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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 ∧ 𝑆 ⊆ 𝑋 ) → ( 𝑆 ∈ 𝐽 ↔ ( 𝑋 ∖ 𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ) )

Proof

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