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I’m stuck on the following lemma from Kunen’s Set Theory:

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More specifically, I don’t see why he concludes “Then $A$ also satisfies the Axiom of Extensionality.” Said axiom is

$$\forall x \forall y \left(\forall z (z\in x\leftrightarrow z\in y)\right)\rightarrow x=y$$

So the relativization to $B$ (which is true, by hypothesis), is

$$\forall x\in B \forall y\in B \left(\forall z \in B(z\in x\leftrightarrow z\in y)\right)\rightarrow x=y$$

And we need to show the relativization of the axiom to $A$, namely

$$\forall x\in A \forall y\in A \left(\forall z \in A(z\in x\leftrightarrow z\in y)\right)\rightarrow x=y$$

Fixing $x,y\in A$ such that $\forall z \in A (z\in x \leftrightarrow z\in y)$, we’d be done if we can see $\forall z\in B (z\in x \leftrightarrow z\in y)$, but I don’t see why that is the case. I’m guessing it has something to do with the fact that $A\preccurlyeq B$, but I’m not sure.

EDIT: Kunen defines $A\preccurlyeq B$ as follows:

Let $\mathfrak{A}$ and $\mathfrak{B}$ be structures for $\mathcal{L}$ with $\mathfrak{A}\subset \mathfrak{B}$. If $\phi$ is a formula of $\mathcal{L}$, then $\mathfrak{A}\preccurlyeq_\phi \mathfrak{B}$ means than $\mathfrak{A}\models \phi[\sigma]$ iff $\mathfrak{B}\models \phi[\sigma]$ for all assignments $\sigma$ for $\phi$ in $A$. $\mathfrak{A}\preccurlyeq \mathfrak{B}$ (elementary substructure or elementary submodel) means that $\mathfrak{A}\preccurlyeq_\phi \mathfrak{B}$ for all formulas $\phi$ of $\mathcal{L}$.

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    That comes from the fact that $A$ is an elementary submodel of $B$, doesn't it ? Or doesn't $\preceq$ denote elementarity ? – 2017-01-26
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    @Max added the definition of $\preccurlyeq$ given in the text. – 2017-01-26
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    Well then it is elementarity; that's what Eric Wofsey said in his answer – 2017-01-26

1 Answers 1

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This is immediate from the fact that $A\preceq B$: just let $\phi$ be the sentence $$\forall x \forall y \left(\forall z (z\in x\leftrightarrow z\in y)\right)\rightarrow x=y.$$ Since $B\models \phi$, $A\models\phi$ as well.

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    Just to clarify, the definition only cares about assignment of free variables in $A$? I think I may have been confusing relativizing a sentence with assignment of free variables. – 2017-01-26
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    There are no free variables in $\phi$, so there is only one "assignment" in this case (which does nothing). – 2017-01-26
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    Ok, I think I get it now. Thank you very much! – 2017-01-26