$\pi$-Systems, $\lambda$-Systems, And Semirings

The collection $\bigcap_{\alpha \in A} \mathcal{P}_{\alpha }$ is a $\pi$-system on $\Omega$.

Let $\mathcal{A} \subseteq 2^{\Omega }$. Suppose that $\left\{ \mathcal{P}_{\alpha }: \alpha \in A\right\}$ is the "exhaustive list" of all the $\pi$ system that contain $\mathcal{A}$. In other words, each $\mathcal{P}_{\alpha } \supseteq \mathcal{A}$, and any $\pi$-system that contains $\mathcal{A}$ coincides with some $\mathcal{P}_{\alpha }$. Then $\bigcap_{\alpha \in A} \mathcal{P}_{\alpha }$ is a $\pi$-system that contains A. If $Q$ is a $\pi$-system containing $\mathcal{A}$, then $\bigcap_{\alpha \in A} \mathcal{P}_{\alpha } \subseteq Q$. The minimal $\pi$-system generated by A always exists.

Suppose that $\mathcal{P}$ is a $\pi$-system with $\mathcal{P} \supseteq \mathcal{A}$, and suppose that $\mathcal{P}$ is contained in any other $\pi$-system that contains $\mathcal{A}$. Then $\mathcal{P}=\bigcap_{\alpha \in A} \mathcal{P}_{\alpha }$, with notation as in (b). The minimal $\pi$-system containing A [which always exists] is also unique.

$\mathscr {L}$ is a $\lambda$-system iff $\mathscr {L}$ satisfies ( $\lambda_{1}$ ), ( $\lambda_{2}^{\prime }$ ), and ( $\lambda_{3}$ ).

Every $\lambda$-system additionally satisfies ( $\lambda_{4}$ ), ( $\lambda_{5}$ ), and ( $\lambda_{6}$ ).

$\mathscr {L}$ is a $\lambda$-system iff $\mathscr {L}$ satisfies ( $\lambda_{1}$ ), ( $\lambda_{2}^{\prime }$ ), and ( $\lambda_{5}$ ).

${ }^{1}$ The conditions are: $\left(\boldsymbol {\lambda }_{\mathbf{1}}\right) \quad \Omega \in \mathscr {L}$; ( $\boldsymbol {\lambda }_{\mathbf{2}}$) $A \in \mathscr {L} \Longrightarrow A^{c} \in \mathscr {L}$; ( $\boldsymbol {\lambda }_{\mathbf{2}}^{\prime }$) $\quad A, B \in \mathscr {L} \& A \subseteq B \Longrightarrow B \backslash A \in \mathscr {L}$; ( $\boldsymbol {\lambda }_{\mathbf{3}}$) For any disjoint $\left\{ A_{n}\right\}_{n=1}^{\infty } \subseteq \mathscr {L}, \bigcup_{n=1}^{\infty } A_{n} \in \mathscr {L}$; (λ4) $\quad A, B \in \mathscr {L} \& A \cap B=\varnothing \Longrightarrow A \cup B \in \mathscr {L}$; ( h $_{\text{5 }}$) $\forall \left\{ A_{n}\right\}_{n=1}^{\infty } \subseteq \mathscr {L}, A_{n} \uparrow \Longrightarrow \bigcup_{n=1}^{\infty } A_{n} \in \mathscr {L}$; ( $\boldsymbol {\lambda }_{\mathbf{6}}$) $\quad \forall \left\{ A_{n}\right\}_{n=1}^{\infty } \subseteq \mathscr {L}, A_{n} \downarrow \Longrightarrow \bigcap_{n=1}^{\infty } A_{n} \in \mathscr {L}$.

If a collection $\mathscr {L}$ is nonempty and satisfies ( $\lambda_{2}$ ) and ( $\lambda_{3}$ ), then $\mathscr {L}$ is a $\lambda$-system.

The collection $\bigcap_{\alpha \in A} \mathscr {L}_{\alpha }$ is a $\lambda$-system on $\Omega$.

Suppose that $\mathcal{A} \subseteq 2^{\Omega }$ is such that $\mathcal{A}$ is contained in each $\mathscr {L}_{\alpha }$, and suppose that $\left\{ \mathscr {L}_{\alpha }: \alpha \in A\right\}$ is the "exhaustive list" of all the $\lambda$-system that contain $\mathcal{A}$. Then $\bigcap_{\alpha \in A} \mathscr {L}_{\alpha }$ is a $\lambda$-system that contains $\mathcal{A}$. If $\mathcal{Q}$ is a $\lambda$-system on $\Omega$ that contains $\mathcal{A}$, then $\bigcap_{\alpha \in A} \mathscr {L}_{\alpha } \subseteq \mathcal{Q}$. The minimal $\lambda$-system generated by $\mathcal{A}$ always exists.

Let $\mathscr {L}$ denote a $\lambda$-system over $\Omega$ with $\mathscr {L} \supseteq \mathscr {A}$ and where $\mathscr {L}$ is contained in any other $\lambda$-system also containing $\mathcal{A}$. Then $\mathscr {L}=\bigcap_{\alpha \in A} \mathscr {L}_{\alpha }$, with notation as in (b). Therefore, the $\lambda$-system generated by Always exists and is unique.

