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Metamath Proof Explorer

Theorem clsss2

Description: If a subset is included in a closed set, so is the subset's closure. (Contributed by NM, 22-Feb-2007)

Ref Expression
Hypothesis clscld.1 ⊢ 𝑋 = ∪ 𝐽
Assertion clsss2 ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝐶 )

Proof

Step Hyp Ref Expression
1 clscld.1 ⊢ 𝑋 = ∪ 𝐽
2 cldrcl ⊢ ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) → 𝐽 ∈ Top )
3 2 adantr ⊢ ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → 𝐽 ∈ Top )
4 1 cldss ⊢ ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) → 𝐶 ⊆ 𝑋 )
5 4 adantr ⊢ ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → 𝐶 ⊆ 𝑋 )
6 simpr ⊢ ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → 𝑆 ⊆ 𝐶 )
7 1 clsss ⊢ ( ( 𝐽 ∈ Top ∧ 𝐶 ⊆ 𝑋 ∧ 𝑆 ⊆ 𝐶 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ ( ( cls ‘ 𝐽 ) ‘ 𝐶 ) )
8 3 5 6 7 syl3anc ⊢ ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ ( ( cls ‘ 𝐽 ) ‘ 𝐶 ) )
9 cldcls ⊢ ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) → ( ( cls ‘ 𝐽 ) ‘ 𝐶 ) = 𝐶 )
10 9 adantr ⊢ ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → ( ( cls ‘ 𝐽 ) ‘ 𝐶 ) = 𝐶 )
11 8 10 sseqtrd ⊢ ( ( 𝐶 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆 ⊆ 𝐶 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ⊆ 𝐶 )