集合半環
(S1)$ \varnothing\in\mathcal S
(S2)$ \forall S_1,S_2\in\mathcal S:S_1\cap S_2\in\mathcal S
(S3)$ \forall A,B\in\mathcal S\exist n\in\N\exist S_\bullet:\N_{\le n}\to\mathcal S:\begin{dcases}\forall i,j\in\N_{\le n}:S_i\cap S_j\neq\varnothing\implies i=j\\A\setminus B=\bigcup_{1\le i\le n}S_i\end{dcases}
$ A\setminus B自体は$ \mathcal Sの元でなくていい
性質
$ \forall A\in\mathcal S\exist n\in\N\exist S_\bullet:\N_{\le n}\to\mathcal S:\begin{dcases}\forall i,j\in\N_{\le n}:S_i\cap S_j\neq\varnothing\implies i=j\\A=\bigcup_{1\le i\le n}S_i\end{dcases}
$ \because(S3)に$ B=\varnothingを代入
ここから、$ \mathcal Sの元$ Aは全て有限非交和$ A=\bigsqcup_{1\le i\le n}S_iで表せるとわかる
任意の集合半環$ \mathcal S\subseteq2^X,\mathcal T\subseteq2^Yにて、$ \mathcal S⨳\mathcal T:=\{R|\exist(S,T)\in\mathcal S\times\mathcal T:R=S\times T\}は$ X\times Y上の集合半環になる (S1)$ \varnothing=\varnothing\times\varnothing\in\mathcal S⨳\mathcal T
(S2)$ \forall R_1,R_2:
$ R_1,R_2\in\mathcal S⨳\mathcal T
$ \iff\exist (S_1,T_1),(S_2,T_2)\in\mathcal S\times\mathcal T: R_1=S_1\times T_1\land R_2=S_2\times T_2
$ \implies\exist (S_1,T_1),(S_2,T_2)\in\mathcal S\times\mathcal T: R_1\cap R_1=(S_1\cap S_2)\times(T_1\cap T_2)
$ \underline{\implies R_1\cap R_2\in\mathcal S⨳\mathcal T\quad}_\blacksquare
$ \because (S_1\cap S_2,T_1\cap T_2)\in\mathcal S\times\mathcal T
(S3)$ \forall A,B:
$ A,B\in\mathcal S⨳\mathcal T
$ \iff\exist (S_1,T_1),(S_2,T_2)\in\mathcal S\times\mathcal T: A=S_1\times T_1\land B=S_2\times T_2
$ \implies\exist (S_1,T_1),(S_2,T_2)\in\mathcal S\times\mathcal T:
$ A\setminus B=(S_1\times(T_1\setminus T_2))\cup((S_1\setminus S_2)\times T_1)
$ =(S_1\times(T_1\setminus T_2))\sqcup((S_1\setminus S_2)\times T_1\cap T_2)
$ \implies\exist (S',T')\in\mathcal S\times\mathcal T\exist m,n\in\N\exist S_\bullet:\N_{\le m}\to\mathcal S\exist T_\bullet:\N_{\le n}\to\mathcal T:
$ A\setminus B=\left(S'\times\bigsqcup_{1\le i\le m}T_i\right)\sqcup\left(\left(\bigsqcup_{1\le i\le n}S_i\right)\times T'\right)
$ =\left(\bigsqcup_{1\le i\le m}S'\times T_i\right)\sqcup\left(\bigsqcup_{1\le i\le n}S_i\times T'\right)
$ \underline{\implies\exist n\in\N\exist R_\bullet:\N_{\le n}\to\mathcal S⨳\mathcal T:A\setminus B=\bigsqcup_{1\le i\le n}R_i\quad}_\blacksquare
$ \mathcal R=\Set{R|\exist n\in\N\exist S_\bullet:\N_{\le n}\to\mathcal S:R=\bigsqcup_{1\le i\le n}S_i}
References
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