Some define $A$ to be a semiring iff $A$ is a nonempty $\pi$-system such that both $E, F \in \mathcal{A}$ and $E \subseteq F$ imply the existence of a finite collection $C_{0}, C_{1}, \ldots , C_{n} \in$ A with $E=C_{0} \subseteq C_{1} \subseteq \ldots \subseteq C_{n} \subseteq F$ and $C_{i} \backslash C_{i-1} \in \mathcal{A}$ for $i=1, \ldots , n$. This definition of a semiring is equivalent to our definition of a semiring.

Some define $\mathcal{A}$ to be a semiring by stipulating (SR1), (SR2), and the following property: $A, B \in \mathcal{A}$ implies the existence of disjoint $\mathcal{A}$-sets $C_{0}, C_{1}, \ldots , C_{n}$ with $B \backslash A=\bigcup_{i=0}^{n} C_{i}$. Note that here $B \backslash A$ is not necessarily a proper difference. If $\mathcal{A}$ is a semiring by this definition, then $\mathcal{A}$ is a semiring by our definition, but the converse is not necessarily true.

$\varnothing$ is in every ring.

$\mathcal{R}$ is a ring iff $\mathcal{R}$ satisfies (R1), (R2), and (R4) $A, B \in \mathcal{R}$ with $A \subseteq B$ implies $B \backslash A \in \mathcal{R}$.

Every ring satisfies (R5) $A, B \in \mathcal{R}$ implies $A \Delta B \in \mathcal{R}$.

Every ring is a $\pi$-system.

Every ring is closed under finite unions and finite intersections.

$\mathcal{R}$ is a ring iff $\mathcal{R}$ a nonempty $\pi$-system that satisfies ( $R 4$ ) along with (R6) $A, B \in \mathcal{R}$ and $A \cap B=\varnothing$ imply $A \cup B \in \mathcal{R}$.

$\mathcal{R}$ is a ring iff $\mathcal{R}$ is a nonempty $\pi$-system that satisfies (R5).

Suppose that $\left\{ \mathcal{R}_{\alpha }: \alpha \in A\right\}$ is the "exhaustive list" of all rings that contain $\mathcal{A}$. Then $\bigcap_{\alpha \in A} \mathcal{R}_{\alpha }$ is a ring that contains $\mathcal{A}$, and $\bigcap_{\alpha \in A} \mathcal{R}_{\alpha }$ is contained in any ring that contains $\mathcal{A}$. The minimal ring containing $\mathcal{A}$ is always exists and is unique.

The collection of finite unions of rsc intervals is a ring on $\mathbb {R}$.

Let $\Omega$ be uncountable. The collection of all amc subsets of $\Omega$ is a ring on $\Omega$.

Every ring is a semiring. However, not every semiring is a ring.

Let $\mathcal{A}$ denote a semiring on $\Omega$, and let $\mathcal{R}$ consist of the finite disjoint unions of $A$-sets. Then $\mathcal{R}$ is closed under finite intersections and disjoint unions.

If $A, B \in \mathcal{A}$ and $A \subseteq B$, then $B-A \in \mathcal{R}$.

$A \in \mathcal{A}, B \in \mathcal{R}$, and $A \subseteq B$ imply $B-A \in \mathcal{R}$.

$A, B \in \mathcal{R}$ and $A \subseteq B$ imply $B-A \in \mathcal{R}$.

$\mathcal{R}$ is the minimal ring generated by $\mathcal{A}$.

A semiring that satisfies (R2) is a ring.

Given $\mathscr {C} \subseteq 2^{\Omega }$, let $\mathscr {C}^{*}$ denote the collection of all finite unions of differences of $\mathscr {C}$-sets. If $\operatorname {card}(\mathscr {C})=\aleph_{0}$, then $\operatorname {card}\left(\mathscr {C}^{*}\right)=\aleph_{0}$. Also, $\varnothing \in \mathscr {C}$ implies $\mathscr {C} \subseteq \mathscr {C}^{*}$.

Let $\mathcal{A}_{0}=\mathcal{A}$, and define $\mathcal{A}_{n}=\mathcal{A}_{n-1}^{*}$ for $n \geqslant 1$. Then $\mathcal{A} \subseteq \bigcup_{n=0}^{\infty } \mathcal{A}_{n} \subseteq \mathfrak {R}(\mathcal{A})$, where $\mathfrak {R}(\mathcal{A})$ is the minimal ring generated by $\mathcal{A}$, and where [without loss of generality] $\varnothing \in \mathcal{A}$. Also, $\operatorname {card}\left(\bigcup_{n=0}^{\infty } \mathcal{A}_{n}\right)=\aleph_{0}$.

$\bigcup_{n=0}^{\infty } \mathcal{A}_{n}$ is a ring on $\Omega$, and from the fact that $\mathfrak {R}(\mathcal{A})$ is the minimal ring containing $\mathcal{A}$, we have $\bigcup_{n=0}^{\infty } \mathcal{A}_{n}=\mathfrak {R}(\mathcal{A})$, and thus $\operatorname {card}(\mathfrak {R}(\mathcal{A}))=\aleph_{0}$.

We may generalize: if $\mathcal{A}$ is infinite, then $\operatorname {card}(\mathcal{A})=\operatorname {card}(\mathfrak {R}(\mathcal{A}))$.