Metamath Proof Explorer


Theorem ntreq0

Description: Two ways to say that a subset has an empty interior. (Contributed by NM, 3-Oct-2007) (Revised by Mario Carneiro, 11-Nov-2013)

Ref Expression
Hypothesis clscld.1 𝑋 = 𝐽
Assertion ntreq0 ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( ( int ‘ 𝐽 ) ‘ 𝑆 ) = ∅ ↔ ∀ 𝑥𝐽 ( 𝑥𝑆𝑥 = ∅ ) ) )

Proof

Step Hyp Ref Expression
1 clscld.1 𝑋 = 𝐽
2 1 ntrval ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( int ‘ 𝐽 ) ‘ 𝑆 ) = ( 𝐽 ∩ 𝒫 𝑆 ) )
3 2 eqeq1d ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( ( int ‘ 𝐽 ) ‘ 𝑆 ) = ∅ ↔ ( 𝐽 ∩ 𝒫 𝑆 ) = ∅ ) )
4 neq0 ( ¬ ( 𝐽 ∩ 𝒫 𝑆 ) = ∅ ↔ ∃ 𝑦 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) )
5 4 con1bii ( ¬ ∃ 𝑦 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ( 𝐽 ∩ 𝒫 𝑆 ) = ∅ )
6 ancom ( ( 𝑦𝑥𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ) ↔ ( 𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ∧ 𝑦𝑥 ) )
7 elin ( 𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ( 𝑥𝐽𝑥 ∈ 𝒫 𝑆 ) )
8 7 anbi1i ( ( 𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ∧ 𝑦𝑥 ) ↔ ( ( 𝑥𝐽𝑥 ∈ 𝒫 𝑆 ) ∧ 𝑦𝑥 ) )
9 anass ( ( ( 𝑥𝐽𝑥 ∈ 𝒫 𝑆 ) ∧ 𝑦𝑥 ) ↔ ( 𝑥𝐽 ∧ ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ) )
10 6 8 9 3bitri ( ( 𝑦𝑥𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ) ↔ ( 𝑥𝐽 ∧ ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ) )
11 10 exbii ( ∃ 𝑥 ( 𝑦𝑥𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ) ↔ ∃ 𝑥 ( 𝑥𝐽 ∧ ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ) )
12 eluni ( 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ∃ 𝑥 ( 𝑦𝑥𝑥 ∈ ( 𝐽 ∩ 𝒫 𝑆 ) ) )
13 df-rex ( ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ↔ ∃ 𝑥 ( 𝑥𝐽 ∧ ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ) )
14 11 12 13 3bitr4i ( 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) )
15 14 exbii ( ∃ 𝑦 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ∃ 𝑦𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) )
16 rexcom4 ( ∃ 𝑥𝐽𝑦 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ↔ ∃ 𝑦𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) )
17 19.42v ( ∃ 𝑦 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ↔ ( 𝑥 ∈ 𝒫 𝑆 ∧ ∃ 𝑦 𝑦𝑥 ) )
18 17 rexbii ( ∃ 𝑥𝐽𝑦 ( 𝑥 ∈ 𝒫 𝑆𝑦𝑥 ) ↔ ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 ∧ ∃ 𝑦 𝑦𝑥 ) )
19 15 16 18 3bitr2i ( ∃ 𝑦 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 ∧ ∃ 𝑦 𝑦𝑥 ) )
20 19 notbii ( ¬ ∃ 𝑦 𝑦 ( 𝐽 ∩ 𝒫 𝑆 ) ↔ ¬ ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 ∧ ∃ 𝑦 𝑦𝑥 ) )
21 5 20 bitr3i ( ( 𝐽 ∩ 𝒫 𝑆 ) = ∅ ↔ ¬ ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 ∧ ∃ 𝑦 𝑦𝑥 ) )
22 ralinexa ( ∀ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 → ¬ ∃ 𝑦 𝑦𝑥 ) ↔ ¬ ∃ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 ∧ ∃ 𝑦 𝑦𝑥 ) )
23 velpw ( 𝑥 ∈ 𝒫 𝑆𝑥𝑆 )
24 neq0 ( ¬ 𝑥 = ∅ ↔ ∃ 𝑦 𝑦𝑥 )
25 24 con1bii ( ¬ ∃ 𝑦 𝑦𝑥𝑥 = ∅ )
26 23 25 imbi12i ( ( 𝑥 ∈ 𝒫 𝑆 → ¬ ∃ 𝑦 𝑦𝑥 ) ↔ ( 𝑥𝑆𝑥 = ∅ ) )
27 26 ralbii ( ∀ 𝑥𝐽 ( 𝑥 ∈ 𝒫 𝑆 → ¬ ∃ 𝑦 𝑦𝑥 ) ↔ ∀ 𝑥𝐽 ( 𝑥𝑆𝑥 = ∅ ) )
28 21 22 27 3bitr2i ( ( 𝐽 ∩ 𝒫 𝑆 ) = ∅ ↔ ∀ 𝑥𝐽 ( 𝑥𝑆𝑥 = ∅ ) )
29 3 28 bitrdi ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( ( ( int ‘ 𝐽 ) ‘ 𝑆 ) = ∅ ↔ ∀ 𝑥𝐽 ( 𝑥𝑆𝑥 = ∅ ) ) )