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Here is the proof:

Let $A$ be a set, and $\mathcal{P}$ be a property not mentioning the set $B$. Consider the formula $\phi(x, y)$ that means $y=x$ and $\mathcal{P}(x)$ is true. Clearly $(\forall x \in A) \forall y_1 \forall y_2$, $\phi(x, y_1) \wedge\phi(x, y_2)$ implies $y_1=x=y_2$. Therefore, by the replacement axiom schema on the formula $\phi(x, y)$, there exists a set $B$ such that $\forall y$, $y\in B \iff \exists x\in A : \phi(x, y) \iff \exists x\in A : y=x \wedge \mathcal{P}(x) \text{ is true}$.

The part that most worries me about this proof is wether I have defined the formula $\phi(x, y)$ correctly, because I have only assigned it meaning for the $x$'s such that $\mathcal{P}(x)$ is true.

Asaf Karagila
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roxingby
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  • Maybe have a look here: https://math.stackexchange.com/questions/32483/how-do-the-separation-axioms-follow-from-the-replacement-axioms – D.Q. Nov 05 '21 at 17:55

1 Answers1

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If I understand the issue correctly, it depends a bit on the underlying formulation of replcament you are working with. If all you need is that $\phi(x,y)$ is a functional formula, then your proof suffices (this is the version of replcament that you can find e.g. in Jech's Set Theory). In this case you can let $\phi(x,x):=P(x)$, which defines the class function $F=\{(x,x): P(x) \}$. Then since $A$ is a set we have that $F(A)=\{x\in A : P(x) \}$ is also set.

Differently, if your formulation of replacement requires $F$ to be total over $A$ we let $\psi(x,y):= (P(x)\to y=\{ x\} ) \land(\neg P(x)\to y=\emptyset) $. Then $\psi$ defines a total class function $G$, which is in turn total over $A$. By replacement we then get again that $G(A)= \{ \emptyset \}\cup\{ \{x\}: x\in A \land P(x) \}$ or $G(A)= \{ \{x\}: x\in A \land P(x) \}$. In both cases, $\bigcup G(A)=\{ x\in A : P(x) \}$, which is what we wanted.

D.Q.
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