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Metamath Proof Explorer

Theorem eltg

Description: Membership in a topology generated by a basis. (Contributed by NM, 16-Jul-2006) (Revised by Mario Carneiro, 10-Jan-2015)

Ref Expression
Assertion eltg ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ ( topGen ‘ 𝐵 ) ↔ 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 tgval ⊢ ( 𝐵 ∈ 𝑉 → ( topGen ‘ 𝐵 ) = { 𝑥 ∣ 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) } )
2 1 eleq2d ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ ( topGen ‘ 𝐵 ) ↔ 𝐴 ∈ { 𝑥 ∣ 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) } ) )
3 elex ⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) } → 𝐴 ∈ V )
4 3 adantl ⊢ ( ( 𝐵 ∈ 𝑉 ∧ 𝐴 ∈ { 𝑥 ∣ 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) } ) → 𝐴 ∈ V )
5 inex1g ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐵 ∩ 𝒫 𝐴 ) ∈ V )
6 5 uniexd ⊢ ( 𝐵 ∈ 𝑉 → ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ∈ V )
7 ssexg ⊢ ( ( 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ∧ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ∈ V ) → 𝐴 ∈ V )
8 6 7 sylan2 ⊢ ( ( 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ∧ 𝐵 ∈ 𝑉 ) → 𝐴 ∈ V )
9 8 ancoms ⊢ ( ( 𝐵 ∈ 𝑉 ∧ 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ) → 𝐴 ∈ V )
10 id ⊢ ( 𝑥 = 𝐴 → 𝑥 = 𝐴 )
11 pweq ⊢ ( 𝑥 = 𝐴 → 𝒫 𝑥 = 𝒫 𝐴 )
12 11 ineq2d ⊢ ( 𝑥 = 𝐴 → ( 𝐵 ∩ 𝒫 𝑥 ) = ( 𝐵 ∩ 𝒫 𝐴 ) )
13 12 unieqd ⊢ ( 𝑥 = 𝐴 → ∪ ( 𝐵 ∩ 𝒫 𝑥 ) = ∪ ( 𝐵 ∩ 𝒫 𝐴 ) )
14 10 13 sseq12d ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) ↔ 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ) )
15 14 elabg ⊢ ( 𝐴 ∈ V → ( 𝐴 ∈ { 𝑥 ∣ 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) } ↔ 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ) )
16 4 9 15 pm5.21nd ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝑥 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝑥 ) } ↔ 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ) )
17 2 16 bitrd ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ ( topGen ‘ 𝐵 ) ↔ 𝐴 ⊆ ∪ ( 𝐵 ∩ 𝒫 𝐴 ) ) )