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 ∈ 𝐶 → ( 𝐵 ∈ TopBases ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		ineq1	⊢ ( 𝑧 = 𝐵 → ( 𝑧 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) = ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) )
2	1	unieqd	⊢ ( 𝑧 = 𝐵 → ∪ ( 𝑧 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) = ∪ ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) )
3	2	sseq2d	⊢ ( 𝑧 = 𝐵 → ( ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝑧 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ↔ ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ) )
4	3	raleqbi1dv	⊢ ( 𝑧 = 𝐵 → ( ∀ 𝑦 ∈ 𝑧 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝑧 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ↔ ∀ 𝑦 ∈ 𝐵 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ) )
5	4	raleqbi1dv	⊢ ( 𝑧 = 𝐵 → ( ∀ 𝑥 ∈ 𝑧 ∀ 𝑦 ∈ 𝑧 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝑧 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ) )
6		df-bases	⊢ TopBases = { 𝑧 ∣ ∀ 𝑥 ∈ 𝑧 ∀ 𝑦 ∈ 𝑧 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝑧 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) }
7	5 6	elab2g	⊢ ( 𝐵 ∈ 𝐶 → ( 𝐵 ∈ TopBases ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑥 ∩ 𝑦 ) ⊆ ∪ ( 𝐵 ∩ 𝒫 ( 𝑥 ∩ 𝑦 ) ) ) )