Show that $\{A_{\alpha}: \alpha\in I\}$ is a set using the axiom of replacement in ZFC I'm reading Enderton's Elements of Set Theory.
Let $I$ be a set, and for every $\alpha\in I$, let $A_\alpha$ be a set. In order to apply the axiom schema of replacement to construct the set $\{A_{\alpha}: \alpha\in I\}$, we need a formula $φ(x,y)$（a formula is defined to be a string of symbles constructed from the atomic formulas $x\in y$ and $x=y$ by repeated applications of logical operators $\vee$, $\wedge$, etc). I know that we generally don't write out the formula itself (for the sake of convenience), but the fact of the matter is that we should be able to "translate" what we write into a legal formula.
For instance, if one wished to prove the existence of the collection $\{\mathscr{P}(x):x\in A\}$ of all power sets of members of $A$, one could use the "illegal" statement $P(x,y)：y=\mathscr{P}(x)$, which can be rewritten as a legal formula $φ(x,y):\forall t(t\in y\leftrightarrow (\forall u(u\in t \rightarrow u\in x)))$.
When constructing the set $\{A_{\alpha}: \alpha\in I\}$, the "illegal" statement we use is $P(\alpha,y): y=A_\alpha$. But I have no idea how I shoud rewrite this statement as a legal formula.
We can rewrite it first as $\forall t(t\in y\leftrightarrow t\in A_\alpha)$. But how can I rewrite "$t\in A_\alpha$" as a formula?
Here is an example of how $\alpha\mapsto A_\alpha$ might be defined:$I=\{1,2,3\}, A_1:=\{2,3\},A_2:=\{3,4\},A_3:=\{4,5\}$.
Sorry for the language barrier.
 A: The point you're missing here is what does the statement "Let $I$ be a set and  for every $\alpha\in I$ let $A_\alpha$ be a set". There are two ways to read this statement.

*

*The statement is saying that there is a function $A$ with domain $I$, such that $A(\alpha)$, denoted by $A_\alpha$ is also a set.
In this case, $A$, being a function in the universe, is already a set, so it is enough to prove that if $F$ is a function $\operatorname{rng}(F)$ is a set. The way to prove this depends on the specific encoding of "function", but generally this will involve some kind of Separation axiom and possibly some unions.


*The statement is saying that there is a definable property $\varphi(x,y)$ (possibly with parameters, which I will omit) which defines the function $A$ on the set $I$. In that case, either apply Replacement directly, or first prove that if $I$ is a set and $\varphi$ defines a function on $I$, then there is a set which equals that function, and then fallback to the previous case. (Note that the latter option will require you to first do the former, so in a sense it is full of redundancies here.)
