Metamath Proof Explorer


Theorem kur14lem3

Description: Lemma for kur14 . A closure is a subset of the base set. (Contributed by Mario Carneiro, 11-Feb-2015)

Ref Expression
Hypotheses kur14lem.j ⊢ 𝐽 ∈ Top
kur14lem.x ⊢ 𝑋 = ∪ 𝐽
kur14lem.k ⊢ 𝐾 = ( cls ‘ 𝐽 )
kur14lem.i ⊢ 𝐼 = ( int ‘ 𝐽 )
kur14lem.a ⊢ 𝐴 ⊆ 𝑋
Assertion kur14lem3 ( 𝐾 ‘ 𝐴 ) ⊆ 𝑋

Proof

Step Hyp Ref Expression
1 kur14lem.j ⊢ 𝐽 ∈ Top
2 kur14lem.x ⊢ 𝑋 = ∪ 𝐽
3 kur14lem.k ⊢ 𝐾 = ( cls ‘ 𝐽 )
4 kur14lem.i ⊢ 𝐼 = ( int ‘ 𝐽 )
5 kur14lem.a ⊢ 𝐴 ⊆ 𝑋
6 3 fveq1i ⊢ ( 𝐾 ‘ 𝐴 ) = ( ( cls ‘ 𝐽 ) ‘ 𝐴 )
7 2 clsss3 ⊢ ( ( 𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ) → ( ( cls ‘ 𝐽 ) ‘ 𝐴 ) ⊆ 𝑋 )
8 1 5 7 mp2an ⊢ ( ( cls ‘ 𝐽 ) ‘ 𝐴 ) ⊆ 𝑋
9 6 8 eqsstri ⊢ ( 𝐾 ‘ 𝐴 ) ⊆ 𝑋