# Is Global Choice conservative over Zermelo with Choice?

To be explicit, by Zermelo set theory with Choice, ZC, I mean the theory with the same language and axioms as ZFC except not Foundation (also called Regularity) and with the axiom scheme of Separation instead of Replacement. By Global Choice I mean adding a new function symbol $$F$$ and an axiom:$$\forall v[v\neq\emptyset \rightarrow F(v)\in v]$$, and extending the Separation scheme to include formulas using $$F$$.

It is known that the axiom of Global Choice gives a conservative extension of Zermelo Frankel set theory with Choice (ZFC). The proof I know is by Haim Gaifman in his "Local and Global Choice Functions" (Israel J. of Math v. 22 nos. 3-4, 1975, pp. 257--265. And there is one I have not worked through which uses forcing by Ulrich Felgner "Comparison of the axioms of local and universal choice" Fund. Math. 71, 1971, pp. 43-62. Both seem to require the idea of the rank of a set. Maybe one can be adapted to work for ZC, but I do not see it.

Or is there some other proof? Or a disproof?

• Do you mean ZF+Global Choice, since you are referring to Replacement at the end of your first paragraph. Or do you mean Z+Global Choice, and Separation can refer to the choice function $F$? Mar 19, 2019 at 6:57
• @DavidRoberts Sorry, that "replacement" was a typo for "separation." Corrected. Mar 19, 2019 at 11:26
• thanks, that makes more sense. Mar 19, 2019 at 11:34
• The proof by Felgner is not hard: take a model of ZFC and define a (proper class) forcing consisting of 'all partial well-orderings of the universe'. This forcing will add no new sets, but (by genericity) G will be a global well-ordering. Furthermore, Replacement with respect to G holds due to the Forcing Theorem (which holds for this particular class forcing) Mar 20, 2019 at 15:19
• @JohannesSchürz Yes, I believe Gaifman's proof is essentially the same. He just uses the fact that ZFC proves existence of enough partial well-orderings of the universe, that you do not really need forcing. Mar 20, 2019 at 15:52

The known proofs of conservativity of ZF + GC (Global Choice) over ZFC make significant use of replacement, and as far as I know the problem of conservativity of Z + GC over ZC is wide open.

Let me add that I have discussed the problem with a number of experts over the past two decades, and also posed it on FOM in 2006 in this posting.

• This persuades me but I will leave the question open a little longer to see if it produces any surprises. Mar 20, 2019 at 15:53

Global Choice is not conservative over ZC. We'll build a model of ZC which satisfies a sentence disprovable by Global Choice. Warning: it is hideous, and I've been struggling to come up with a clean way to present it.

We work in ZF + GCH below $$\aleph_{\omega}$$ + existence of countable sets $$(X_n)_{n<\omega}$$ such that there is a surjection $$f: \mathcal{P}(\bigcup_{n<\omega} X_n) \rightarrow \omega_2.$$ This holds in a symmetric extension of $$L$$, see Asaf Karagila's Iterated failures of choice, section 4.

The model we construct will satisfy ZC + GCH and that every infinite set has cardinality some $$\aleph_n,$$ yet will code $$(X_n)_{n<\omega}$$ and $$f$$ by definable classes. Note that we are identifying $$\aleph_n$$ with the canonical prewellordering of $$\mathcal{P}^n(\omega)$$ of length $$\omega_n$$ since the von Neumann construction of the $$\aleph$$'s doesn't work in ZC.

Let $$X = \bigcup_{n<\omega} X_n.$$ There will be Quine atoms $$a_x$$ for each $$x \in X$$ and $$a_Y$$ for each $$Y \subset X.$$

Let $$B = \{V_{\omega}\} \cup \{\{a_x: x \in X_n\}: n<\omega\}\cup \{\{a_Y\}: Y \subset X\}.$$

Consider this model:

$$M_1=\bigcup_{n<\omega}\bigcup_{S \in [B]^{<\omega}} \mathcal{P}^n(\bigcup S).$$

Then $$M_1 \models ZC + GCH + \forall S \exists n (|S| \le \aleph_n).$$

Let $$P = \{(n, a_x): x \in X_n\} \cup \{(a_x,a_Y): x \in Y \subset X\} \cup \{(a_Y, f(Y)): Y \subset X\}.$$ We will build an extension of $$M_1$$ in which $$P$$ is a definable predicate. For each $$p \in P,$$ let $$b_p = \{p, b_p\}.$$

Let $$C = B \cup \{\{b_p\}: p \in P\}.$$

Let $$M = \bigcup_{n<\omega}\bigcup_{S \in [C]^{<\omega}} \mathcal{P}^n(\bigcup S).$$

Then $$M \models ZC + \varphi,$$ where $$\varphi$$ is the conjunction CH $$\wedge$$ $$$$the class $$\{p: \exists b (b=\{b, p\})\}$$ codes a countable sequence of countable sets of atoms $$X_n,$$ a class of atoms which each relate to a different subclass of $$\bigcup_{n<\omega} X_n,$$ and a surjection from the latter class onto $$\aleph_2."$$

Finally, we see that Global Choice proves $$\neg \varphi,$$ since from a global choice function, we can choose enumerations of each $$X_n,$$ enumerate $$\bigcup_{n<\omega} X_n,$$ and thus define a surjection from $$\mathcal{P}(\omega)$$ onto $$\omega_2,$$ violating CH.

Also note that we can adjust the construction of $$M$$ so that it satisfies Foundation by several applications of the trick used here: Is $\in$-induction provable in first order Zermelo set theory?

• Is $c$ in the definition of $B$ a typo of $X$, or does it mean another set? Nov 28, 2021 at 0:39
• Also, the definition of $b_p$ baffled me. Do you assume some kind of anti-foundation axioms to construct the model? (like Aczel's anti-foundation axiom, albeit that would be overkill.) Nov 28, 2021 at 0:53
• You’re right about the typo. I’m not assuming ill-foundedness in $V,$ I’m defining $b_p$ in terms of how they’re intended to be interpreted in $M.$ A formal definition would be by a quotient structure, same as how Quine atoms are included in the model. Nov 28, 2021 at 1:21
• So, do you construct a well-founded model whose appropriate (and maybe definable) quotient is ill-founded under the manner you claimed? Nov 28, 2021 at 1:33
• Yes, first build a well-founded model of Zermelo with Urelements, where the $a$'s and $b$'s are the urelements, and then quotient out the intended relationship in the resulting model Nov 28, 2021 at 2:07