Terminology confusion: Why is $z$ considered a free variable in the set $z^*=\{x \in z: \exists u (x \in u \land u \in z)\}$ In my book "The Foundations of Mathematics" by Kenneth Kunen, the following sentence is written:

In Comprehension [referencing the Comprehension Schema axiom], $\varphi$ can even have $z$ free - for example, it's legitimate to form $z^*=\{x \in z: \exists u (x \in u \land u \in z)\}$

I am slightly confused as to why $z$ is considered free in this set. The Comprehension Schema reads as follows:
Comprehension Schema: $\forall z \Big ( \exists y \forall x (x \in y \leftrightarrow x\in z \land \varphi(x) \big) \Big)$.
Swapping out $\varphi (x)$ for $\exists u (x \in u \land u \in z)$ we would create:
$\forall z \Big ( \exists y \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big ) \Big)$
The $z$ in the $\varphi$ formula is in the scope of the first universal quantifier at the beginning of this sentence. So why exactly is $z$ "free", as stated by Kunen? Does "free" mean different things depending on the context?
Thank you~

Edit: I figured I would put this here in case anyone else had a similar question.
The terms "bound" and "free" are always relative to a particular formula.
Consider the following:
Let $\psi_1 := \forall z \Big ( \exists y \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big ) \Big)$. In $\psi_1$, all variables are bound (i.e. no variable is free).
Let $\psi_2: = \exists y \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big )$. In $\psi_2$, variables $y,x,$ and $u$ are all bound but variable $z$ is free.
Let $\psi_3: = \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big )$. In $\psi_3$, variables $x$ and $u$ are bound, but variables $z$ and $y$ are free.
Let $\psi_4: = x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z)$. In $\psi_4$, variable $u$ is bound, but variable $z, y,$ and $x$ are free.
 A: Variable $z$ is free in formula $∃u(x∈u ∧ u∈z)$ because there is no quantifier binding it, while the two occurrences of variable $u$ are bound because they are in the scope of the quantifier $∃u$
The general form of the axiom schema is $A = \{ x \mid \varphi(x) \}$ where $x$ must be free in the formula $\varphi(x)$ specifying the condition that defines the set.
Thus, Kunen's formula amounts to:

$A_z = \{x \mid x∈z \land ∃u(x∈u ∧ u∈z) \}$.

The formula $\varphi(x)$ is $(x∈z \land ∃u(x∈u ∧ u∈z))$ with $x$ free.
The formula has an additional free variables (a parameter): $z$.
This means that we are defining a "families" of sets $A_z$: one for each value of $z$.

The set-forming operator $\{ x \mid \varphi(x) \}$ maps a formula ($\varphi$) into a term (a "name") denoting a set.
It binds the variable $x$. Thus, if there are no other free variables in the formula, it is a closed term.
The key-property of the operator is:

$\forall z [z \in \{ x \mid \varphi(x) \} \leftrightarrow \varphi (z)]$.


Having said that, how to apply it to Kunen's example ?
We have:

$∀z(∃y∀x(x∈y ↔ x∈z∧∃u(x∈u ∧ u∈z)))$.

We have a "specifying condition": $\psi(x,z) := x∈z ∧ ∃u(x∈u ∧ u∈z)$, with two free variables.
The axioms schema asserts that there is a set $y$ with all and only those elements $x$ that satisfy the condition:

$y = \{ x \mid  x∈z ∧ ∃u(x∈u ∧ u∈z) \}$.

Thus, applying the above formula:


$\forall x [x \in y \leftrightarrow (x∈z ∧ ∃u(x∈u ∧ u∈z))]$.


