Theorem isbasisg

Description: Express the predicate "the set B is a basis for a topology". (Contributed by NM, 17-Jul-2006)

	Ref	Expression
Assertion	isbasisg	$$⊢B\in C\to \left(B\in \mathrm{TopBases}\leftrightarrow \forall x\in B\forall y\in Bx\cap y\subseteq \bigcup \left(B\cap 𝒫\left(x\cap y\right)\right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		ineq1	$$⊢z=B\to z\cap 𝒫\left(x\cap y\right)=B\cap 𝒫\left(x\cap y\right)$$
2	1	unieqd	$$⊢z=B\to \bigcup \left(z\cap 𝒫\left(x\cap y\right)\right)=\bigcup \left(B\cap 𝒫\left(x\cap y\right)\right)$$
3	2	sseq2d	$$⊢z=B\to \left(x\cap y\subseteq \bigcup \left(z\cap 𝒫\left(x\cap y\right)\right)\leftrightarrow x\cap y\subseteq \bigcup \left(B\cap 𝒫\left(x\cap y\right)\right)\right)$$
4	3	raleqbi1dv	$$⊢z=B\to \left(\forall y\in zx\cap y\subseteq \bigcup \left(z\cap 𝒫\left(x\cap y\right)\right)\leftrightarrow \forall y\in Bx\cap y\subseteq \bigcup \left(B\cap 𝒫\left(x\cap y\right)\right)\right)$$
5	4	raleqbi1dv	$$⊢z=B\to \left(\forall x\in z\forall y\in zx\cap y\subseteq \bigcup \left(z\cap 𝒫\left(x\cap y\right)\right)\leftrightarrow \forall x\in B\forall y\in Bx\cap y\subseteq \bigcup \left(B\cap 𝒫\left(x\cap y\right)\right)\right)$$
6		df-bases	$$⊢\mathrm{TopBases}=\left\{z|\forall x\in z\forall y\in zx\cap y\subseteq \bigcup \left(z\cap 𝒫\left(x\cap y\right)\right)\right\}$$
7	5 6	elab2g	$$⊢B\in C\to \left(B\in \mathrm{TopBases}\leftrightarrow \forall x\in B\forall y\in Bx\cap y\subseteq \bigcup \left(B\cap 𝒫\left(x\cap y\right)\right)\right)$$