閉集合系の公理→Kuratowskiの閉包公理系

閉集合系の公理$ \land\overline\bullet:2^X\ni A\mapsto\bigcap\{C\in\mathcal C\mid A\subseteq C\}\in\mathcal CからKuratowskiの閉包公理系$ \land\mathcal C=\{C\in2^X\mid\overline{C}=C\}を導く

$ \bar\bulletがKuratowskiの閉包公理系を満たすことを示す

(K1)$ \bar\varnothing=\varnothing

$ \because\bar\varnothing=\bigcap\{C\in\mathcal C\mid\varnothing\subseteq C\}

$ =\bigcap\mathcal C

$ =\varnothing

$ \because閉集合系の公理(C1)より$ \bigcap\mathcal C=\varnothing\cap\cdots=\varnothing

(K2)$ \forall A\in2^X:A\subseteq\bar A

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

$ x\in A

$ \implies\forall C:(A\subseteq C\implies x\in A\subseteq C)

$ \implies\forall C\in\mathcal C:(A\subseteq C\implies x\in C)

$ \iff x\in\bar A

(K3)$ \forall A\in2^X:\bar{\bar{A}}=\bar A

$ \because\forall A:

$ A\in2^X

$ \implies \bar A\subseteq\bar{\bar{A}}

$ \because(K2)

$ \implies \bar A\subseteq\bar{\bar{A}}\land\forall C\in\mathcal C:(\bar A\subseteq C\implies\bar{\bar{A}}\subseteq C)

$ \because閉集合系の公理(C1)

$ \implies\bar{\bar{A}}=\bar A

$ \because\bar A\in\mathcal C

(K4)$ \forall A,B\in 2^X:\overline{A\cup B}=\bar A\cup\bar B

$ \because\forall A,B\in2^X\forall x:

$ x\in\overline{A\cup B}

$ \iff\forall C\in\mathcal C:(A\cup B\subseteq C\implies x\in C)

$ \iff\forall C\in\mathcal C:((A\subseteq C\implies x\in C)\lor(B\subseteq C\implies x\in C))

$ \implies x\in\bar A\cup\bar B

$ \iff x\in\bar A\cup\bar B\land\bar A\subseteq\overline{A\cup B}\land\bar B\subseteq\overline{A\cup B}

$ \because 閉集合系の公理(C1)

$ \implies x\in\overline{A\cup B}

閉集合系の構成

$ \overline{C}=C

$ \iff C=\min\{C'\in\mathcal C\mid C\subseteq C'\}

$ \iff C\in\mathcal C\land\forall C'\in\mathcal C:(C\subseteq C'\implies C\subseteq C')

$ \iff C\in\mathcal C

$ \underline{\therefore\mathcal C=\{C\in2^X\mid\overline{C}=C\}\quad}_\blacksquare

#2025-02-06 21:45:27

#2025-02-02 11:31:30

#2025-01-29 18:33:46