Metamath Proof Explorer


Theorem cncfmet

Description: Relate complex function continuity to metric space continuity. (Contributed by Paul Chapman, 26-Nov-2007) (Revised by Mario Carneiro, 7-Sep-2015)

Ref Expression
Hypotheses cncfmet.1 ⊢ C = abs ∘ − ↾ A × A
cncfmet.2 ⊢ D = abs ∘ − ↾ B × B
cncfmet.3 ⊢ J = MetOpen ⁡ C
cncfmet.4 ⊢ K = MetOpen ⁡ D
Assertion cncfmet ⊢ A ⊆ ℂ ∧ B ⊆ ℂ → A ⟶cn B = J Cn K

Proof

Step Hyp Ref Expression
1 cncfmet.1 ⊢ C = abs ∘ − ↾ A × A
2 cncfmet.2 ⊢ D = abs ∘ − ↾ B × B
3 cncfmet.3 ⊢ J = MetOpen ⁡ C
4 cncfmet.4 ⊢ K = MetOpen ⁡ D
5 simplll ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → A ⊆ ℂ
6 simprl ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → x ∈ A
7 simprr ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → w ∈ A
8 1 oveqi ⊢ x C w = x abs ∘ − ↾ A × A w
9 ovres ⊢ x ∈ A ∧ w ∈ A → x abs ∘ − ↾ A × A w = x abs ∘ − w
10 8 9 eqtrid ⊢ x ∈ A ∧ w ∈ A → x C w = x abs ∘ − w
11 10 ad2ant2l ⊢ A ⊆ ℂ ∧ x ∈ A ∧ A ⊆ ℂ ∧ w ∈ A → x C w = x abs ∘ − w
12 ssel2 ⊢ A ⊆ ℂ ∧ x ∈ A → x ∈ ℂ
13 ssel2 ⊢ A ⊆ ℂ ∧ w ∈ A → w ∈ ℂ
14 eqid ⊢ abs ∘ − = abs ∘ −
15 14 cnmetdval ⊢ x ∈ ℂ ∧ w ∈ ℂ → x abs ∘ − w = x − w
16 12 13 15 syl2an ⊢ A ⊆ ℂ ∧ x ∈ A ∧ A ⊆ ℂ ∧ w ∈ A → x abs ∘ − w = x − w
17 11 16 eqtrd ⊢ A ⊆ ℂ ∧ x ∈ A ∧ A ⊆ ℂ ∧ w ∈ A → x C w = x − w
18 5 6 5 7 17 syl22anc ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → x C w = x − w
19 18 breq1d ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → x C w < z ↔ x − w < z
20 ffvelcdm ⊢ f : A ⟶ B ∧ x ∈ A → f ⁡ x ∈ B
21 20 ad2ant2lr ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ x ∈ B
22 ffvelcdm ⊢ f : A ⟶ B ∧ w ∈ A → f ⁡ w ∈ B
23 22 ad2ant2l ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ w ∈ B
24 2 oveqi ⊢ f ⁡ x D f ⁡ w = f ⁡ x abs ∘ − ↾ B × B f ⁡ w
25 ovres ⊢ f ⁡ x ∈ B ∧ f ⁡ w ∈ B → f ⁡ x abs ∘ − ↾ B × B f ⁡ w = f ⁡ x abs ∘ − f ⁡ w
26 24 25 eqtrid ⊢ f ⁡ x ∈ B ∧ f ⁡ w ∈ B → f ⁡ x D f ⁡ w = f ⁡ x abs ∘ − f ⁡ w
27 21 23 26 syl2anc ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ x D f ⁡ w = f ⁡ x abs ∘ − f ⁡ w
28 simpllr ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → B ⊆ ℂ
29 28 21 sseldd ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ x ∈ ℂ
30 28 23 sseldd ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ w ∈ ℂ
31 14 cnmetdval ⊢ f ⁡ x ∈ ℂ ∧ f ⁡ w ∈ ℂ → f ⁡ x abs ∘ − f ⁡ w = f ⁡ x − f ⁡ w
