No, that's not true at all.

Let's start with the most basic of meta-theories: $\sf PRA$, the theory of Primitive Recursive Arithmetic. Here we have the natural numbers, and the primitive recursive functions.

This theory is strong enough to internalise FOL, so we can just assume that we're manipulating strings. The point of an *axiom schema* is that it is a predicate that lets us recognise all the axioms which have a certain form. And any reasonable schema, including those of $\sf ZFC$, are in fact primitive recursive. In other words, there is a primitive recursive function $f_{\rm Sep}(n)$ which takes in $n$, and if $n$ is the Gödel number of a formula with some basic properties $\varphi$, then $f_{\rm Sep}(n)$ is the Gödel number of the axiom obtained from the schema by putting $\varphi$ into it. If $n$ is not the Gödel number of a suitable formula, just return the axiom for the formula $\varphi$ given by $x=x$ or something like that.

Since $\sf ZFC$ is presented as finitely many axioms and one or two schemata (Separation and Replacement, but Replacement is generally enough to prove Separation, making it redundant), then the collection of Gödel numbers of axioms of $\sf ZFC$ is in fact primitive recursive. So we can *really* talk about the first-order theory that is $\sf ZFC$.

To recap, the point of the schemata is to allow us to have infinitely many axioms with the same structure—that we can recognise mechanically—so that when we need to use any such axiom in a proof, we can always be sure that it is part of this or not.

Another way to approach this is by saying that our foundation is in fact $\sf ZFC$. We use set theory to discuss set theory. This sounds circular, but how is this any different than using $\sf PRA$ to study the logical consequences of $\sf PRA$? It's not. Mathematics is not something that we do in vacuum, some assumptions are needed. And it is perfectly fine to study $\sf ZFC$ inside $\sf ZFC$.

There, we have the notions of sets already existing, so we can talk about a set of axioms. Of course, we need to argue why a certain set exists, which is to say that we need to be able to prove that the set of these axioms actually exists. And again we use the fact that a schema is a function which takes formulas and returns axioms, so we may replace the schema with the axioms, as it is the range of this function.

So again, we have that $\sf ZFC$ is a set of first-order axioms in the language of set theory.

Not in any sense. $T'$ is in fact a conservative extension of $\mathsf{ZF}$-$\mathsf{Inf}$; this is a consequence of the compactness theorem.

Now any *well-founded* model (or even $\omega$-model) of $T'$ restricts to a model of $\mathsf{ZF}$, but in the absence of such a semantic restriction $T'$ is no better than $\mathsf{ZF}$.

## Best Answer

I was pretty sure this was a duplicate, but right now I can't find this exact question having been asked earlier, so here goes:Yes, you can do this - the key is the axiom scheme of

replacement(orcollectiondepending on how you've seen $\mathsf{ZF}$ presented).Consider the following (English shorthand for a) formula $\varphi(x,y,z)$:

Then "$\varphi(x,y,z)$" should be interpreted as "$x=\mathcal{P}^y(z)$." A quick argument shows that we can apply replacement (and infinity) to prove in $\mathsf{ZF}$ the following:

Applying the union axiom to $C_z$ then gives the desired set.

There are two key things worth noting here:

The use of the "coding sequence" $b$ exactly parallels a similar technique in the context of arithmetic for talking about computations.

By taking unions at limits there's no difficulty in extending this to arbitrary ordinals instead of just natural numbers, where $y$ is concerned - in $\mathsf{ZF}$ we can in fact make sense of $\mathcal{P}^\alpha(S)$ and $\bigcup_{\beta<\alpha}\mathcal{P}^\beta(S)$ for any set $S$ and ordinal $\alpha$.