Re: R^R = R^2 = R

From: G. Frege <nomail@invalid>
Date: Thu, 29 Nov 2007 00:52:44 +0100

On Wed, 28 Nov 2007 23:47:14 +0000 (UTC), Dave Seaman
<dseaman@xxxxxxxxxxxx> wrote:

[...] ZF, which is the set theory that I and most others are using.

Not so sure about that. I'd guess that many mathematicians just use ZFC
without making a fuss. (Of course, in many cases a weaker axiom than CH
would do, still...) No?

F.

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