Metamath Proof Explorer


Theorem sub1cncf

Description: Subtracting a constant is a continuous function. (Contributed by Jeff Madsen, 2-Sep-2009) (Proof shortened by Mario Carneiro, 12-Sep-2015)

Ref Expression
Hypothesis sub1cncf.1 ⊢ F = x ∈ ℂ ⟼ x − A
Assertion sub1cncf ⊢ A ∈ ℂ → F : ℂ ⟶cn ℂ

Proof

Step Hyp Ref Expression
1 sub1cncf.1 ⊢ F = x ∈ ℂ ⟼ x − A
2 eqid ⊢ TopOpen ⁡ ℂ fld = TopOpen ⁡ ℂ fld
3 2 subcn ⊢ − ∈ TopOpen ⁡ ℂ fld × t TopOpen ⁡ ℂ fld Cn TopOpen ⁡ ℂ fld
4 3 a1i ⊢ A ∈ ℂ → − ∈ TopOpen ⁡ ℂ fld × t TopOpen ⁡ ℂ fld Cn TopOpen ⁡ ℂ fld
5 eqid ⊢ x ∈ ℂ ⟼ x = x ∈ ℂ ⟼ x
6 5 idcncf ⊢ x ∈ ℂ ⟼ x : ℂ ⟶cn ℂ
7 6 a1i ⊢ A ∈ ℂ → x ∈ ℂ ⟼ x : ℂ ⟶cn ℂ
8 ssid ⊢ ℂ ⊆ ℂ
9 cncfmptc ⊢ A ∈ ℂ ∧ ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ → x ∈ ℂ ⟼ A : ℂ ⟶cn ℂ
10 8 8 9 mp3an23 ⊢ A ∈ ℂ → x ∈ ℂ ⟼ A : ℂ ⟶cn ℂ
11 2 4 7 10 cncfmpt2f ⊢ A ∈ ℂ → x ∈ ℂ ⟼ x − A : ℂ ⟶cn ℂ
12 1 11 eqeltrid ⊢ A ∈ ℂ → F : ℂ ⟶cn ℂ