Metamath Proof Explorer


Theorem neindisj

Description: Any neighborhood of an element in the closure of a subset intersects the subset. Part of proof of Theorem 6.6 of Munkres p. 97. (Contributed by NM, 26-Feb-2007)

Ref Expression
Hypothesis neips.1 ⊢ X = ⋃ J
Assertion neindisj ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ∈ nei ⁡ J ⁡ P → N ∩ S ≠ ∅

Proof

Step Hyp Ref Expression
1 neips.1 ⊢ X = ⋃ J
2 1 clsss3 ⊢ J ∈ Top ∧ S ⊆ X → cls ⁡ J ⁡ S ⊆ X
3 2 sseld ⊢ J ∈ Top ∧ S ⊆ X → P ∈ cls ⁡ J ⁡ S → P ∈ X
4 3 impr ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S → P ∈ X
5 1 isneip ⊢ J ∈ Top ∧ P ∈ X → N ∈ nei ⁡ J ⁡ P ↔ N ⊆ X ∧ ∃ g ∈ J P ∈ g ∧ g ⊆ N
6 4 5 syldan ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S → N ∈ nei ⁡ J ⁡ P ↔ N ⊆ X ∧ ∃ g ∈ J P ∈ g ∧ g ⊆ N
7 3anass ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ↔ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S
8 1 clsndisj ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ g ∈ J ∧ P ∈ g → g ∩ S ≠ ∅
9 7 8 sylanbr ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ g ∈ J ∧ P ∈ g → g ∩ S ≠ ∅
10 9 anassrs ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ g ∈ J ∧ P ∈ g → g ∩ S ≠ ∅
11 10 adantllr ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ⊆ X ∧ g ∈ J ∧ P ∈ g → g ∩ S ≠ ∅
12 11 adantrr ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ⊆ X ∧ g ∈ J ∧ P ∈ g ∧ g ⊆ N → g ∩ S ≠ ∅
13 ssdisj ⊢ g ⊆ N ∧ N ∩ S = ∅ → g ∩ S = ∅
14 13 ex ⊢ g ⊆ N → N ∩ S = ∅ → g ∩ S = ∅
15 14 necon3d ⊢ g ⊆ N → g ∩ S ≠ ∅ → N ∩ S ≠ ∅
16 15 ad2antll ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ⊆ X ∧ g ∈ J ∧ P ∈ g ∧ g ⊆ N → g ∩ S ≠ ∅ → N ∩ S ≠ ∅
17 12 16 mpd ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ⊆ X ∧ g ∈ J ∧ P ∈ g ∧ g ⊆ N → N ∩ S ≠ ∅
18 17 rexlimdva2 ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ⊆ X → ∃ g ∈ J P ∈ g ∧ g ⊆ N → N ∩ S ≠ ∅
19 18 expimpd ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S → N ⊆ X ∧ ∃ g ∈ J P ∈ g ∧ g ⊆ N → N ∩ S ≠ ∅
20 6 19 sylbid ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S → N ∈ nei ⁡ J ⁡ P → N ∩ S ≠ ∅
21 20 exp32 ⊢ J ∈ Top → S ⊆ X → P ∈ cls ⁡ J ⁡ S → N ∈ nei ⁡ J ⁡ P → N ∩ S ≠ ∅
22 21 imp43 ⊢ J ∈ Top ∧ S ⊆ X ∧ P ∈ cls ⁡ J ⁡ S ∧ N ∈ nei ⁡ J ⁡ P → N ∩ S ≠ ∅