Proof of Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra

In the proofs of (vi), (vii), (viii), (ix), (x) and (xi), $\mathcal{F}$ is the $\sigma$-algebra on $X$ fixed before these clauses. By σ-Algebra and Measurable Space (i), $\mathcal{F}\subseteq\mathcal{P}(X)$, $X\in\mathcal{F}$, $X\setminus A\in\mathcal{F}$ for every $A\in\mathcal{F}$, and $\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{F}$ for every sequence $(A_{n})$ in $\mathcal{F}$. By The Union Set and the Power Set of a Set (ii), every element of $\mathcal{F}$ is a subset of $X$. By Indexed Families of Sets and Their Union, Intersection and Product (ii), a set $u$ lies in the union $\bigcup_{n\in\mathbb{N}}A_{n}$ of a sequence $(A_{n})$ if and only if $u\in A_{n}$ for some $n\in\mathbb{N}$.

Proof of (vi). As $X\in\mathcal{F}$, also $X\setminus X\in\mathcal{F}$. The set $X\setminus X$ has no element, so it is $\emptyset$ by Elements Are Sets, and the Empty Set and the Pair of Two Sets Exist and Are Unique (ii).

Proof of (vii). Let $A,B\in\mathcal{F}$. By Maps Defined by Cases (i) there is a map $C:\mathbb{N}\to\mathcal{F}$ with $C_{n}=A$ for $n=1$ and $C_{n}=B$ for $n\neq1$; it is a sequence in $\mathcal{F}$. Every term $C_{n}$ is $A$ or $B$, so $\bigcup_{n\in\mathbb{N}}C_{n}\subseteq A\cup B$. Conversely, $1\in\mathbb{N}$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (ii), and $m=1+1$ satisfies $m\in\mathbb{N}$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (vi) and $m\neq1$ by Arithmetic and Order of the Natural Numbers (xi); as $C_{1}=A$ and $C_{m}=B$, we get $A\cup B\subseteq\bigcup_{n\in\mathbb{N}}C_{n}$. Hence $A\cup B=\bigcup_{n\in\mathbb{N}}C_{n}\in\mathcal{F}$.

Proof of (viii). Let $(A_{n})$ be a sequence in $\mathcal{F}$. Since $X\setminus A_{n}\in\mathcal{F}$ for every $n$, the map $n\mapsto X\setminus A_{n}$ is a sequence $(D_{n})$ in $\mathcal{F}$ (a sequence in $\mathcal{P}(X)$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (i), $(A_{n})$ being a sequence in $\mathcal{P}(X)$ as $\mathcal{F}\subseteq\mathcal{P}(X)$, and hence a sequence in $\mathcal{F}$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v)), so $U=\bigcup_{n\in\mathbb{N}}D_{n}\in\mathcal{F}$ and $X\setminus U\in\mathcal{F}$. As $1\in\mathbb{N}$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (ii), the intersection $\bigcap_{n\in\mathbb{N}}A_{n}$ is formed as in Indexed Families of Sets and Their Union, Intersection and Product (iii), and a set $u$ lies in it if and only if $u\in A_{n}$ for every $n\in\mathbb{N}$. We show $\bigcap_{n\in\mathbb{N}}A_{n}=X\setminus U$. If $u\in A_{n}$ for every $n$, then $u\in A_{1}\subseteq X$, and $u\notin D_{n}$ for every $n$, so $u\notin U$; thus $u\in X\setminus U$. If $u\in X\setminus U$, then for every $n$ we have $u\in X$ and $u\notin X\setminus A_{n}$, so $u\in A_{n}$. Hence $\bigcap_{n\in\mathbb{N}}A_{n}=X\setminus U\in\mathcal{F}$.

Proof of (ix). Let $A,B\in\mathcal{F}$. Then $X\setminus A$ and $X\setminus B$ lie in $\mathcal{F}$, so does their union by (vii), and so does $E=X\setminus\bigl((X\setminus A)\cup(X\setminus B)\bigr)$. A set $u$ lies in $E$ if and only if $u\in X$, $u\in A$ and $u\in B$; as $A\subseteq X$, this says $u\in A\cap B$. So $A\cap B=E\in\mathcal{F}$.

Proof of (x). Let $A,B\in\mathcal{F}$. As $A\subseteq X$, $A\setminus B=A\cap(X\setminus B)$, and $X\setminus B\in\mathcal{F}$, so $A\setminus B\in\mathcal{F}$ by (ix).

Proof of (i). We check σ-Algebra and Measurable Space (i) for $\mathcal{P}(X)$, using The Union Set and the Power Set of a Set (ii): a set lies in $\mathcal{P}(X)$ if and only if it is a subset of $X$. Clearly $\mathcal{P}(X)\subseteq\mathcal{P}(X)$, and $X\in\mathcal{P}(X)$ since $X\subseteq X$. For every set $A$, $X\setminus A\subseteq X$, so $X\setminus A\in\mathcal{P}(X)$. For a sequence $(A_{n})$ in $\mathcal{P}(X)$, the union $\bigcup_{n\in\mathbb{N}}A_{n}$ is a set by The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (i), and each of its elements lies in some $A_{n}\subseteq X$; so it is a subset of $X$ and lies in $\mathcal{P}(X)$.

