You might want to try and use \{ for the braces in math mode :)

Just to be nitpicky: $\mathcal{P}(\mathbb{R}) \backslash \{\emptyset\}$.

I assumed that if someone knows what a measurable cardinal is, he'll know that it's a fixed-point of the Aleph function. :) Many thanks.

Thanks for the answer, firstly I'd like to stress that in my re-reading the question on Jech I found several mistakes in my question here and corrected them, one of those was weakening the equality to a weak inequality. Secondly, my biggest problem with this question is that the intended result is not to have some transitive model in which $2^\kappa = \aleph_{\kappa + \beta}$ but rather the original model.

Andres, I think so. It means that $f(\gamma) = \alpha$ for almost all $\alpha < \kappa$, right?

I think that I didn't fully understand the idea of some function $f$ such that $[f] = \alpha$ for some ordinal $\alpha$. The so-called confusing points which you mention are actually fairly clear to me.

When you say "ill-founded $\omega$" what exactly do you mean?

The cardinality of the continuum is NOT $\aleph_1$. This is the continuum hypothesis which is independent of ZFC.

@Pete: So there are no real closed fields which are Dedekind-closed except the real numbers?