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Metamath Proof Explorer

Theorem iscld4

Description: A subset is closed iff it contains its own closure. (Contributed by NM, 31-Jan-2008)

Ref Expression
Hypothesis clscld.1 ⊢ 𝑋 = ∪ 𝐽
Assertion iscld4 ( ( 𝐽 ∈ Top ∧ 𝑆 ⊆ 𝑋 ) → ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) ↔ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 clscld.1 ⊢ 𝑋 = ∪ 𝐽
2 1 iscld3 ⊢ ( ( 𝐽 ∈ Top ∧ 𝑆 ⊆ 𝑋 ) → ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) ↔ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) = 𝑆 ) )
3 eqss ⊢ ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) = 𝑆 ↔ ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝑆 ∧ 𝑆 ⊆ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ) )
4 1 sscls ⊢ ( ( 𝐽 ∈ Top ∧ 𝑆 ⊆ 𝑋 ) → 𝑆 ⊆ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) )
5 4 biantrud ⊢ ( ( 𝐽 ∈ Top ∧ 𝑆 ⊆ 𝑋 ) → ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝑆 ↔ ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝑆 ∧ 𝑆 ⊆ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ) ) )
6 3 5 bitr4id ⊢ ( ( 𝐽 ∈ Top ∧ 𝑆 ⊆ 𝑋 ) → ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) = 𝑆 ↔ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝑆 ) )
7 2 6 bitrd ⊢ ( ( 𝐽 ∈ Top ∧ 𝑆 ⊆ 𝑋 ) → ( 𝑆 ∈ ( Clsd ‘ 𝐽 ) ↔ ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝑆 ) )