If $A$ is a set then $A/R$ is a set.

Question

I just started reading some notes on set theory, and I am still a confused as how to use the different axioms to determine that something is a set.

For example, I wan't to say that given a set A and an equivalence relation $R$, $A/R$ is a set. So I start by saying that from the comprehension axiom (if $z$ is a set and $\phi$ a formula, then $\{x\in z \mid \phi \}$ is a set) the equivalence classes is a set because $[x]=\{y\in A \mid y\}$. Then using that by the power set axiom $\mathcal{P}(A)$ is a set and again by comprehension I have that $\{[x]\in \mathcal{P}(A) \mid x\in A\}$ is a set. Is this the correct reasoning?

Answer 1

That looks good. Except a typo. I imagine you meant

$$[x]=\{y\in A \mid (x,y)\in R\}$$

Answer 2

First comprehension tells us that for each $x \in A$

$$[x]:=\{y \in A \mid (x,y) \in R\}$$ is a well defined set.

Then $$A{/}R = \{B \in \mathscr{P}(A): \exists x \in A: B=[x]\}$$

is well-defined with the power set axiom and comprehension again.

Your way of writing $$A{/}R = \{[x]: x \in A\}$$

appeals to the axiom of replacement instead (using that $x \to [x]$ is a "set function", applying it to the "domain" $A$). It depends on what you want, really. In ZF both are OK.