, 2, 3, ...}
with the indicated parenthesis ABSOLUTELY necessary.


> {1} U {1, 2} U {1, 2, 3} U ... = {1} U {1, 2} U {1, 2, 3} U ... U {1,
> 2, 3, ...}


{1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...}, without any
parentheses, is mathematically meaningless


>
> There is no space in the binary tree to contain infinite paths in
> addition to the sequeneces of all finite paths.


Neither finite paths nor sequences of finite paths are themselves paths
in infinite trees.

> The sequence 0.0, 0.00, 0.000, ... when being completely constructed,
> is already the path 0.000...
> You cannot add it separately.


If you consider it only as a sequence of nodes then that set (or
sequence) of all of those nodes can be identified with a path in the
infinite tree.
>
> Therefore: When all finite paths have been constructed within aleph_0
> steps, then all paths have been constructed.


There are absolutely no such things as finite paths in an infinite tree.

No finite sequence can do more than identify a single node, or finite
set of single nodes, in such a tree, and a node is not a path in any
infinite tee.



> There is no chance to
> construct the path
> 0.000...
> after all finite initial segments have been constructed.


There is no chance to even define, much less construct, an infinite
binary tree by any of the false methods that WM proposes. Though in ZF
it is trivial to construct such a tree.

Example of an infinite binary tree, which exists in ZF:
Starting with the infinite set N = {1,2,3,...}, certain infinite
subsets are paths: a subset that contains 1 and for each n in N,
contains one and only one of 2*n or 2*n+1 is a path, and no
other subsets of N are paths. Each natural is a node but sets of
naturals are never nodes and finite sets of naturals are never paths.

WM has absolutely no model for an infinite binary tree that is even
self-consistent, much less compatible with any acceptable set theory.


Date Subject Author
11/23/09
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11/27/09
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11/27/09
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Virgil
11/27/09
Read Re: Another AC anomaly?
mueckenh@rz.fh-augsburg.de
11/27/09
Read Re: Another AC anomaly?
William Hughes
11/27/09
Read Re: Another AC anomaly?
mueckenh@rz.fh-augsburg.de
11/27/09
Read Re: Another AC anomaly?
Virgil

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