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

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1. Jun 8, 2016

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 PDef(.) is taking the constructible power set (i.e., subsets are formed using Def), then
(VRamsey card,∈) |= PDef(L) is countable & PDef(PDef(L)) is uncountable.

2. Jun 9, 2016

### TeethWhitener

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.

3. Jun 9, 2016

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 Lα for some α.

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

4. Jun 9, 2016

### 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 Lκ 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: PDef(Lκ) ⊆ P(Lκ). You're asking whether PDef(Lκ) is countable or uncountable based on the existence of Ramsey cardinals. Is that right?

5. Jun 9, 2016

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

Last edited: Jun 9, 2016
6. Jun 16, 2016

### 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 PDef(Lκ) ⊂ P(Lκ); 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.

7. Jun 16, 2016