Shoenfield absoluteness and choice

Absoluteness of wellfoundedness and Shoenfield absoluteness are two absoluteness results in set theory that are both used incredibly often. But what if we want to apply the result to absoluteness between arbitrary models M and N, rather than absoluteness between V and L? It turns out that our models have to satisfy dependent choice in both absoluteness results, and in Shoenfield absoluteness we have to ensure that the models are of “similar height”.


We start with the absoluteness of wellfoundedness, where the following standard proof turns out to not require power set, but do require DC.

Theorem (Absoluteness of wellfoundedness). Let R\subseteq X\times X be a binary relation on some set X. Then wellfoundedness of R is absolute between transitive models of \textsf{ZF}^-+\textsf{DC}.

Proof. It suffices to show that wellfoundedness of a binary relation R can be described in a \Delta_1^{\textsf{ZF}^-+\textsf{DC}} fashion.

Firstly wellfoundedness of R is equivalent (over \textsf{ZF}^-+\textsf{DC}) to there being no function f:\omega\to X such that f(n+1)Rf(n) for all n<\omega, which is clearly a \Pi_1 statement. Secondly, wellfoundedness of R is also equivalent (over \textsf{ZF}^-) to there existing a ranking function r:X\to\textsf{On}; i.e., that r(x)=\sup\{r(y)+1\mid yRx\}, which is a \Sigma_1 statement. QED

In the above proof we only needed DC to show the \Pi_1 part, meaning that wellfoundedness is always downwards absolute between transitive models of \textsf{ZF}^-. The next few lemmas lead up to Shoenfield absoluteness, where they treat \Pi^0_1(x)=\Pi^1_0(x) sets and \Pi^1_1(x) sets, respectively.

Lemma (ZF+DC). Let x\in\mathbb R. Then for every \Pi^0_1(x) set A\subseteq\mathbb R^n there’s a tree T\in L[x] on \omega^n such that A=[T].

Proof. For notational ease let’s assume that n=1. Let A=\{z\in\mathbb R\mid\forall k<\omega:\varphi(z\upharpoonright k,x)\} for \varphi(u,v) a \Pi^0_1 formula, and define

T:=\{s\in{^{<\omega}\omega}\mid\exists y\in A: s\subseteq y\}.

Note that T\in L[x] and clearly A\subseteq[T], so let y\in [T]. Then y\upharpoonright n\in T for all n<\omega, so that \textsf{DC} implies that there’s a sequence \langle y_k\mid k<\omega\rangle of y_k\in A such that y\upharpoonright k\subseteq y_k, which implies that \varphi(y\upharpoonright k,x) holds for all k<\omega, making y\in AQED

Lemma (ZF+DC). Let x\in\mathbb R. Then every \Pi^1_1(x) set A\subseteq\mathbb R^n is the projection of a tree S\in L[x] on \omega^n\times\omega_1.

Proof. Assume for notational simplicity that n=1. Let A\subseteq\mathbb R be \Pi^1_1(x), so that \mathbb R\setminus A=pB for some \Pi^0_1(x) set B. By the above lemma there’s a tree T\in L[x] such that B=[T], so that \mathbb R\setminus A=p[T], meaning that y\in A iff T_y is wellfounded.

Note that, using replacement, T_y is wellfounded iff it can be ranked by a function f:T_y\to\omega_1, meaning that s\supsetneq t implies that f(s)<f(t) for all s,t\in T_y. We can pick such a ranking function with codomain \omega_1 since wellfoundedness of T_y implies that |T_y|=\aleph_0.

Now define a tree S\in L[x] on \omega\times\omega_1 as follows. Fix a bijection e:\omega\to{^{<\omega}\omega} such that, for every s\in{^{<\omega}\omega}, n<\text{lh}(s) implies that e^{-1}(s\upharpoonright n)<e^{-1}(s). For s\in{^{<\omega}\omega} define

T_s:=\{t\in{^{<\omega}\omega}\mid\text{lh}(t)\leq\text{lh}(s)\land(s\upharpoonright\text{lh}(t),t)\in T\}

and then define S\subseteq{^{<\omega}\omega}\times{^{<\omega}\omega_1} as (s,t)\in S iff \text{lh}(s)=\text{lh}(t) and that

\forall k,l<\text{lh}(s)[(e(k)\in T_s\land e(l)\in T_s\land e(k)\supsetneq e(l))\Rightarrow t(k)<t(l)].

Note that S\in L[x] since T\in L[x]. For (s,t),(s',t')\in S we set (s,t)\leq_S(s',t') iff s\supseteq s' and t\supseteq t'. Then y\in A iff S_y is illfounded iff y\in p[S]QED

Theorem (Shoenfield absoluteness). Let x\in\mathbb R. Then every \Sigma^1_2(x) sentence is absolute between transitive models M\subseteq N of \textsf{ZF}+\textsf{DC} such that \omega_1^N,x\in M.

Proof. Let \varphi(v) be a \Sigma^1_2 formula, so that \varphi(v)\equiv\exists y\in\mathbb R\psi(v,y) for a \Pi^1_1-formula \psi(v,w). If M\models\varphi[x] then trivially N\models\varphi[x] by upwards absoluteness of \Pi_1-formulas, so assume instead that N\models\varphi[x]. Define

A:=\{y\in\mathbb R^N\mid N\models\psi[x,y]\}\in N,

so that A\neq\emptyset by assumption. As A is a \Pi^1_1(x) set in N, the above lemma implies that we get that A=p[T] for a tree T\in L[x]^N on \omega\times\omega_1^N, so that y\in A iff T_y is illfounded in N.

Pick some y\in A. Then T_y, and in particular also T, is illfounded in N. Since \omega_1^N,x\in M we get that T\in M as well, so that absoluteness of wellfoundedness yields some y'\in A\cap M\neq\emptyset, concluding M\models\varphi[x]QED

We see that the only thing we needed \omega_1^N,x\in M for was to ensure that the resulting tree was in indeed in M, to make sure that we could use absoluteness of wellfoundedness.