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.

  • $\begingroup$ Your question is discussed in this answer (without a resolution) $\endgroup$
    – Wojowu
    Dec 29 '19 at 19:30
  • $\begingroup$ @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). $\endgroup$ Dec 29 '19 at 19:32
  • $\begingroup$ Is this finally helpful for anything? $\endgroup$
    – Asaf Karagila
    Dec 29 '19 at 21:04
  • $\begingroup$ Is the statement "$x$ codes an ordinal which in $L$ is of size $\aleph_n$" for some natural number $n$ a projective statement? $\endgroup$
    – Asaf Karagila
    Dec 29 '19 at 21:08
  • $\begingroup$ @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). $\endgroup$ Dec 29 '19 at 23:09

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