Metamath Proof Explorer


Theorem ulm0

Description: Every function converges uniformly on the empty set. (Contributed by Mario Carneiro, 3-Mar-2015)

Ref Expression
Hypotheses ulm0.z ⊢ Z = ℤ ≥ M
ulm0.m ⊢ φ → M ∈ ℤ
ulm0.f ⊢ φ → F : Z ⟶ ℂ S
ulm0.g ⊢ φ → G : S ⟶ ℂ
Assertion ulm0 ⊢ φ ∧ S = ∅ → F ⇝u ⁡ S G

Proof

Step Hyp Ref Expression
1 ulm0.z ⊢ Z = ℤ ≥ M
2 ulm0.m ⊢ φ → M ∈ ℤ
3 ulm0.f ⊢ φ → F : Z ⟶ ℂ S
4 ulm0.g ⊢ φ → G : S ⟶ ℂ
5 uzid ⊢ M ∈ ℤ → M ∈ ℤ ≥ M
6 2 5 syl ⊢ φ → M ∈ ℤ ≥ M
7 6 1 eleqtrrdi ⊢ φ → M ∈ Z
8 7 ne0d ⊢ φ → Z ≠ ∅
9 ral0 ⊢ ∀ z ∈ ∅ F ⁡ k ⁡ z − G ⁡ z < x
10 simpr ⊢ φ ∧ S = ∅ → S = ∅
11 10 raleqdv ⊢ φ ∧ S = ∅ → ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x ↔ ∀ z ∈ ∅ F ⁡ k ⁡ z − G ⁡ z < x
12 9 11 mpbiri ⊢ φ ∧ S = ∅ → ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
13 12 ralrimivw ⊢ φ ∧ S = ∅ → ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
14 13 ralrimivw ⊢ φ ∧ S = ∅ → ∀ j ∈ Z ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
15 r19.2z ⊢ Z ≠ ∅ ∧ ∀ j ∈ Z ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x → ∃ j ∈ Z ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
16 8 14 15 syl2an2r ⊢ φ ∧ S = ∅ → ∃ j ∈ Z ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
17 16 ralrimivw ⊢ φ ∧ S = ∅ → ∀ x ∈ ℝ + ∃ j ∈ Z ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
18 2 adantr ⊢ φ ∧ S = ∅ → M ∈ ℤ
19 3 adantr ⊢ φ ∧ S = ∅ → F : Z ⟶ ℂ S
20 eqidd ⊢ φ ∧ S = ∅ ∧ k ∈ Z ∧ z ∈ S → F ⁡ k ⁡ z = F ⁡ k ⁡ z
21 eqidd ⊢ φ ∧ S = ∅ ∧ z ∈ S → G ⁡ z = G ⁡ z
22 4 adantr ⊢ φ ∧ S = ∅ → G : S ⟶ ℂ
23 0ex ⊢ ∅ ∈ V
24 10 23 eqeltrdi ⊢ φ ∧ S = ∅ → S ∈ V
25 1 18 19 20 21 22 24 ulm2 ⊢ φ ∧ S = ∅ → F ⇝u ⁡ S G ↔ ∀ x ∈ ℝ + ∃ j ∈ Z ∀ k ∈ ℤ ≥ j ∀ z ∈ S F ⁡ k ⁡ z − G ⁡ z < x
26 17 25 mpbird ⊢ φ ∧ S = ∅ → F ⇝u ⁡ S G