Proof of Unions and Intersections over a Countable Index Set in a σ-Algebra

Proof of (i). If $I=\emptyset$, then by Indexed Families of Sets and Their Union, Intersection and Product (ii) the class $\bigcup_{i\in I}A_{i}$ has no element, so it is the empty set by Elements Are Sets, and the Empty Set and the Pair of Two Sets Exist and Are Unique (ii), and it lies in $\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vi).

Let now $I\neq\emptyset$. By Countable Set (i), $I=\emptyset$ or there is a surjective map $g:\mathbb{N}\to I$; as $I\neq\emptyset$, such a map $g$ exists. The composite $A\circ g$ is a map $\mathbb{N}\to\mathcal{F}$ with value $A_{g(k)}$ at $k\in\mathbb{N}$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)); so $k\mapsto A_{g(k)}$ is a sequence in $\mathcal{F}$. Its union $\bigcup_{k\in\mathbb{N}}A_{g(k)}$ lies in $\mathcal{F}$ by σ-Algebra and Measurable Space (i). We show $\bigcup_{k\in\mathbb{N}}A_{g(k)}=\bigcup_{i\in I}A_{i}$. By Indexed Families of Sets and Their Union, Intersection and Product (ii), a set $u$ lies in the left side if and only if $u\in A_{g(k)}$ for some $k\in\mathbb{N}$, and in the right side if and only if $u\in A_{i}$ for some $i\in I$. If $u\in A_{g(k)}$ with $k\in\mathbb{N}$, then $i=g(k)$ is an element of $I$ with $u\in A_{i}$. If $u\in A_{i}$ with $i\in I$, then, $g$ being surjective, there is $k\in\mathbb{N}$ with $g(k)=i$ by Injective, Surjective and Bijective Functions between Classes (ii), and $u\in A_{g(k)}$. So the two classes have the same elements and are equal, and $\bigcup_{i\in I}A_{i}\in\mathcal{F}$.

Proof of (ii). Let $I\neq\emptyset$, and let $g:\mathbb{N}\to I$ be a surjective map, which exists by Countable Set (i) as in the proof of (i). As there, the composite $A\circ g$ is a map $\mathbb{N}\to\mathcal{F}$ with value $A_{g(k)}$ at $k\in\mathbb{N}$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)), so $k\mapsto A_{g(k)}$ is a sequence in $\mathcal{F}$, and its intersection $\bigcap_{k\in\mathbb{N}}A_{g(k)}$ lies in $\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (viii). We show $\bigcap_{k\in\mathbb{N}}A_{g(k)}=\bigcap_{i\in I}A_{i}$. By Indexed Families of Sets and Their Union, Intersection and Product (iii), a set $u$ lies in the left side if and only if $u\in A_{g(k)}$ for every $k\in\mathbb{N}$, and in the right side if and only if $u\in A_{i}$ for every $i\in I$. If $u\in A_{i}$ for every $i\in I$, then $u\in A_{g(k)}$ for every $k\in\mathbb{N}$, since $g(k)\in I$. Conversely, let $u\in A_{g(k)}$ for every $k\in\mathbb{N}$, and let $i\in I$; $g$ being surjective, there is $k\in\mathbb{N}$ with $g(k)=i$ by Injective, Surjective and Bijective Functions between Classes (ii), so $u\in A_{g(k)}=A_{i}$. So the two classes have the same elements and are equal, and $\bigcap_{i\in I}A_{i}\in\mathcal{F}$.