Re: Is this equivalent to the Axiom of Choice?
From: Stephen J. Herschkorn (sjherschko_at_netscape.net)
Date: 12/27/04
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Date: 27 Dec 2004 15:44:00 -0800
Butch Malahide wrote:
> Dan Christensen wrote:
> > I have been wrestling with this problem, or variants of it, for
> several
> > weeks now. Is the following equivalent to the Axiom of Choice? If
so,
> how
> > might I prove it? Or can I prove it the following using AC? Any
hints
> would
> > be appreciated.
> >
> > Given non-empty sets X and Y.
> > If for all functions f: X->Y, f(X) is a proper subset of Y,
> > then there exists a 1-1 (injective) function g:X -> Y.
>
> That's equivalent to the Axiom of Choice. Use Hartogs' theorem: for
any
> set Y there is a well-orderable set X such that there is no injective
> function from X to Y.
Using Hartogs's theorem, Dan's axiom implies that for any set, there is
a surjection from a cardinal to that set. How does that imply AC?
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