Formula of $\bigcap X$ when $X$ is a class When $X$ is a set, I know we could define $\bigcap X$ as$Z$ such that $z \in Z \to \forall x \in X(z \in x)$ 
If $X$ is a class,  we may run into problem in the process of determining whether an element belongs to $X$. Thus, it seems to me it's not legitimate to write $\forall x \in X$. How could we overcome such difficulty?
ADDED: I want to show "$x$ is a natural number" is $\Sigma_0$ in Levy hierarchy(a lemma without proof on page 28, Constructibility, K.J.Devlin). So I have to define natural number in the first place, which is the intersection of all inductive sets, but all inductive sets constitute a proper class.
 A: This is not a real difficulty. If $X$ is a formula then there is some $\varphi(x,p)$ such that for some fixed parameter $p$ we have $X=\{x\mid\varphi(x,p)\}$.
Now what is the intersection of a class, any class even a set? It is the collection of all elements which belong to all members of the class. So we write:
$$\bigcap X=\{y\mid\forall x(\varphi(x,p)\rightarrow y\in x\}$$
"Every element which satisfies $\varphi(x,p)$ includes $y$" (note that this is still our fixed $p$ from before).
One caveat is that if $X=\varnothing$ then this is not well-defined, at all. Vacuously $\bigcap\varnothing$ includes every element of the universe, but some authors require the elements in $\bigcap X$ to be elements from $\bigcup X$, in which case $\bigcap\varnothing=\varnothing$.
So as long as you know that there is at least one inductive set, the class $X$ of all inductive sets is non-empty and we can talk about its intersection well.

As for your difficulty, think of $\forall x\in X:\psi(x)$ as a shorthand for $\forall x(\varphi(x,p)\rightarrow\psi(x))$.
A: This answer only deals with the added note in the OP.
An alternate way to show that "$x$ is a natural number" is expressible as a $\Sigma_0$-formula is to actually go through the work and do it.  The following will not depend on the Axiom of Infinity.  Note that $x$ is a natural number iff $x$ either $x = \emptyset$, or $x$ is a successor ordinal and all elements of $x$ are either $\emptyset$ or successor ordinals.  Now it is just a matter of going through the steps:


*

*assuming enough of ZFC, "$x$ is an ordinal" iff "$x$ is transitive and well-ordered by $\in$":

*

*"$x$ is a transitive set" may be expressed as $( \forall y \in x ) ( \forall z \in y ) ( z \in x )$;

*"all pairs of elements of $x$ are $\in$-comparable" may be expressed as $( \forall y \in x ) ( \forall z \in x ) ( y = z \vee y \in z \vee z \in y )$;

*note that if all pairs of elements of $x$ are $\in$-comparable, then by Foundation (you need a small bit more, but not much) it will follow that $\in$ is a transitive relation on $x$;

*by Foundation (essentially) $\in$ is an asymmetric relation on all sets;

*by Foundation $\in$ is well-founded;


*"$x$ is a successor ordinal" is expressed as "$x$ is an ordinal $\wedge ( \exists y \in x ) ( \forall z \in x ) ( z \in y \vee z = y )$";

*"$x = \emptyset$ is expressed as $( \forall z \in x ) ( z \neq z )$.


Note that all of the formulas above are $\Sigma_0$, and so boolean combinations and restricted quantifications of these are also $\Sigma_0$.  Now just put all these parts together to express "$x$ is a natural number" as a $\Sigma_0$ formula.
A: Note that writing something like $x\in X$ or $\forall x\in X$ is not in itself the major problem as element relations with a class (on the right side, of course) can be viewed as abbreviations for a more general predicate definig the class $X$.The problem is rather that what we try to define per
$$\bigcap X := \{x\mid \forall b\in X\colon x\in b\}$$
could be a proper class.
To be on the safe side, if $X$ is a nonempty class and $a\in X$, we might define
$$\bigcap X := \{x\in a\mid \forall b\in X\colon x\in b\}$$
and this is definitely a set. One must of course (readily) verify that this is well-defined, i.e. if also $a'\in X$ then 
$$\{x\in a'\mid \forall b\in X\colon x\in b\}= \{x\in a\mid \forall b\in X\colon x\in b\}.$$
