Re: Question: What conditions are sufficient to prove a subset does not exist?

In article <1185418167.824559.42040@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Scott <ToaTerra@xxxxxxxxx> wrote:
On Jul 25, 4:13 pm, LauLuna <laureanol...@xxxxxxxx> wrote:
When a subset defined in such way raises contradiction, then what you
can deduce is that either S or at least f does not exist. For example,
in Cantor's theorem this is used to show that f (bijective) does not
exist.

Perhaps I am asking the wrong question, then. What can I conclude if
the following is true:

xeS & (Ef)(f(x)=A) & (xeA <-> ~xeA)

This is also probably the wrong question.

First, remember that your set A depends on f. Better to call it A(f)
or A_f to keep that in mind.

That said, as I noted: if from the assumption that A is a set you can
derive a proposition of the form P & not(P), then you would be able to
conclude that "A is not a set" from the simple rules of logic.

If you can prove that there exists an f and there exists an x such
that x in A(f) and not(x in A(f)), then this would be a statement of
the form P & not(P).

The rest is obfuscation.


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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