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In first-order logic with the Zermelo-Fraenkel axioms, it is convenient to introduce notations for sets that we prove exist and are unique. For example the union of two sets, ZF proves that: $$\forall a \forall b\; \exists! u \;\forall t, \; t\in u \Leftrightarrow (t \in a \lor t \in b) $$ So we note $u = a \cup b$. That means we introduce a binary operator symbol $\cup$ with an axiom derived from the theorem above. It is easy to show that any model of ZF can be extended to a model of ZF $+\cup$, by interpreting $\cup$ as a function that maps $(a,b)$ to the unique $u$ above.

However there is a glitch in the replacement axiom scheme. With the newly introduced symbol $\cup$, there are more formulas that can go into the replacement scheme, to produce more axioms. The previous reasoning didn't check that these new axioms are satisfied by the extended model.

If we drop the unicity and start with this other ZF theorem : $\forall a, \; a\neq \emptyset \Rightarrow \exists u, u \in a$, then introduce the associated symbol Choice$(a)$ with the following axiom, $$ \forall a, \; a\neq\emptyset \Rightarrow \text{Choice}(a) \in a $$ it is easy to derive the axiom of choice from that. Adding the symbol Choice and its axiom to ZF is consistent, but not conservative.

Is there a proof that ZF plus usual operations (empty set, union, intersection, powerset, pairs of sets, tuples, cartesian products, ...) is a conservative extension of ZF with only the membership symbol $\in$ ?

V. Semeria
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  • I don't think you can derive the axiom of choice from just adding the symbol: you'd have to change the axioms as well (for instance the separation schema, or the replacement schema should include instances where the sentences involve said function) – Maxime Ramzi May 01 '18 at 15:17
  • But making that change makes it indeed nonconservative as you noticed (because you can make that change for $\emptyset, \cup,...$ without altering the theorems): the essential difference is then that these other "operators" name a unique element of the universe; i.e. there is a unique interpretation of them. I assume there is somewhere in the literature some proof that this does produce a conservative extension – Maxime Ramzi May 01 '18 at 15:20

1 Answers1

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This is straightforward: any formula can just be replaced by one without any of the new symbols. For instance, given a formula in the enlarged language with $\cup$, you can get an equivalent formula by making the following substitutions:

  • Every time $s\cup t=u$ or $u=s\cup t$ appears (for terms $s,t,$ and $u$), replace it with $\forall x (x\in u\leftrightarrow(x\in s\vee x\in t))$.
  • Every time $s\in t\cup u$ appears, replace it with $s\in t\vee s\in u$.
  • Every time $s\cup t\in u$ appears, replace it with $\exists x(x=s\cup t\wedge x\in u)$.

Here $s,t,$ and $u$ are terms, and $x$ is a variable not appearing in $s,t,$ or $u$. Note that these substitutions may need to be iterated to eliminate all uses of $\cup$ (for instance, the third rule introduces $x=s\cup t$ which then needs to be eliminated using the first rule, and if a nested term like $(x\cup y)\cup z$ appears in the formula the rules will be used first to remove the outer $\cup$ and then to remove the inner $\cup$). A straightforward induction on formulas shows that each formula in the enlarged language with $\cup$ is equivalent to the replacement formula.

So, in particular, if $\varphi$ is any formula in the enlarged language with $\cup$, the instance of Replacement using $\varphi$ is equivalent to instance of Replacement using the formula $\varphi'$ obtained by removing $\cup$ from $\varphi$ as above. Since Replacement for $\varphi'$ is included in ZF, this means that the new Replacement axiom for $\varphi$ is true in the model with $\cup$ extended from any model of ZF.

(Another way to say this is that Replacement is really a statement about all definable functions on any model. Adding new symbols to the language for functions that were already definable will not make any new functions definable.)

Eric Wofsey
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