Metamath Proof Explorer


Theorem opncldf3

Description: The values of the converse/inverse of the open-closed bijection. (Contributed by Jeff Hankins, 27-Aug-2009) (Proof shortened by Mario Carneiro, 1-Sep-2015)

Ref Expression
Hypotheses opncldf.1 ⊢ X = ⋃ J
opncldf.2 ⊢ F = u ∈ J ⟼ X ∖ u
Assertion opncldf3 ⊢ B ∈ Clsd ⁡ J → F -1 ⁡ B = X ∖ B

Proof

Step Hyp Ref Expression
1 opncldf.1 ⊢ X = ⋃ J
2 opncldf.2 ⊢ F = u ∈ J ⟼ X ∖ u
3 cldrcl ⊢ B ∈ Clsd ⁡ J → J ∈ Top
4 1 2 opncldf1 ⊢ J ∈ Top → F : J ⟶ 1-1 onto Clsd ⁡ J ∧ F -1 = x ∈ Clsd ⁡ J ⟼ X ∖ x
5 4 simprd ⊢ J ∈ Top → F -1 = x ∈ Clsd ⁡ J ⟼ X ∖ x
6 3 5 syl ⊢ B ∈ Clsd ⁡ J → F -1 = x ∈ Clsd ⁡ J ⟼ X ∖ x
7 6 fveq1d ⊢ B ∈ Clsd ⁡ J → F -1 ⁡ B = x ∈ Clsd ⁡ J ⟼ X ∖ x ⁡ B
8 1 cldopn ⊢ B ∈ Clsd ⁡ J → X ∖ B ∈ J
9 difeq2 ⊢ x = B → X ∖ x = X ∖ B
10 eqid ⊢ x ∈ Clsd ⁡ J ⟼ X ∖ x = x ∈ Clsd ⁡ J ⟼ X ∖ x
11 9 10 fvmptg ⊢ B ∈ Clsd ⁡ J ∧ X ∖ B ∈ J → x ∈ Clsd ⁡ J ⟼ X ∖ x ⁡ B = X ∖ B
12 8 11 mpdan ⊢ B ∈ Clsd ⁡ J → x ∈ Clsd ⁡ J ⟼ X ∖ x ⁡ B = X ∖ B
13 7 12 eqtrd ⊢ B ∈ Clsd ⁡ J → F -1 ⁡ B = X ∖ B