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I've been pondering this old question Is the Subset Axiom Schema in ZF necessary? and seem to have an answer which didn't come up, but I'm wondering if there's a mistake in my reasoning.

Axiom of specification: given any set $x$ and formula $\phi(y)$ there exists a set $z$ whose members are exactly those members of $x$ which satisfy $\phi$: $$\forall \phi \forall x \exists z \forall y(y \in z \leftrightarrow y \in x \land \phi(y)).$$

So let $x$ be a set and $z \subseteq x$ a subcollection. Define the formula $\phi(y) = (y \in z)$. Then by the axiom of specification there exists a set $z'$ such that $\forall y(y \in z' \leftrightarrow y \in z)$. I thought about applying the axiom of extensionality here to deduce that $z = z'$, but obviously you can't do that because $z$ is not necessarily a set. Still, to every subcollection I've associated a set with exactly the same members. One of the answers https://math.stackexchange.com/q/2383 says it's possible there may be subsets which aren't expressible by a formula, but if I know that $z$ is a set then I seem to have a formula. Did I make a mistake? Maybe my $\phi$ isn't a proper formula because it refers to something which may not be a set?

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George Skelton
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    Since $z$ isn't originally confirmed to be a set, using it in a formula makes it an invalid formula. Also, the $\in$ operation only works on sets, so what does $y\in z$ even mean if $z$ isn't a set? – Arthur Dec 18 '15 at 19:57
  • Ah ok I thought $\epsilon$ and $\subseteq$ worked for non-sets and ZF was just about restricting the universe of possible sets. But I take your point that $\phi(y)$ isn't a valid formula hence can't apply specification. Still, do we really think there may be subcollections of a set which aren't sets? – George Skelton Dec 18 '15 at 20:08

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The "formula" "$y\in z$" . . . isn't really a formula. It's not expressible, unless $z$ is definable from parameters in the model somehow. So in general the axiom of separation won't apply.

For example, take (by the Lowenheim-Skolem theorem) a model $M$ of ZFC containing $\omega$ but which is countable. Then most subsets of $\omega$ - that is, most reals - won't be in $M$, but will of course be sub-collections of an element of $M$ (namely $\omega$). Do you see why this isn't a contradiction?

Noah Schweber
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  • I do ... except, doesn't the power set of $\omega$ have to belong to $M$? – George Skelton Dec 18 '15 at 20:34
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    @GeorgeSkelton Not the way you want it to. $M$ has to contain an element it thinks is the powerset of $\omega$. If $M$ is an elementary submodel of the "true" universe, then yes, $M$ contains the powerset of $\omega$ as an element, but $M$ won't be transitive - just because $a\in M$ we won't have $b\in M$ for $b\in a$. Meanwhile, if $M$ is transitive, then the thing $M$ thinks is the powerset of $\omega$ actually won't be. – Noah Schweber Dec 18 '15 at 20:46
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The subset schema (or replacement or collection schemata) allow parameters.

So the formula $y\in z$, while can be used to define a collection, can only appeal to $z$ that already exists in the model. Namely, if $z$ is already a set. Also note that this is a very semantic way of looking at things. We work in a fixed model. Existence of things is already determined.

So we ask whether or not a set - which already exists - can be defined with parameters. And yes, it can, because it can be a parameter as well.

Of course it is possible that there are models of set theory which do not contain every set. So it is possible that $x$ is a set in such model, and $y$ is a subset of $x$ in the universe, but $y$ is not in the model. In that case, however, you cannot use $y$ as a parameter to define a subset of $x$ inside of your model.

Asaf Karagila
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