Re: ZFC? countable?uncountable?

On Feb 27, 5:01 pm, "MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
On Feb 27, 1:13 pm, "zuhair" <zaljo...@xxxxxxxxx> wrote:

AxEpAy(yep <-> Az(zey <-> (zex & Q))),where 'Q' is any formula in
which 'y' does not occur free.

Now let Q(z)<->z=z, x=w
we have p={w}. which is countable!

Wonderful!!! So what?!!!

In a theory which has this axiom instead of the standard power axiom,
you cannot have an uncountable set, that's what!


MoeBlee


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