A quick question, just to make sure: assuming ~(V=L), then L is countable when viewed as a set inside V, correct?
Ah, sorry. I will put it in two ways, hoping that one of them makes sense.
(1) Is this true or false: Given a sufficiently strong cardinal $\displaystyle \kappa$ so that the axiom of constructibility fails, then <$\displaystyle V_\kappa$, $\displaystyle \epsilon$> $\displaystyle \models$ "there exists a set L closed under Def, contains the empty set, and is equinumerable with the set of natural numbers" ?
(2) The definition of L as a stepwise process in which each step depends on the number of finite first-order formulas; this number would be countable in the above-mentioned model. Or, looked at another way, since no first-order formula can assure that a set is uncountable, it would seem that, in a larger-than-L universe, L would be countable. If one claims uncountability for L from the fact that the process is taken over the ordinals seems like a vicious circle.
Hm, that's still a bit vague. I am hoping that the direction of my question is clearer, so that someone's answer can point me in the right direction to make it more precise.
L is a proper class, so it doesn't have a cardinality.
I don't know what you have in mind with "when viewed as a set inside V".
V is a proper class and L too is a proper class. L is a subclass of V. With the axiom of constructibility, L is not just a subclass of V but L is V. With the negation of the axiom of constructibility, L is a proper subclass of V.
But in any case, neither V nor L have a cardinality.