$\bigcup_{i \in I} A_i := \bigcup \{A_i : i \in I\}$, and $\bigcap_{i \in I} A_i := \bigcap \{A_i : i \in I\}$ for $I \neq \varnothing$

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 $I \neq \varnothing$ its indexed intersection is

$\bigcap_{i \in I} A_i \;:=\; \bigcap \{A_i : i \in I\},$

which is legitimate because a family with $I \neq \varnothing$ has a value at some index, so its range is nonempty (Relation, $\operatorname{dom} R$, $\operatorname{ran} R$, $\operatorname{fld} R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$") and 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 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. $\bigcup_{i \in \varnothing} A_i = \varnothing$, because the empty family has empty range and $\bigcup \varnothing = \varnothing$ (There is exactly one set with no elements, written $\varnothing$). The indexed intersection is left undefined there, for the reason recorded at There is no set $y$ with $x \in y \leftrightarrow \forall s\,(s \in \varnothing \to x \in s)$, so $\bigcap \varnothing$ 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 $X \setminus \bigcap_{i \in I} A_i = \bigcup_{i \in I} (X \setminus A_i)$.