If Ramsey cardinals exist, all powers using Def are countable?

[nomadreid]

I came across some old notes that seem to be dubious, with no references. So please correct:
The notes say that if Ramsey cardinals exist, and the constructible universe L is regarded as a set, and P_Def(.) is taking the constructible power set (i.e., subsets are formed using Def), then
(V_{Ramsey card},∈) |= P_Def(L) is countable & P_Def(P_Def(L)) is uncountable.
Thanks for corrections or comments.

 

[TeethWhitener]

the constructible universe L is regarded as a set

In what's becoming my standard caveat when I post about logic and set theory: It's been a long time since I've done any of this. That said, the constructible universe L is not a set, but a proper class. This is seen most easily by considering the fact that the ordinals are a "subset" (really a subclass) of the constructible universe, and the collection of all ordinals can't be a set because of the Burali-Forti paradox. So I'm not sure notion of the power set of L makes any sense.

EDIT: All of this assumes ZF set theory, of course.

 

[nomadreid]

Thank you, TeethWhitener. You are correct, L cannot be a set, and therefore it is nonsense to talk about its powerset, and even nonsense to ask whether L can be countable. Sorry; I think that the original was referring to an L_α for some α.

The question would now make sense if κ was such that L_κ were the universe of a minimal model of ZFC. Then L_κ would be a countable set (with respect to a larger model), but I am not sure whether the powerset of L_κ would be uncountable wrt to that larger model.

 

[TeethWhitener]

I didn't want to leave you hanging, even though I'm not sure I can answer your question. You're saying that, if L_κ is an (infinite) countable set, its power set (set of all subsets) is uncountable, by Cantor's theorem. However, you're wondering about the "definable power set"--that is, the set of all definable subsets, which is a subset of the power set: P_Def(L_κ) ⊆ P(L_κ). You're asking whether P_Def(L_κ) is countable or uncountable based on the existence of Ramsey cardinals. Is that right?

 

[nomadreid]

Yes, so that if P_Def(L_κ) turns out to be countable when Ramsey cardinals exist, then of course P_Def(P_Def(L_κ) ) is also, and so on. Many thanks for following this up.

 

[TeethWhitener]

Sorry for the delay. My knowledge in this area is limited, but the one thing I can tell you is that the existence of Ramsey cardinals implies the existence of non-constructible sets. This directly implies that P_Def(L_κ) ⊂ P(L_κ); that is, the definable power set is a proper subset of the power set: they aren't equal. In contrast, if all sets are constructible, then the definable power set is equal to the power set (I'm pretty sure this is true). Of course, the non-existence of Ramsey cardinals does not necessarily imply the non-existence of non-constructible sets. But this might be one direction to think about with your initial statement.

 

[nomadreid]

Thanks, TeethWhitener.

if all sets are constructible, then the definable power set is equal to the power set (I'm pretty sure this is true).

Correct.

Of course, the non-existence of Ramsey cardinals does not necessarily imply the non-existence of non-constructible sets.

That is also correct, probably, since it is enough to have the Erdös cardinal κ(ω₁) to insure that V≠L. However, I say "probably" because I could not find any reference later than 1977 on the connection between the existence of Ramsey cardinals and the existence of κ(ω₁). (At that time, there was no known connection, but that was almost 40 years ago. Perhaps some progress has been made since then)