開核公理系→Hausdorffの公理系

from Hausdorffの公理系

開核公理系(I1)~(I5)と(I6)

(I1) $ X^\circ =XX^∘=X

(I2) $ \forall A\in2^X:A^\circ\subseteq A∀A∈2^X(A^∘⊆A)

(I3) $ \forall A,B\in2^X:A\subseteq B\implies A^\circ\subseteq B^\circ∀A,B∈2^X(A⊆B⟹A^∘⊆B^∘)

(I4) $ \forall A,B\in2^X:(A\cap B)^\circ=A^\circ\cap B^\circ∀A,B∈2^X((A∩B)^∘=A^∘∩B^∘)

(I5) $ \forall A\in2^X:{A^\circ}^\circ=A^\circ∀A∈2^X(A^∘∘=A^∘)

(I6)$ \mathcal N(x)=\{N\in2^X\mid x\in N^\circ\}

からHausdorffの公理系(N1)~(N5)と(N6)

(N1)$ \forall x\in X:X\in\mathcal N(x)∀x∈X(X∈N(x))

(N2)$ \forall x\in X\forall N\in\mathcal N(x):x\in N∀x∈X∀N∈N(x)(x∈N)

(N3)$ \forall x\in X\forall N_1,N_2\in\mathcal N(x):N_1\cap N_2\in\mathcal N(x)∀x∈X∀N1,N2∈N(x)(N1∩N2∈N(x))

(N4)$ \forall x\in X\forall N:(2^N\cap\mathcal N(x)\neq\varnothing\implies N\in\mathcal N(x))∀x∈X∀N(2^N∩N(x)≠∅⇒N∈N(x))

(N5)$ \forall x\in X\forall N_1\in\mathcal N(x)\exist N_2\in\mathcal N(x)\forall y\in N_2:N_1\in\mathcal N(y)∀x∈X∀N1∈N(x)∃N2∈N(x)∀y∈N2(N1∈N(y))

(N6)$ A^\circ=\{x\in X\mid A\in\mathcal N(x)\}

を導く

(N1)$ \forall x:

$ x\in X

$ \implies x\in X=X^\circ

$ \becauseX^∘=X

$ \implies X\in\mathcal N(x)

$ \because(I6)

$ \underline{\therefore\forall x\in X:X\in\mathcal N(x)\quad}_\blacksquare

(N2)$ \forall x\in X\forall N:

$ N\in\mathcal N(x)

$ \iff x\in N^\circ

$ \because(I6)

$ \subseteq N

$ \because∀A∈2^X(A^∘⊆A)

$ \implies x\in N

$ \underline{\therefore\forall x\in X\forall N\in\mathcal N(x):x\in N\quad}_\blacksquare

(N3)$ \forall x\in X\forall N_1,N_2:

$ N_1,N_2\in\mathcal N(x)

$ \iff x\in {N_1}^\circ\cap{N_2}^\circ

$ \because(I6)

$ =(N_1\cap N_2)^\circ

$ \because∀A,B∈2^X((A∩B)^∘=A^∘∩B^∘)

$ \implies N_1\cap N_2\in\mathcal N(x)

$ \because(I6)

$ \underline{\therefore\forall x\in X\forall N_1,N_2\in\mathcal N(x):N_1\cap N_2\in\mathcal N(x)\quad}_\blacksquare

(N4)$ \forall x\in X\forall N:

$ 2^N\cap\mathcal N(x)\neq\varnothing

$ \iff\exist N'\in2^N:N'\in\mathcal N(x)

$ \iff \exist N':x\in {N'}^\circ\land N'\subseteq N

$ \because(I6)

$ \implies\exist N':x\in {N'}^\circ\subseteq{N}^\circ

$ \because∀A,B∈2^X(A⊆B⟹A^∘⊆B^∘)

$ \implies x\in{N}^\circ

$ \iff N\in\mathcal N(x)

$ \because(I6)

$ \underline{\therefore\forall x\in X\forall N:(2^N\cap\mathcal N(x)\neq\varnothing\implies N\in\mathcal N(x))\quad}_\blacksquare

これは$ \forall x\in X\forall N_2\in\mathcal N(x):\{N_1\in2^X|N_2\subseteq N_1\}\subseteq\mathcal N(x)とも表せる

(N5)$ \forall x\in X\forall N_1:

$ N_1\in\mathcal N(x)

$ \iff x\in {N_1}^\circ

$ \because(I6)

$ \iff x\in {{N_1}^\circ}^\circ\land{N_1}^\circ\subseteq{N_1}^\circ

$ \because∀A∈2^X(A^∘∘=A^∘)

$ \implies\exist N_2\in2^X: x\in{N_2}^\circ\land N_2\subseteq{N_1}^\circ

$ \because N_2={N_1}^\circとした

$ \iff\exist N_2\in\mathcal N(x):N_2\subseteq{N_1}^\circ

$ \because(I6)

$ \iff\exist N_2\in\mathcal N(x)\forall y\in N_2:N_1\in\mathcal N(y)

$ \because(I6)

$ \underline{\therefore\forall x\in X\forall N_1\in\mathcal N(x)\exist N_2\in\mathcal N(x)\forall y\in N_2:N_1\in\mathcal N(y)\quad}_\blacksquare

(N6)

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

$ x\in A^\circ

$ \iff A\in\mathcal N(x)

$ \because(I6)

$ \iff A\in\mathcal N(x)\land x\in A^\circ\subseteq A\subseteq X

$ \because ∀A∈2^X(A^∘⊆A)

$ \iff A\in\mathcal N(x)\land x\in X

$ \iff x\in\{x\in X\mid A\in\mathcal N(x)\}

$ \underline{\therefore\forall A\in 2^X:A^\circ=\{x\in X\mid A\in\mathcal N(x)\}\quad}_\blacksquare

#2025-03-02 17:04:49

#2025-02-06 16:42:56

#2025-01-29 19:08:35