4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Yizheng Zhu Oct 5 '16 at 19:27
  • $\begingroup$ @YizhengZhu No, I do not. I believe the "boldface" version of $FA(\Sigma^1_3)$ is equivalent to "every real has a sharp," correct? $\endgroup$ – Noah Schweber Oct 5 '16 at 19:53
  • $\begingroup$ To be precise, "every set has a sharp". $\endgroup$ – Yizheng Zhu Oct 6 '16 at 14:31
  • $\begingroup$ @YizhengZhu Aargh, I even knew that - that was a silly typo. But yes, I am interested in only the "lightface" version of $FA(\Sigma^1_3)$. $\endgroup$ – Noah Schweber Oct 6 '16 at 14:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.