${\bigcup }_{i\in I}{A}_i:=\bigcup \{A_i:i\in I\}$, and ${\bigcap }_{i\in I}{A}_i:=\bigcap \{A_i:i\in I\}$ for $I\mathrm{\ne }\mathrm{\varnothing }$

Let $(A_i{)}_{i\in I}$ be an indexed family (An indexed family $(A_i{)}_{i\in I}$ is a function with domain $I$; $\{A_i:i\in I\}$ is its range). Its indexed union is

${\bigcup }_{i\in I}{A}_i:=\bigcup \{A_i:i\in I\},$

the union of its range (The union $\bigcup x$ of a set, and the binary union $a\cup b:=\bigcup \{a,b\}$); so $z\in {\bigcup }_{i\in I}{A}_i$ holds if and only if $z\in A_i$ for some $i\in I$.

When $I\mathrm{\ne }\mathrm{\varnothing }$ its indexed intersection is

${\bigcap }_{i\in I}{A}_i:=\bigcap \{A_i:i\in I\},$

which is legitimate because a family with $I\mathrm{\ne }\mathrm{\varnothing }$ has a value at some index, so its range is nonempty (Relation, $\mathrm{dom}R$, $\mathrm{ran}R$, $\mathrm{fld}R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$") and 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 applies (The intersection $\bigcap x$ of a nonempty set, the binary intersection $a\cap b:=\bigcap \{a,b\}$, and disjointness); so $z\in {\bigcap }_{i\in I}{A}_i$ holds if and only if $z\in A_i$ for every $i\in I$.

• The empty index set. ${\bigcup }_{i\in \mathrm{\varnothing }}{A}_i=\mathrm{\varnothing }$, because the empty family has empty range and $\bigcup \mathrm{\varnothing }=\mathrm{\varnothing }$ (There is exactly one set with no elements, written $\mathrm{\varnothing }$). The indexed intersection is left undefined there, for the reason recorded at 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: no set has every set as an element.

• Indexing changes nothing about the operations. Both notations are the primitive $\bigcup$ and $\bigcap$ applied to the range of the family, so every law proved for $\bigcup F$ and $\bigcap F$ transfers verbatim; that transfer is For a nonempty index set $I$: $B\cap {\bigcup }_{i\in I}{A}_i={\bigcup }_{i\in I}(B\cap A_i)$, $B\cup {\bigcap }_{i\in I}{A}_i={\bigcap }_{i\in I}(B\cup A_i)$, $X\setminus {\bigcup }_{i\in I}{A}_i={\bigcap }_{i\in I}(X\setminus A_i)$, and $X\setminus {\bigcap }_{i\in I}{A}_i={\bigcup }_{i\in I}(X\setminus A_i)$.