Metamath Proof Explorer


Theorem dnsconst

Description: If a continuous mapping to a T_1 space is constant on a dense subset, it is constant on the entire space. Note that ( ( clsJ )A ) = X means " A is dense in X " and A C_ (`' F " { P } ) means " F is constant on A ` " (see funconstss ). (Contributed by NM, 15-Mar-2007) (Proof shortened by Mario Carneiro, 21-Aug-2015)

Ref Expression
Hypotheses dnsconst.1 ⊢ X = ⋃ J
dnsconst.2 ⊢ Y = ⋃ K
Assertion dnsconst ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → F : X ⟶ P

Proof

Step Hyp Ref Expression
1 dnsconst.1 ⊢ X = ⋃ J
2 dnsconst.2 ⊢ Y = ⋃ K
3 simplr ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → F ∈ J Cn K
4 1 2 cnf ⊢ F ∈ J Cn K → F : X ⟶ Y
5 ffn ⊢ F : X ⟶ Y → F Fn X
6 3 4 5 3syl ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → F Fn X
7 simpr3 ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → cls ⁡ J ⁡ A = X
8 simpll ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → K ∈ Fre
9 simpr1 ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → P ∈ Y
10 2 t1sncld ⊢ K ∈ Fre ∧ P ∈ Y → P ∈ Clsd ⁡ K
11 8 9 10 syl2anc ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → P ∈ Clsd ⁡ K
12 cnclima ⊢ F ∈ J Cn K ∧ P ∈ Clsd ⁡ K → F -1 P ∈ Clsd ⁡ J
13 3 11 12 syl2anc ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → F -1 P ∈ Clsd ⁡ J
14 simpr2 ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → A ⊆ F -1 P
15 1 clsss2 ⊢ F -1 P ∈ Clsd ⁡ J ∧ A ⊆ F -1 P → cls ⁡ J ⁡ A ⊆ F -1 P
16 13 14 15 syl2anc ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → cls ⁡ J ⁡ A ⊆ F -1 P
17 7 16 eqsstrrd ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → X ⊆ F -1 P
18 fconst3 ⊢ F : X ⟶ P ↔ F Fn X ∧ X ⊆ F -1 P
19 6 17 18 sylanbrc ⊢ K ∈ Fre ∧ F ∈ J Cn K ∧ P ∈ Y ∧ A ⊆ F -1 P ∧ cls ⁡ J ⁡ A = X → F : X ⟶ P