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