Question about the definition of intersection in Zermelo–Fraenkel set theory I was following the "Lectures on the Geometric Anatomy of Theoretical Physics" of Frederick Schuler and at some point he gives as a homework to define intersection $\cap x$, where $x$ is a set. At that point in the lecture he didn't finish stating all the axioms yet. Now, my first instinct was to write
$$\bigcap x =: \{y \in x \mid \forall w \in x : y \in w\},$$
which I also later found in many places when I searched on the internet... (for example on wiki: https://en.wikipedia.org/wiki/Intersection_(set_theory)#Definition)
But then we got to the axiom of foundation, which implies that we cannot write $x \in x$. So, wouldn't the correct definition for intersection be
$$\bigcap x =: \{y \in x \mid \forall w \in x : y \subseteq w\},$$
since we want for example: $\bigcap\{y,\{y,z\}\} = y$, which has a problem with the first definition of intersection (as $y\in \cap\{y,\{y,z\}\}$ implies that $y\in\{y,z\}$, but also $y\in y$)?
 A: The correct definition of intersection of an arbitrary set $x$ in a theory that doesn't allow for the existence of the universal set (like $\sf ZFC$) is:
$\bigcap x = \{y \mid \exists z \, (z \in x) \land  \forall w \, (w \in x \to y \in w)\}$
or equivalently (in $\sf ZFC$):
$\bigcap x = \{y \in \bigcup x \mid  \forall w \, (w \in x \to y \in w)\}$
If the universe is allowed (like in $\sf NF(U)$), then it is simply:
$\bigcap x = \{y \mid  \forall w \, (w \in x \to y \in w)\}$
Foundation has nothing to do with allowing us to write $x \in x$, this is a well formed formula of first order logic with equality, and it can be written whether foundation holds or not. Foundation can tell us that $x \in x$ is false, but it doesn't prevent us writting it.
In $\sf ZFC$ it is a theorem that: $$\bigcap \{y, \{y,z\}\} =y \iff y=\{z\} \lor y=\varnothing $$; and also   $x \in  \{\{x\}\}$ is always false in $\sf ZFC$.
A: In $\mathsf{ZFC}$ (in fact just because of Axiom of Foundation), we have $\forall x(x\notin x)$ as you said.
First, the definition of generalized intersection which you give is clearly circular and hence false. Both of the definitions for intersection $\cap$ and generalized intersection $\bigcap$ are underlying Axiom of Comprehension. We can define intersection first and generalized intersection then.
Axiom of Comprehension. Suppose $\varphi(y)$ is a first order formula of the set theory language $\mathscr{L}_\in$, without $x$ free. Then for any set $z$ there is a set $x$ such that for all $y\in z$ we have $y\in x$ if and only if $\varphi(y)$ holds, i.e., $x=\{y\in z\mid \varphi(y)\}$.
Definition 1. Suppose $x,y$ are sets. Set
$$\begin{array}{rll}
x\cap y&=:&\{z\in x\mid z\in y\}\\
&=&\{z\mid z\in x\wedge z\in y\}.
\end{array}$$
Definition 2. Suppose $x\neq\varnothing$, and pick one $z\in x$. Set
$$\begin{array}{rll}
\displaystyle\bigcap x&=:&\{y\in z\mid \forall w(w\in x\to y\in w)\}\\
&=&\{y\mid \forall w(w\in x\to y\in w)\}.
\end{array}$$
Then we find them to be consistent:
Fact 3. For any sets $x,y$, we have $x\cap y=\bigcap\{x,y\}$.
Further more, we could also define the generalized intersection first and intersection then by the generalized intersection: $x\cap y=:\bigcap\{x,y\}$.
There are two reasons why we set $x\neq\varnothing$ in Definition 2: (1) since then we can pick one $z\in x$ to make the definition for generalized intersection be well-defined by Axiom of Comprehension; (2) even if you define $\bigcap\varnothing$ by the second line instead of the first line, then $\bigcap\varnothing$ would be the set $V$ of all sets (or equivalently $V=\{x\mid x=x\}$)，and this means that $V$ is a set. But $V$ can't be a set since otherwise we would have $V\in V$ by the definition of $V$.
At last, suppose $\bigcap \{y, \{y,z\}\} =y$, then we would have
$$
\begin{array}{rcll}
\displaystyle\bigcap \{y, \{y,z\}\} =y &\Longleftrightarrow&  y\cap \{y,z\}=y&\\
&\Longleftrightarrow& y\subseteq \{y,z\}&\\
&\Longleftrightarrow& \forall x(x\in y\leftrightarrow x=y\vee x=z)&\\
&\Longleftrightarrow& \forall x(x\in y\leftrightarrow x=z)&\text{by}~\forall x(x\notin x)\\
&\Longleftrightarrow& y=\varnothing~\text{or}~y=\{z\}.&
\end{array}
$$
So if you set $\bigcap \{y, \{y,z\}\} =y$, it wouldn't bring about any contradiction.
