Re: Cantor's diagonal proof wrong?

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 11/18/04 

Date: 18 Nov 2004 09:48:12 -0800

Hi,

Basically from the empty set all possible constructions, which means
basically any group of balanced brackets, is a set. As well, as there
is no foundation axiom, then you would be able to arbitrarily label a
set and then insert that label anywhere within that or any other set
definition.

Each of these is formed because each is unique, and thus through
excluded middle, not your axiom of difference, it is generated.

The sets by themselves can be quite meaningless. It's just the set of
all possible constructions of the empty set, the set of all sets.

Then, each set is claimed to fall together in a consistent way to form
ordinals. Take any literal, and assign each set to be a given
ordinal. Thus, each set is unique, and also, each set is among many
other sets representing the same ordinal. It doesn't matter how you
select the ordinals, it's just a sequence, any sequence.

Then are ascribed certain properties of the ordinals that they
represent the natural integers in most regards. They're well ordered.

Of the many-to-one mappings of sets to ordinals, one has the property
that it is basically the von Neumann form, or another form that has
each ordinal as least set being an element of its order type as
ordinal. The direct sum of infinitely copies of N is the empty set.

So there aren't any axioms, just an assertion of empty set and
excluded middle, and a definition of ordinal.

In a generic extension V^G of V, there exists the set N and a
bijection between N and P(N). Is that not so? Why or why not? What
elements of N are not elements of N? The answer to that is none.

Warm regards,

Ross F.

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