Thread: cardinality of L inside V

A quick question, just to make sure: assuming ~(V=L), then L is countable when viewed as a set inside V, correct?

Originally Posted by nomadreid

A quick question, just to make sure: assuming ~(V=L), then L is countable when viewed as a set inside V, correct?

I don't see the sense of the question. Could you reword it?.

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 so that the axiom of constructibility fails, then < , > "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.

Ah, right, I should have thought of that when mentioning the bit about On. Silly of me, so thanks for the correction. I should stick to some to get a countable model.