The intersection $\bigcap x$ of a nonempty set, the binary intersection $a \cap b := \bigcap\{a,b\}$, and disjointness

Let $x$ be a set with $x \neq \varnothing$. For a set $x \neq \varnothing$ the collection $\{\, z : \forall s\,(s \in x \to z \in s) \,\}$ is a set, and it does not depend on the member of $x$ used to separate it shows that the sets belonging to every member of $x$ form a set, and that it does not depend on the member of $x$ used to build it; it is written $\bigcap x$. Thus for $x \neq \varnothing$, $\bigcap x$ is the set whose elements are exactly the sets belonging to every element of $x$, and $a \cap b := \bigcap\{a,b\}$ is the binary intersection of $a$ and $b$:

$z \in \bigcap x \leftrightarrow \forall s\,(s \in x \to z \in s) \quad (x \neq \varnothing).$

The binary case is legitimate because $a \in \{a,b\}$, so the unordered pair of The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$ is never empty. Two sets $a$ and $b$ are disjoint when $a \cap b = \varnothing$ (There is exactly one set with no elements, written $\varnothing$).

• The hypothesis $x \neq \varnothing$ is not a convenience. For $x = \varnothing$ the condition "$z$ belongs to every member of $x$" is satisfied by every set whatever, and no set has every set as an element: There is no set $y$ with $x \in y \leftrightarrow \forall s\,(s \in \varnothing \to x \in s)$, so $\bigcap \varnothing$ is undefined proves this. So $\bigcap \varnothing$ is left undefined, and every statement below about $\bigcap$ of a family carries the nonemptiness hypothesis in its own statement.

• Membership criterion for the binary case. That $z \in a \cap b$ holds exactly when $z \in a$ and $z \in b$ is proved at $\bigcup \varnothing = \varnothing$, $\bigcup \{a\} = a$, $\bigcup \{a,b\} = a \cup b$, $\bigcap \{a\} = a$, and $\bigcap \{a,b\} = a \cap b$, not assumed here.