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.

• @MauroALLEGRANZA but isn't $z$ quantified by the $\forall z$ quantifier in the comprehension schema?
– S.C.
Oct 30, 2020 at 13:15
• Now I'm even more confused hah. How is $x$ a free variable when it is within the scope of the $\forall x$ quantifier?
– S.C.
Oct 30, 2020 at 13:18

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:

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

• Could you please clarify if the following is correct? Consider two formulas. $\psi_1 = \forall x \varphi (x)$ and $\psi_2 = \varphi(x)$ (where the $\varphi$ in $\psi_2$ refers to the same $\varphi$ in $\psi_1$). When I am speaking about $\psi_1$, $x$ is not free. However, when I speak about $\psi_2$, even though I am still referring to the same $\varphi$, I would say that that x in $\varphi(x)$ is free. Is that right?
– S.C.
Oct 30, 2020 at 13:29
• @S.Cramer - correct. See my first line in the answer above. Oct 30, 2020 at 13:35
• Great. Thank you very much. I added something to my post (Edit section) that I would love for you to comment on. Is my edit correct? (i.e. that the terms "free" and "bound" are always relative to a particular formula, even if that particular formula is "embedded" within another 'larger' formula)
– S.C.
Oct 30, 2020 at 13:51
• @S.Cramer - to be precise, free and bound are always relative to a particular occurrence in a particular formula. In $\forall x (z \in x \land \exists z (z=z))$ we have that the 1st (from left) occurrence of $z$ in the complete formula is free while the remaining are bound. Oct 30, 2020 at 13:58
• And yes, in the sub-formula ∃z(z=z) (of the complete formula) the occurrences of $z$ are bound. Oct 30, 2020 at 13:58