# Projective conservativity of choice?

Shoenfield's absoluteness theorem says that every $$\Sigma^1_3$$ statement is upwards absolute between models of ZF with the same ordinals. As a consequence, it implies that ZF and ZFC+V=L prove the same $$\Sigma^1_3$$ sentences. Meanwhile, this is sharp since "Every real is constructible" is a $$\Pi^1_3$$ theorem of ZFC+V=L which is independent of ZFC.

In the realm between ZFC and ZF, however, we can find significantly more absoluteness. For example, suppose $$M$$ is a model of ZF+DC. Then countably closed forcing adds no new reals, so when we force over $$M$$ with partial bijections $$2^\omega\rightarrow\Theta$$ with range bounded below $$\Theta$$ the resulting extension $$M[G]$$ has the same reals as $$M$$. But so does $$L[G]^{M[G]}$$, and this latter model satisfies choice since $$G$$ is appropriately equivalent to a set of ordinals (consider the set of pairs $$\langle n,\alpha\rangle\in\omega\times\Theta^M$$ such that $$G^{-1}(\alpha)(n)=1$$). Since $$M$$ and $$L[G]^{M[G]}$$ have the same reals, they satisfy the same projective sentences.

However, there are models of ZF without DC which do not have the same reals as any model of ZFC: for example, any model of ZF + "$$2^\omega$$ is a countable union of countable sets" has this property. So even ignoring the forcing details, the "model-swithcing" idea above breaks down when we try to answer the natural follow-up question:

Is ZFC conservative over ZF for projective sentences?

A natural candidate counterexample is "For every uniformly $$\bf \Sigma^1_k$$ sequence of countable sets of reals $$(A_n)_{n\in\omega}$$, there is an injection from $$\bigcup_{n\in\omega}A_n$$ to $$\omega$$" for some large enough $$k$$, but that would rely on building a model of ZF + "$$2^\omega$$ is projectively definably a countable union of countable sets which I don't see how to do. Variations on this idea hit the same problem.

On the other hand, I don't see how to make any nontrivial positive progress; in particular, I can't even resolve the following: is ZFC at least $$\Pi^1_3$$ conservative over ZF?

EDIT: Actually, unless I'm missing something the naive idea above does show that ZFC is conservative over ZF for $$\Pi^1_3$$ sentences since we still wind up building for a given model $$M$$ of ZF a model $$N$$ of ZFC with the same ordinals as $$M$$ and with $$\mathbb{R}^M\subseteq\mathbb{R}^N$$. Now for each $$r\in\mathbb{R}^M$$ and each $$\Sigma^1_2$$ formula $$\varphi$$ we have by Shoenfield that $$M\models\varphi(r)$$ iff $$N\models\varphi(r)$$, and this means that $$\Pi^1_3$$ facts holding in $$N$$ also hold in $$N$$.

• In fact, what's really going on is the following. Shoenfield immediately says that if $$(*)$$ is any sentence in the language of set theory with the property that for every model $$M$$ of ZF there is some model $$N$$ of ZF+$$(*)$$ such that $$M$$ and $$N$$ have the same ordinals and $$\mathbb{R}^M\subseteq\mathbb{R}^N$$, then ZF+$$(*)$$ and ZF have the same $$\Pi^1_3$$ consequences. The forcing described above shows that AC has this property (and meanwhile V=L obviously doesn't).

So - unless I've made a silly mistake here - it's at the fourth level of the projective hierarchy that things become nontrivial.

• Your question is discussed in this answer (without a resolution) Dec 29 '19 at 19:30
• @Wojowu Aha! I feel both less hopeful and less silly. Glad to highlight the question though (and the $\Pi^1_3$ case might still be easy to answer positively). Dec 29 '19 at 19:32
• Is this finally helpful for anything? Dec 29 '19 at 21:04
• Is the statement "$x$ codes an ordinal which in $L$ is of size $\aleph_n$" for some natural number $n$ a projective statement? Dec 29 '19 at 21:08
• @AsafKaragila Re: your second comment, I don't see how (and it's also not a ZFC theorem). Re: your first, maybe? But not immediately (to me at least). Dec 29 '19 at 23:09