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 $x\mathrm{\ne }\mathrm{\varnothing }$. For a set $x\mathrm{\ne }\mathrm{\varnothing }$ the collection $\{z:\mathrm{\forall }s(s\in x\to z\in s)\}$ 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 $x\mathrm{\ne }\mathrm{\varnothing }$, $\bigcap x$ 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 $b$:

$z\in \bigcap x\leftrightarrow \mathrm{\forall }s(s\in x\to z\in s)(x\mathrm{\ne }\mathrm{\varnothing })\mathrm{.}$

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 $a\cap b=\mathrm{\varnothing }$ (There is exactly one set with no elements, written $\mathrm{\varnothing }$).

• The hypothesis $x\mathrm{\ne }\mathrm{\varnothing }$ is not a convenience. For $x=\mathrm{\varnothing }$ 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 $x\in y\leftrightarrow \mathrm{\forall }s(s\in \mathrm{\varnothing }\to x\in s)$, so $\bigcap \mathrm{\varnothing }$ is undefined proves this. So $\bigcap \mathrm{\varnothing }$ 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 $\bigcup \mathrm{\varnothing }=\mathrm{\varnothing }$, $\bigcup \{a\}=a$, $\bigcup \{a,b\}=a\cup b$, $\bigcap \{a\}=a$, and $\bigcap \{a,b\}=a\cap b$, not assumed here.