Metamath Proof Explorer


Theorem dfac14lem

Description: Lemma for dfac14 . By equipping S u. { P } for some P e/ S with the particular point topology, we can show that P is in the closure of S ; hence the sequence P ( x ) is in the product of the closures, and we can utilize this instance of ptcls to extract an element of the closure of X_ k e. I S . (Contributed by Mario Carneiro, 2-Sep-2015)

Ref Expression
Hypotheses dfac14lem.i ( 𝜑𝐼𝑉 )
dfac14lem.s ( ( 𝜑𝑥𝐼 ) → 𝑆𝑊 )
dfac14lem.0 ( ( 𝜑𝑥𝐼 ) → 𝑆 ≠ ∅ )
dfac14lem.p 𝑃 = 𝒫 𝑆
dfac14lem.r 𝑅 = { 𝑦 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ∣ ( 𝑃𝑦𝑦 = ( 𝑆 ∪ { 𝑃 } ) ) }
dfac14lem.j 𝐽 = ( ∏t ‘ ( 𝑥𝐼𝑅 ) )
dfac14lem.c ( 𝜑 → ( ( cls ‘ 𝐽 ) ‘ X 𝑥𝐼 𝑆 ) = X 𝑥𝐼 ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) )
Assertion dfac14lem ( 𝜑X 𝑥𝐼 𝑆 ≠ ∅ )

