Metamath Proof Explorer


Theorem subspopn

Description: An open set is open in the subspace topology. (Contributed by Jeff Madsen, 2-Sep-2009) (Revised by Mario Carneiro, 15-Dec-2013)

Ref Expression
Assertion subspopn ( ( ( 𝐽 ∈ Top ∧ 𝐴𝑉 ) ∧ ( 𝐵𝐽𝐵𝐴 ) ) → 𝐵 ∈ ( 𝐽t 𝐴 ) )

Proof

Step Hyp Ref Expression
1 elrestr ( ( 𝐽 ∈ Top ∧ 𝐴𝑉𝐵𝐽 ) → ( 𝐵𝐴 ) ∈ ( 𝐽t 𝐴 ) )
2 df-ss ( 𝐵𝐴 ↔ ( 𝐵𝐴 ) = 𝐵 )
3 eleq1 ( ( 𝐵𝐴 ) = 𝐵 → ( ( 𝐵𝐴 ) ∈ ( 𝐽t 𝐴 ) ↔ 𝐵 ∈ ( 𝐽t 𝐴 ) ) )
4 2 3 sylbi ( 𝐵𝐴 → ( ( 𝐵𝐴 ) ∈ ( 𝐽t 𝐴 ) ↔ 𝐵 ∈ ( 𝐽t 𝐴 ) ) )
5 1 4 syl5ibcom ( ( 𝐽 ∈ Top ∧ 𝐴𝑉𝐵𝐽 ) → ( 𝐵𝐴𝐵 ∈ ( 𝐽t 𝐴 ) ) )
6 5 3expa ( ( ( 𝐽 ∈ Top ∧ 𝐴𝑉 ) ∧ 𝐵𝐽 ) → ( 𝐵𝐴𝐵 ∈ ( 𝐽t 𝐴 ) ) )
7 6 impr ( ( ( 𝐽 ∈ Top ∧ 𝐴𝑉 ) ∧ ( 𝐵𝐽𝐵𝐴 ) ) → 𝐵 ∈ ( 𝐽t 𝐴 ) )