Metamath Proof Explorer


Theorem ntrin

Description: A pairwise intersection of interiors is the interior of the intersection. This does not always hold for arbitrary intersections. (Contributed by Jeff Hankins, 31-Aug-2009)

Ref Expression
Hypothesis clscld.1 ⊢ X = ⋃ J
Assertion ntrin ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ B = int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B

Proof

Step Hyp Ref Expression
1 clscld.1 ⊢ X = ⋃ J
2 inss1 ⊢ A ∩ B ⊆ A
3 1 ntrss ⊢ J ∈ Top ∧ A ⊆ X ∧ A ∩ B ⊆ A → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ A
4 2 3 mp3an3 ⊢ J ∈ Top ∧ A ⊆ X → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ A
5 4 3adant3 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ A
6 inss2 ⊢ A ∩ B ⊆ B
7 1 ntrss ⊢ J ∈ Top ∧ B ⊆ X ∧ A ∩ B ⊆ B → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ B
8 6 7 mp3an3 ⊢ J ∈ Top ∧ B ⊆ X → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ B
9 8 3adant2 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ B
10 5 9 ssind ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ B ⊆ int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B
11 simp1 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → J ∈ Top
12 ssinss1 ⊢ A ⊆ X → A ∩ B ⊆ X
13 12 3ad2ant2 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → A ∩ B ⊆ X
14 1 ntropn ⊢ J ∈ Top ∧ A ⊆ X → int ⁡ J ⁡ A ∈ J
15 14 3adant3 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∈ J
16 1 ntropn ⊢ J ∈ Top ∧ B ⊆ X → int ⁡ J ⁡ B ∈ J
17 16 3adant2 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ B ∈ J
18 inopn ⊢ J ∈ Top ∧ int ⁡ J ⁡ A ∈ J ∧ int ⁡ J ⁡ B ∈ J → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ∈ J
19 11 15 17 18 syl3anc ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ∈ J
20 inss1 ⊢ int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ int ⁡ J ⁡ A
21 1 ntrss2 ⊢ J ∈ Top ∧ A ⊆ X → int ⁡ J ⁡ A ⊆ A
22 21 3adant3 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ⊆ A
23 20 22 sstrid ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ A
24 inss2 ⊢ int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ int ⁡ J ⁡ B
25 1 ntrss2 ⊢ J ∈ Top ∧ B ⊆ X → int ⁡ J ⁡ B ⊆ B
26 25 3adant2 ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ B ⊆ B
27 24 26 sstrid ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ B
28 23 27 ssind ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ A ∩ B
29 1 ssntr ⊢ J ∈ Top ∧ A ∩ B ⊆ X ∧ int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ∈ J ∧ int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ A ∩ B → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ int ⁡ J ⁡ A ∩ B
30 11 13 19 28 29 syl22anc ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B ⊆ int ⁡ J ⁡ A ∩ B
31 10 30 eqssd ⊢ J ∈ Top ∧ A ⊆ X ∧ B ⊆ X → int ⁡ J ⁡ A ∩ B = int ⁡ J ⁡ A ∩ int ⁡ J ⁡ B