Metamath Proof Explorer


Theorem clscld

Description: The closure of a subset of a topology's underlying set is closed. (Contributed by NM, 4-Oct-2006)

Ref Expression
Hypothesis clscld.1 𝑋 = 𝐽
Assertion clscld ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∈ ( Clsd ‘ 𝐽 ) )

Proof

Step Hyp Ref Expression
1 clscld.1 𝑋 = 𝐽
2 1 clsval ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) = { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } )
3 1 topcld ( 𝐽 ∈ Top → 𝑋 ∈ ( Clsd ‘ 𝐽 ) )
4 3 anim1i ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( 𝑋 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆𝑋 ) )
5 sseq2 ( 𝑥 = 𝑋 → ( 𝑆𝑥𝑆𝑋 ) )
6 5 elrab ( 𝑋 ∈ { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ↔ ( 𝑋 ∈ ( Clsd ‘ 𝐽 ) ∧ 𝑆𝑋 ) )
7 4 6 sylibr ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → 𝑋 ∈ { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } )
8 7 ne0d ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ≠ ∅ )
9 ssrab2 { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ⊆ ( Clsd ‘ 𝐽 )
10 intcld ( ( { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ≠ ∅ ∧ { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ⊆ ( Clsd ‘ 𝐽 ) ) → { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ∈ ( Clsd ‘ 𝐽 ) )
11 8 9 10 sylancl ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → { 𝑥 ∈ ( Clsd ‘ 𝐽 ) ∣ 𝑆𝑥 } ∈ ( Clsd ‘ 𝐽 ) )
12 2 11 eqeltrd ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∈ ( Clsd ‘ 𝐽 ) )