I'm currently doing a question from the Foundations of Mathematics written by Kenneth Kunen for self-teaching purposes, and I'm currently stuck on a question that asks me to justify the definition of A/R using the power set axiom and not the replacement axiom.
The steps I have done so far (and I'm not sure if I've done them correctly to be honest) are as follows:
We are given $ A/R \subseteq P(A) $
$ [x] $ is defined as $ [x] = \{ y \in A \wedge y R x\}$ and the Power set is defined as $ P(x) = \{x: x \subseteq A \} $
$\forall x \in A/R \rightarrow x \in P(A) $ as $A/R \subseteq P(A)$
$x$ is then a subset for which $\forall a \in x \rightarrow a \in A$ as $x \in P(A)$
$[x] \in A/R \rightarrow [x] \in P(A) $ by definition
Which leads to $\forall y \in [x] \rightarrow y \in A $ as $[x] \in P(A)$
From here I'm lost as to how to proceed. I think I can use restricted comprehension to prove that the set containing [x] exists since I have one condition fulfilled as $[x] \in P(A)$, but I don't know what the corresponding property would be. Any pointers or critics (regarding proof or structuring of question) would be greatly appreciated. Thanks!
I don't have Kunen's book to hand. I take it that $A/R$ is intended to represent the set of equivalence classes of the (equivalence) relation $R$ on a set $A$. Then you have:
$$ A/R = \{ C \in P(A) : \exists x \in A . \forall y . (y \in C \Leftrightarrow x \mathrel{R} y)\} $$
I.e., $A/R$ comprises all sets $C$ that, for some $x$, comprise precisely the elements $y$ that are equivalent under $R$ to $x$.
The power set axiom and the axiom of comprehension (which Kunen may be calling restricted comprehension) imply the existence of the set on the right-hand side of the above equation - the replacement axiom is not required.