Proof

Step Hyp Ref Expression
1 dfac14lem.i ( 𝜑𝐼𝑉 )
2 dfac14lem.s ( ( 𝜑𝑥𝐼 ) → 𝑆𝑊 )
3 dfac14lem.0 ( ( 𝜑𝑥𝐼 ) → 𝑆 ≠ ∅ )
4 dfac14lem.p 𝑃 = 𝒫 𝑆
5 dfac14lem.r 𝑅 = { 𝑦 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ∣ ( 𝑃𝑦𝑦 = ( 𝑆 ∪ { 𝑃 } ) ) }
6 dfac14lem.j 𝐽 = ( ∏t ‘ ( 𝑥𝐼𝑅 ) )
7 dfac14lem.c ( 𝜑 → ( ( cls ‘ 𝐽 ) ‘ X 𝑥𝐼 𝑆 ) = X 𝑥𝐼 ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) )
8 eleq2w ( 𝑦 = 𝑧 → ( 𝑃𝑦𝑃𝑧 ) )
9 eqeq1 ( 𝑦 = 𝑧 → ( 𝑦 = ( 𝑆 ∪ { 𝑃 } ) ↔ 𝑧 = ( 𝑆 ∪ { 𝑃 } ) ) )
10 8 9 imbi12d ( 𝑦 = 𝑧 → ( ( 𝑃𝑦𝑦 = ( 𝑆 ∪ { 𝑃 } ) ) ↔ ( 𝑃𝑧𝑧 = ( 𝑆 ∪ { 𝑃 } ) ) ) )
11 10 5 elrab2 ( 𝑧𝑅 ↔ ( 𝑧 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ∧ ( 𝑃𝑧𝑧 = ( 𝑆 ∪ { 𝑃 } ) ) ) )
12 3 adantr ( ( ( 𝜑𝑥𝐼 ) ∧ 𝑧 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ) → 𝑆 ≠ ∅ )
13 ineq1 ( 𝑧 = ( 𝑆 ∪ { 𝑃 } ) → ( 𝑧𝑆 ) = ( ( 𝑆 ∪ { 𝑃 } ) ∩ 𝑆 ) )
14 ssun1 𝑆 ⊆ ( 𝑆 ∪ { 𝑃 } )
15 sseqin2 ( 𝑆 ⊆ ( 𝑆 ∪ { 𝑃 } ) ↔ ( ( 𝑆 ∪ { 𝑃 } ) ∩ 𝑆 ) = 𝑆 )
16 14 15 mpbi ( ( 𝑆 ∪ { 𝑃 } ) ∩ 𝑆 ) = 𝑆
17 13 16 eqtrdi ( 𝑧 = ( 𝑆 ∪ { 𝑃 } ) → ( 𝑧𝑆 ) = 𝑆 )
18 17 neeq1d ( 𝑧 = ( 𝑆 ∪ { 𝑃 } ) → ( ( 𝑧𝑆 ) ≠ ∅ ↔ 𝑆 ≠ ∅ ) )
19 12 18 syl5ibrcom ( ( ( 𝜑𝑥𝐼 ) ∧ 𝑧 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ) → ( 𝑧 = ( 𝑆 ∪ { 𝑃 } ) → ( 𝑧𝑆 ) ≠ ∅ ) )
20 19 imim2d ( ( ( 𝜑𝑥𝐼 ) ∧ 𝑧 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ) → ( ( 𝑃𝑧𝑧 = ( 𝑆 ∪ { 𝑃 } ) ) → ( 𝑃𝑧 → ( 𝑧𝑆 ) ≠ ∅ ) ) )
21 20 expimpd ( ( 𝜑𝑥𝐼 ) → ( ( 𝑧 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ∧ ( 𝑃𝑧𝑧 = ( 𝑆 ∪ { 𝑃 } ) ) ) → ( 𝑃𝑧 → ( 𝑧𝑆 ) ≠ ∅ ) ) )
22 11 21 syl5bi ( ( 𝜑𝑥𝐼 ) → ( 𝑧𝑅 → ( 𝑃𝑧 → ( 𝑧𝑆 ) ≠ ∅ ) ) )
23 22 ralrimiv ( ( 𝜑𝑥𝐼 ) → ∀ 𝑧𝑅 ( 𝑃𝑧 → ( 𝑧𝑆 ) ≠ ∅ ) )
24 snex { 𝑃 } ∈ V
25 unexg ( ( 𝑆𝑊 ∧ { 𝑃 } ∈ V ) → ( 𝑆 ∪ { 𝑃 } ) ∈ V )
26 2 24 25 sylancl ( ( 𝜑𝑥𝐼 ) → ( 𝑆 ∪ { 𝑃 } ) ∈ V )
27 ssun2 { 𝑃 } ⊆ ( 𝑆 ∪ { 𝑃 } )
28 uniexg ( 𝑆𝑊 𝑆 ∈ V )
29 pwexg ( 𝑆 ∈ V → 𝒫 𝑆 ∈ V )
30 2 28 29 3syl ( ( 𝜑𝑥𝐼 ) → 𝒫 𝑆 ∈ V )
31 4 30 eqeltrid ( ( 𝜑𝑥𝐼 ) → 𝑃 ∈ V )
32 snidg ( 𝑃 ∈ V → 𝑃 ∈ { 𝑃 } )
33 31 32 syl ( ( 𝜑𝑥𝐼 ) → 𝑃 ∈ { 𝑃 } )
34 27 33 sseldi ( ( 𝜑𝑥𝐼 ) → 𝑃 ∈ ( 𝑆 ∪ { 𝑃 } ) )
35 epttop ( ( ( 𝑆 ∪ { 𝑃 } ) ∈ V ∧ 𝑃 ∈ ( 𝑆 ∪ { 𝑃 } ) ) → { 𝑦 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ∣ ( 𝑃𝑦𝑦 = ( 𝑆 ∪ { 𝑃 } ) ) } ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) )
