Metamath Proof Explorer


Theorem topfneec2

Description: A topology is precisely identified with its equivalence class. (Contributed by Jeff Hankins, 12-Oct-2009)

Ref Expression
Hypothesis topfneec2.1 = ( Fne ∩ Fne )
Assertion topfneec2 ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → ( [ 𝐽 ] = [ 𝐾 ] 𝐽 = 𝐾 ) )

Proof

Step Hyp Ref Expression
1 topfneec2.1 = ( Fne ∩ Fne )
2 1 fneval ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → ( 𝐽 𝐾 ↔ ( topGen ‘ 𝐽 ) = ( topGen ‘ 𝐾 ) ) )
3 1 fneer Er V
4 3 a1i ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → Er V )
5 elex ( 𝐽 ∈ Top → 𝐽 ∈ V )
6 5 adantr ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → 𝐽 ∈ V )
7 4 6 erth ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → ( 𝐽 𝐾 ↔ [ 𝐽 ] = [ 𝐾 ] ) )
8 tgtop ( 𝐽 ∈ Top → ( topGen ‘ 𝐽 ) = 𝐽 )
9 tgtop ( 𝐾 ∈ Top → ( topGen ‘ 𝐾 ) = 𝐾 )
10 8 9 eqeqan12d ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → ( ( topGen ‘ 𝐽 ) = ( topGen ‘ 𝐾 ) ↔ 𝐽 = 𝐾 ) )
11 2 7 10 3bitr3d ( ( 𝐽 ∈ Top ∧ 𝐾 ∈ Top ) → ( [ 𝐽 ] = [ 𝐾 ] 𝐽 = 𝐾 ) )