Metamath Proof Explorer


Theorem cldopn

Description: The complement of a closed set is open. (Contributed by NM, 5-Oct-2006) (Revised by Stefan O'Rear, 22-Feb-2015)

Ref Expression
Hypothesis iscld.1 𝑋 = 𝐽
Assertion cldopn ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) → ( 𝑋𝑆 ) ∈ 𝐽 )

Proof

Step Hyp Ref Expression
1 iscld.1 𝑋 = 𝐽
2 cldrcl ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) → 𝐽 ∈ Top )
3 1 iscld ( 𝐽 ∈ Top → ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) ↔ ( 𝑆𝑋 ∧ ( 𝑋𝑆 ) ∈ 𝐽 ) ) )
4 3 simplbda ( ( 𝐽 ∈ Top ∧ 𝑆 ∈ ( Clsd ‘ 𝐽 ) ) → ( 𝑋𝑆 ) ∈ 𝐽 )
5 2 4 mpancom ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) → ( 𝑋𝑆 ) ∈ 𝐽 )