@Andrej: Isn't this just cherrypicking an interpretation to make the argument? This is like saying that since I can encode strings as vectors in Javascript, they are inherently distinct from integers. Even though that residing in your memory, even a vector is just a big number...

If my memory serves me right, there's a fun little way of presenting forcing, from a model theoretic point of view (i.e., without worrying out well-founded models and adding ordinals) where we use compactness and type omitting to produce generic extensions. And in a way, if you study the way AC is involved in compactness, you see that compactness, to an extent, is not too far from forcing-style arguments.

@LaveCave: Recall that $\Bbb R$ is really just $V_{\omega+1}$ in disguise; once that's established to be a part of $\rm HOD$, we can continue by induction. But alternatively, just use induction to show that $\cal P(\alpha)\subseteq\rm HOD$ for all ordinals.

Assuming that the Principle of Dependent Choice and that every set of reals has the Baire Property (or Lebesgue measurable), every group homomorphism between completely metrisable groups is continuous. Therefore, in that case, any linear map between Banach spaces is continuous, and therefore is bounded.

@SarcasticSully: In ZF the statement "The power set of an ordinal can be well-ordered" implies the Axiom of Choice. Yes. It is not the naive and "obvious" proof that one can think of, and in fact the proof relies on the Axiom of Regularity in a nontrivial way.

@SarcasticSully: I'm not sure what the question you're asking is.

@Joel: Elliot has given what is the gist of Specker's argument. (In fact, a slightly better one!)

I'm not sure that 3 power sets is really unnecessary. mathoverflow.net/questions/98365/…