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Topic:
Cantor's absurdity, once again, why not?
Replies:
77
Last Post:
Mar 19, 2013 11:02 PM



Virgil
Posts:
8,833
Registered:
1/6/11


Re: Cantor's absurdity, once again, why not?
Posted:
Mar 15, 2013 3:10 PM


In article <e539ed586e9549a7b52a58ed70317e11@o5g2000vbp.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 14 Mrz., 23:36, fom <fomJ...@nyms.net> wrote: > > On 3/14/2013 5:15 PM, WM wrote: > > > > > > > > distinguishable, that means definable by finite words > > > > How does a definition "distinguish"? > > A definition is a name. For instance one of many definitions of pi. > That can be used to distinguish it from three. > > Extensionality says that sets with same elements are identical. > Therefore it must be possible, in ZF, to fix whether elements are same > or not.
It is equally possible to show that the two test sets equal without looking at any single member of either.
The set of naturals is the same as the union of the set of even naturals and the set of odd naturals, which can be known, at least outside of WMytheology, without looking at any individual member of either set
> Therefore it must be possible to compare and to recognize > elements.
Not necessarily. If the properties of the members of on set description exactly overlap those of another, the two descriptions can be know not describe the same set without any individual member or either set needing to be identified.
***********************************************************************
WM has frequently claimed that a mapping from the set of all infinite binary sequences to the set of paths of a CIBT is a linear mapping. In order to show that such a mapping is a linear mapping, WM must first show that the set of all binary sequences is a vector space and that the set of paths of a CIBT is also a vector space, which he has not done and apparently cannot do, and then show that his mapping satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of a field of scalars and x and y are f(x) and f(y) are vectors in suitable linear spaces.
By the way, WM, what are a, b, ax, by and ax+by when x and y are binary sequences?
If a = 1/3 and x is binary sequence, what is ax ? and if f(x) is a path in a CIBT, what is af(x)?
Until these and a few other issues are settled, WM will still have failed to justify his claim of a LINEAR mapping from the set (but not yet proved to be vector space) of binary sequences to the set (but not yet proved to be vector space) of paths ln a CIBT.
Just another of WM's many wild claims of what goes on in his WMytheology that he cannot back up. 