Proof of (ii). Let $\mathcal{G}_{0}\in\mathcal{S}$ and $\mathcal{I}=\bigcap_{\mathcal{G}\in\mathcal{S}}\mathcal{G}$. By The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (ii), $\mathcal{I}$ is a set, and by Indexed Families of Sets and Their Union, Intersection and Product (iii), a set $u$ lies in $\mathcal{I}$ if and only if $u\in\mathcal{G}$ for every $\mathcal{G}\in\mathcal{S}$. We check σ-Algebra and Measurable Space (i) for $\mathcal{I}$, using that every $\mathcal{G}\in\mathcal{S}$ satisfies it. First, $\mathcal{I}\subseteq\mathcal{G}_{0}\subseteq\mathcal{P}(X)$. Next, $X\in\mathcal{G}$ for every $\mathcal{G}\in\mathcal{S}$, so $X\in\mathcal{I}$. If $A\in\mathcal{I}$ and $\mathcal{G}\in\mathcal{S}$, then $A\in\mathcal{G}$, so $X\setminus A\in\mathcal{G}$; hence $X\setminus A\in\mathcal{I}$. Finally let $(A_{n})$ be a sequence in $\mathcal{I}$ and $\mathcal{G}\in\mathcal{S}$. As $\mathcal{I}\subseteq\mathcal{G}$, $(A_{n})$ is a sequence in $\mathcal{G}$, so $\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{G}$; hence $\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{I}$.

Proof of (xi). Let $\varphi:\mathcal{F}\to\mathcal{P}(X)$ be the map $A\mapsto A\cap Y$, whose values are subsets of $X$ (by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii) there is exactly one map $\mathcal{F}\to\mathcal{P}(Y)$ with these values, and it is a map into $\mathcal{P}(X)$, as $\mathcal{P}(Y)\subseteq\mathcal{P}(X)$ because $Y\subseteq X$). Then $\mathcal{F}_{Y}=\varphi[\mathcal{F}]$ is a set by Images of Sets under Functions Are Sets, and a Function with a Set Domain Is a Set (i), and by The Image and the Preimage of a Class under a Class (i) and Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (i), a set $B$ lies in $\mathcal{F}_{Y}$ if and only if $B=A\cap Y$ for some $A\in\mathcal{F}$. If $B=A\cap Y$ with $A\in\mathcal{F}$, then $B\in\mathcal{F}$ by (ix), as $Y\in\mathcal{F}$, and $B\subseteq Y$. If $A\in\mathcal{F}$ and $A\subseteq Y$, then $A=A\cap Y\in\mathcal{F}_{Y}$. Hence $\mathcal{F}_{Y}=\{A\in\mathcal{F}:A\subseteq Y\}$, and in particular $\mathcal{F}_{Y}\subseteq\mathcal{F}$.

We check σ-Algebra and Measurable Space (i) for $\mathcal{F}_{Y}$ on $Y$. Every element of $\mathcal{F}_{Y}$ is a subset of $Y$, so $\mathcal{F}_{Y}\subseteq\mathcal{P}(Y)$ by The Union Set and the Power Set of a Set (ii). As $Y\in\mathcal{F}$ and $Y\subseteq Y$, $Y\in\mathcal{F}_{Y}$. For $B\in\mathcal{F}_{Y}$, we have $B\in\mathcal{F}$, so $Y\setminus B\in\mathcal{F}$ by (x); as $Y\setminus B\subseteq Y$, $Y\setminus B\in\mathcal{F}_{Y}$. For a sequence $(B_{n})$ in $\mathcal{F}_{Y}$, $(B_{n})$ is a sequence in $\mathcal{F}$, so $\bigcup_{n\in\mathbb{N}}B_{n}\in\mathcal{F}$; each of its elements lies in some $B_{n}\subseteq Y$, so $\bigcup_{n\in\mathbb{N}}B_{n}\subseteq Y$ and $\bigcup_{n\in\mathbb{N}}B_{n}\in\mathcal{F}_{Y}$.

Proof of (iii). By The σ-Algebra Generated by a Set of Subsets (i), $\sigma(\mathcal{C})=\bigcap_{\mathcal{G}\in\mathcal{S}_{\mathcal{C}}}\mathcal{G}$, the intersection of the family $\mathcal{G}\mapsto\mathcal{G}$ indexed by the set $\mathcal{S}_{\mathcal{C}}$, which has an element and whose elements are $\sigma$-algebras on $X$. So $\sigma(\mathcal{C})$ is a $\sigma$-algebra on $X$ by (ii). Let $A\in\mathcal{C}$. Then $A\subseteq X$ as $\mathcal{C}\subseteq\mathcal{P}(X)$, and $A\in\mathcal{G}$ for every $\sigma$-algebra $\mathcal{G}$ on $X$ with $\mathcal{C}\subseteq\mathcal{G}$; so $A\in\sigma(\mathcal{C})$ by The σ-Algebra Generated by a Set of Subsets (ii). Hence $\mathcal{C}\subseteq\sigma(\mathcal{C})$.

Proof of (iv). Let $\mathcal{G}$ be a $\sigma$-algebra on $X$ with $\mathcal{C}\subseteq\mathcal{G}$. Every $A\in\sigma(\mathcal{C})$ lies in $\mathcal{G}$ by The σ-Algebra Generated by a Set of Subsets (ii), so $\sigma(\mathcal{C})\subseteq\mathcal{G}$.

Proof of (v). By (iii) applied to $\mathcal{D}$, $\sigma(\mathcal{D})$ is a $\sigma$-algebra on $X$ with $\mathcal{D}\subseteq\sigma(\mathcal{D})$. If $\mathcal{C}\subseteq\sigma(\mathcal{D})$, then $\sigma(\mathcal{C})\subseteq\sigma(\mathcal{D})$ by (iv) applied with $\mathcal{G}=\sigma(\mathcal{D})$. If $\mathcal{C}\subseteq\mathcal{D}$, then $\mathcal{C}\subseteq\mathcal{D}\subseteq\sigma(\mathcal{D})$, and the first case applies.