36 26 34 35 syl2anc ( ( 𝜑𝑥𝐼 ) → { 𝑦 ∈ 𝒫 ( 𝑆 ∪ { 𝑃 } ) ∣ ( 𝑃𝑦𝑦 = ( 𝑆 ∪ { 𝑃 } ) ) } ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) )
37 5 36 eqeltrid ( ( 𝜑𝑥𝐼 ) → 𝑅 ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) )
38 topontop ( 𝑅 ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) → 𝑅 ∈ Top )
39 37 38 syl ( ( 𝜑𝑥𝐼 ) → 𝑅 ∈ Top )
40 toponuni ( 𝑅 ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) → ( 𝑆 ∪ { 𝑃 } ) = 𝑅 )
41 37 40 syl ( ( 𝜑𝑥𝐼 ) → ( 𝑆 ∪ { 𝑃 } ) = 𝑅 )
42 14 41 sseqtrid ( ( 𝜑𝑥𝐼 ) → 𝑆 𝑅 )
43 34 41 eleqtrd ( ( 𝜑𝑥𝐼 ) → 𝑃 𝑅 )
44 eqid 𝑅 = 𝑅
45 44 elcls ( ( 𝑅 ∈ Top ∧ 𝑆 𝑅𝑃 𝑅 ) → ( 𝑃 ∈ ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ↔ ∀ 𝑧𝑅 ( 𝑃𝑧 → ( 𝑧𝑆 ) ≠ ∅ ) ) )
46 39 42 43 45 syl3anc ( ( 𝜑𝑥𝐼 ) → ( 𝑃 ∈ ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ↔ ∀ 𝑧𝑅 ( 𝑃𝑧 → ( 𝑧𝑆 ) ≠ ∅ ) ) )
47 23 46 mpbird ( ( 𝜑𝑥𝐼 ) → 𝑃 ∈ ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) )
48 47 ralrimiva ( 𝜑 → ∀ 𝑥𝐼 𝑃 ∈ ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) )
49 mptelixpg ( 𝐼𝑉 → ( ( 𝑥𝐼𝑃 ) ∈ X 𝑥𝐼 ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ↔ ∀ 𝑥𝐼 𝑃 ∈ ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ) )
50 1 49 syl ( 𝜑 → ( ( 𝑥𝐼𝑃 ) ∈ X 𝑥𝐼 ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ↔ ∀ 𝑥𝐼 𝑃 ∈ ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ) )
51 48 50 mpbird ( 𝜑 → ( 𝑥𝐼𝑃 ) ∈ X 𝑥𝐼 ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) )
52 51 ne0d ( 𝜑X 𝑥𝐼 ( ( cls ‘ 𝑅 ) ‘ 𝑆 ) ≠ ∅ )
53 37 ralrimiva ( 𝜑 → ∀ 𝑥𝐼 𝑅 ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) )
54 6 pttopon ( ( 𝐼𝑉 ∧ ∀ 𝑥𝐼 𝑅 ∈ ( TopOn ‘ ( 𝑆 ∪ { 𝑃 } ) ) ) → 𝐽 ∈ ( TopOn ‘ X 𝑥𝐼 ( 𝑆 ∪ { 𝑃 } ) ) )
55 1 53 54 syl2anc ( 𝜑𝐽 ∈ ( TopOn ‘ X 𝑥𝐼 ( 𝑆 ∪ { 𝑃 } ) ) )
56 topontop ( 𝐽 ∈ ( TopOn ‘ X 𝑥𝐼 ( 𝑆 ∪ { 𝑃 } ) ) → 𝐽 ∈ Top )
57 cls0 ( 𝐽 ∈ Top → ( ( cls ‘ 𝐽 ) ‘ ∅ ) = ∅ )
58 55 56 57 3syl ( 𝜑 → ( ( cls ‘ 𝐽 ) ‘ ∅ ) = ∅ )
59 52 7 58 3netr4d ( 𝜑 → ( ( cls ‘ 𝐽 ) ‘ X 𝑥𝐼 𝑆 ) ≠ ( ( cls ‘ 𝐽 ) ‘ ∅ ) )
60 fveq2 ( X 𝑥𝐼 𝑆 = ∅ → ( ( cls ‘ 𝐽 ) ‘ X 𝑥𝐼 𝑆 ) = ( ( cls ‘ 𝐽 ) ‘ ∅ ) )
61 60 necon3i ( ( ( cls ‘ 𝐽 ) ‘ X 𝑥𝐼 𝑆 ) ≠ ( ( cls ‘ 𝐽 ) ‘ ∅ ) → X 𝑥𝐼 𝑆 ≠ ∅ )
62 59 61 syl ( 𝜑X 𝑥𝐼 𝑆 ≠ ∅ )