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Virgil
Posts:
4,660
Registered:
1/6/11
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Re: Cantor's absurdity, once again, why not?
Posted:
Mar 14, 2013 6:17 PM
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In article <bc40a766-c250-4652-8533-c5b8ebc12653@r8g2000vbj.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 14 Mrz., 21:45, fom <fomJ...@nyms.net> wrote: > > On 3/14/2013 3:26 PM, Virgil wrote: > > > > > In article > > > <9e11404e-5e55-4f63-bdfe-e2c416345...@o5g2000vbp.googlegroups.com>, > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > > >> I believe, and I hope, that a > > >> future generation will laugh at this hocus pocus. > > > > > And go back to unsolving Zeno's paradoxes? > > there is not the least relation between Cantor and Zenon.
Both of them appear to be original thinkers of an ilk far beyond WM's perpetual pettiness.
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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. --
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