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

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In summary, 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.
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nomadreid
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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.
Thanks for corrections or comments.
 
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nomadreid said:
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.
 
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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.
 
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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?
 
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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.
 
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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.
 
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Thanks, TeethWhitener.
TeethWhitener said:
if all sets are constructible, then the definable power set is equal to the power set (I'm pretty sure this is true).
Correct.

TeethWhitener said:
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:oldconfused:)
 

1. What is a Ramsey cardinal?

A Ramsey cardinal is an infinite cardinal number that is the smallest cardinal number for which a certain combinatorial property, known as Ramsey's property, holds.

2. What is the connection between Ramsey cardinals and the use of Def?

The use of Def, short for the Axiom of Definable Comprehension, is a set-theoretic axiom that allows for the construction of sets based on certain properties. The existence of Ramsey cardinals is closely related to the use of Def, as the existence of Ramsey cardinals implies the truth of Def.

3. How does the existence of Ramsey cardinals impact powers of sets?

If Ramsey cardinals exist, then all powers of sets are countable. This means that for any set, the collection of all its subsets (its power set) is countable, which has significant consequences for set theory and mathematical logic.

4. Are there any known examples of Ramsey cardinals?

Currently, there are no known examples of Ramsey cardinals. In fact, the existence of Ramsey cardinals is a highly debated topic in the mathematical community. Some mathematicians believe that they do exist, while others believe that they do not exist.

5. What implications does the existence of Ramsey cardinals have for mathematics?

If Ramsey cardinals exist, it would have significant implications for mathematics and set theory. It would provide a solution to many open problems and questions, and would greatly impact our understanding of infinite sets and their properties.

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