# Does absoluteness imply a club dichotomy?

My question is about two types of consequence of large cardinals, considered over ZFC on their own.

First, we have statements of the form, "The club filter on $\omega_1$ is an ultrafilter when restricted to 'reasonably nice' sets of ordinals."

For $\Gamma$ a class of formulas, let "$CD(\Gamma)$" denote the statement $$\mbox{Every \Gamma-definable set of countable ordinals contains or is disjoint from a club}$$ (where we say $X\subseteq\omega_1$ is a $\Gamma$-definable set of countable ordinals iff for some $A\subseteq\mathbb{R}$, $A$ is definable by a formula in $\Gamma$ with real parameters, and $A$ consists of reals which code ordinals, and the ordinals coded by elements of $A$ are exactly those in $X$).

Second, we have statements of the form, "Forcing can't change the truth value of 'reasonably simple' propositions."

For $\Gamma$ a class of formulas, let "$FA(\Gamma)$" denote the statement $$\mbox{Sentences in \Gamma with real parameters are absolute under (set) forcing.}$$ (EDIT: To clarify in response to Yizheng Zhu's comment below, only parameters from the ground model are allowed. So, $FA(\Gamma)$ holding in $V$ does not imply that $FA(\Gamma)$ holds in $V$'s set-generic extensions.)

My question is whether there are outright implications over ZFC between these kinds of statements, especially at the projective level.

First, the specific case I'm looking at is:

Does $FA(\Sigma^1_3)$ imply $CD(\Sigma^1_2)$ over ZFC?

More generally, though, I'm interested in:

For which $m, n$ does $FA(\Sigma^1_m)\implies CD(\Sigma^1_n)$ over ZFC? What about conversely?

I don't immediately see a way to deduce any amount of FA from CD, or vice versa, so it's plausible to me that over ZFC they are independent; however, the only way I know to construct models of these principles is via large cardinals, which imply both.

My interest in this question, besides intrinsic, is: I have a claim which I can prove using $FA(\Sigma^1_3)+CD(\Sigma^1_2)$ in a simple way, or using $FA(\Sigma^1_3)$ with a bit more work. Having both proofs is nice - unless $FA(\Sigma^1_3)$ implies $CD(\Sigma^1_2)$, in which case there isn't really any reason to hang on to the second proof.

• Do you allow real parameters in generic extensions? I.e. $\Sigma^1_3$ statements with parameters in any generic extension are absolute in further generic extensions? If so, the answer to your first question is yes. – Yizheng Zhu Oct 5 '16 at 19:27
• @YizhengZhu No, I do not. I believe the "boldface" version of $FA(\Sigma^1_3)$ is equivalent to "every real has a sharp," correct? – Noah Schweber Oct 5 '16 at 19:53
• To be precise, "every set has a sharp". – Yizheng Zhu Oct 6 '16 at 14:31
• @YizhengZhu Aargh, I even knew that - that was a silly typo. But yes, I am interested in only the "lightface" version of $FA(\Sigma^1_3)$. – Noah Schweber Oct 6 '16 at 14:38