Metamath Proof Explorer


Theorem isopn2

Description: A subset of the underlying set of a topology is open iff its complement is closed. (Contributed by NM, 4-Oct-2006)

Ref Expression
Hypothesis iscld.1 𝑋 = 𝐽
Assertion isopn2 ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( 𝑆𝐽 ↔ ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ) )

Proof

Step Hyp Ref Expression
1 iscld.1 𝑋 = 𝐽
2 difss ( 𝑋𝑆 ) ⊆ 𝑋
3 1 iscld2 ( ( 𝐽 ∈ Top ∧ ( 𝑋𝑆 ) ⊆ 𝑋 ) → ( ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ↔ ( 𝑋 ∖ ( 𝑋𝑆 ) ) ∈ 𝐽 ) )
4 2 3 mpan2 ( 𝐽 ∈ Top → ( ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ↔ ( 𝑋 ∖ ( 𝑋𝑆 ) ) ∈ 𝐽 ) )
5 dfss4 ( 𝑆𝑋 ↔ ( 𝑋 ∖ ( 𝑋𝑆 ) ) = 𝑆 )
6 5 biimpi ( 𝑆𝑋 → ( 𝑋 ∖ ( 𝑋𝑆 ) ) = 𝑆 )
7 6 eleq1d ( 𝑆𝑋 → ( ( 𝑋 ∖ ( 𝑋𝑆 ) ) ∈ 𝐽𝑆𝐽 ) )
8 4 7 sylan9bb ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ↔ 𝑆𝐽 ) )
9 8 bicomd ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( 𝑆𝐽 ↔ ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ) )