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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 14, 2013 4:29 PM


In article <1a29638ff1a24a59a07c75e93e7ad53d@k14g2000vbv.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> Every welldefined Cantor diagonal belongs to the countable set of > nameable real numbers (named by the definition of the list and of the > replacement rule). And undefined Cantorlists do not yield real > numbers. > > Regards, WM
Cantor says a complete listing of real numbers cannot exist. WM says a complete listing of real numbers cannot exist, but simultaneously insists that Cantor is wrong.
***********************************************************************
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. 



