単項filter

proof

$ \underline{\therefore\forall X\forall A\in 2^X\setminus\Set{\varnothing}:\lang\{A\}\rang_X=\min\Set{\mathcal F\in\mathscr F_X|A\in\mathcal F}\quad}_\blacksquare

$ \forall\mathcal F\in\mathscr F_X:

$ \lang\{\{x\}\}\rang_X\subseteq\mathcal F

$ \implies\forall F:

$ F\in\mathcal F

$ \implies F\cap\Set{x}\in\mathcal F\land F\in2^X

$ \because\Set{x}\in\lang\{\{x\}\}\rang_X\subseteq\mathcal F

$ \implies F\cap\Set{x}\neq\varnothing\land F\in2^X

$ \iff x\in F\in2^X

$ \iff F\in\lang\{\{x\}\}\rang_X

$ \implies\lang\{\{x\}\}\rang_X=\mathcal F

$ \underline{\therefore\forall X\forall x\in X:\lang\{\{x\}\}\rang_X\in\mathrm{max}^*\mathscr F_X\quad}_\blacksquare

$ \forall X\forall A\in2^X:

$ \lang\Set{A}\rang_X\in\mathrm{max}^*\mathscr F_X

$ \iff\forall\mathcal F\in\mathscr F_X:\left(\Set{F\in2^X|A\subseteq F}\subseteq\mathcal F\implies\forall F\in\mathcal F:A\subseteq F\right)

さすがにこれは無理か？

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