Intersection Paradox in ZFC? Following the axiom of separation, we define the intersection of a set x as:
$$\cap x= \{ y: \forall z (z \in x \Rightarrow y \in z) \}$$
But by the definition of material implication:
$$z \in x \Rightarrow y \in z \equiv z \notin x \lor y \in z$$
If we look at the first part of the OR statement:
$$z \notin x$$
which would be true if and only if z is not in x.
However, z is arbitrary, and spans over the entirety of the domain of discourse. Hence, any set not being a member of x would satisfy the condition of the intersection class, without regards to y.
Hence all objects in the domain of discourse satisfy y, which would obviously NOT be the Intersection Class.
Note: I am NOT referring to the intuitively meaningful members $\cap x$ which lie within x; these satisfy the condition meaningfully, but only those elements which exist within the Universe of Sets, but not in x.
Surely there must be a flaw in my reasoning, but I can't seem to find it. Could someone please clarify?
 A: The case $z \not\in x$ doesn't say anything about $y$ at all. Indeed, the only cases we care about are those for which $z \in x$; then we check whether $y \in z$. As Robert Shore wrote, the only case in which we never get to check the case we care about is if $x$ is empty; but then $\bigcap x$ is undefined (some authors define $\bigcap \emptyset = V$).
So, where did your reasoning go wrong? You're implying that if $z \not\in x$ then that would witness that $y \in \bigcap x$. But that's not true; given $x$, the formula $$ z \in x \Rightarrow y \in z $$ must hold for any $z$. Finding a singular $z$ for which the formula holds is not sufficient.
A: It is true that for any $y$, all $z$ that do not fall within $x$ satisfy the condition.  But the intersection of $x$ is defined as those $y$ for which every last $z$ satisfies the condition (however it happens to be expressed)—not just those that don't happen to fall within $x$.  Therefore the definition does not blow up.
To make this concrete, let $x = \{\{1, 2\}, \{2\}\}$.  We see that $2 \in \cap x$, because for all $z \in x, 2 \in z$.  However, no other element is in $\cap x$.  For example, $1 \not\in \cap x$, because there exists a $z \in x$ such that $1 \not\in z$—namely, $\{2\}$.  Equivalently (and this goes to the point of your question), there exists a $z$ such that neither $z \not\in x$ nor $1 \in z$—again, namely, $\{2\}$.