32 29 30 31 syl2anc ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ x abs ∘ − f ⁡ w = f ⁡ x − f ⁡ w
33 27 32 eqtrd ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ x D f ⁡ w = f ⁡ x − f ⁡ w
34 33 breq1d ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → f ⁡ x D f ⁡ w < y ↔ f ⁡ x − f ⁡ w < y
35 19 34 imbi12d ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → x C w < z → f ⁡ x D f ⁡ w < y ↔ x − w < z → f ⁡ x − f ⁡ w < y
36 35 anassrs ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A ∧ w ∈ A → x C w < z → f ⁡ x D f ⁡ w < y ↔ x − w < z → f ⁡ x − f ⁡ w < y
37 36 ralbidva ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A → ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y ↔ ∀ w ∈ A x − w < z → f ⁡ x − f ⁡ w < y
38 37 rexbidv ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A → ∃ z ∈ ℝ + ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y ↔ ∃ z ∈ ℝ + ∀ w ∈ A x − w < z → f ⁡ x − f ⁡ w < y
39 38 ralbidv ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B ∧ x ∈ A → ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y ↔ ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x − w < z → f ⁡ x − f ⁡ w < y
40 39 ralbidva ⊢ A ⊆ ℂ ∧ B ⊆ ℂ ∧ f : A ⟶ B → ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y ↔ ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x − w < z → f ⁡ x − f ⁡ w < y
41 40 pm5.32da ⊢ A ⊆ ℂ ∧ B ⊆ ℂ → f : A ⟶ B ∧ ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y ↔ f : A ⟶ B ∧ ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x − w < z → f ⁡ x − f ⁡ w < y
42 cnxmet ⊢ abs ∘ − ∈ ∞Met ⁡ ℂ
43 xmetres2 ⊢ abs ∘ − ∈ ∞Met ⁡ ℂ ∧ A ⊆ ℂ → abs ∘ − ↾ A × A ∈ ∞Met ⁡ A
44 42 43 mpan ⊢ A ⊆ ℂ → abs ∘ − ↾ A × A ∈ ∞Met ⁡ A
45 1 44 eqeltrid ⊢ A ⊆ ℂ → C ∈ ∞Met ⁡ A
46 xmetres2 ⊢ abs ∘ − ∈ ∞Met ⁡ ℂ ∧ B ⊆ ℂ → abs ∘ − ↾ B × B ∈ ∞Met ⁡ B
47 42 46 mpan ⊢ B ⊆ ℂ → abs ∘ − ↾ B × B ∈ ∞Met ⁡ B
48 2 47 eqeltrid ⊢ B ⊆ ℂ → D ∈ ∞Met ⁡ B
49 3 4 metcn ⊢ C ∈ ∞Met ⁡ A ∧ D ∈ ∞Met ⁡ B → f ∈ J Cn K ↔ f : A ⟶ B ∧ ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y
50 45 48 49 syl2an ⊢ A ⊆ ℂ ∧ B ⊆ ℂ → f ∈ J Cn K ↔ f : A ⟶ B ∧ ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x C w < z → f ⁡ x D f ⁡ w < y
51 elcncf ⊢ A ⊆ ℂ ∧ B ⊆ ℂ → f : A ⟶cn B ↔ f : A ⟶ B ∧ ∀ x ∈ A ∀ y ∈ ℝ + ∃ z ∈ ℝ + ∀ w ∈ A x − w < z → f ⁡ x − f ⁡ w < y
52 41 50 51 3bitr4rd ⊢ A ⊆ ℂ ∧ B ⊆ ℂ → f : A ⟶cn B ↔ f ∈ J Cn K
53 52 eqrdv ⊢ A ⊆ ℂ ∧ B ⊆ ℂ → A ⟶cn B = J Cn K