Metamath Proof Explorer


Theorem divccn

Description: Division by a nonzero constant is a continuous operation. (Contributed by Mario Carneiro, 5-May-2014)

Ref Expression
Hypothesis expcn.j 𝐽 = ( TopOpen ‘ ℂfld )
Assertion divccn ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 ∈ ℂ ↦ ( 𝑥 / 𝐴 ) ) ∈ ( 𝐽 Cn 𝐽 ) )

Proof

Step Hyp Ref Expression
1 expcn.j 𝐽 = ( TopOpen ‘ ℂfld )
2 divrec ( ( 𝑥 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 / 𝐴 ) = ( 𝑥 · ( 1 / 𝐴 ) ) )
3 2 3expb ( ( 𝑥 ∈ ℂ ∧ ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ) → ( 𝑥 / 𝐴 ) = ( 𝑥 · ( 1 / 𝐴 ) ) )
4 3 ancoms ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ 𝑥 ∈ ℂ ) → ( 𝑥 / 𝐴 ) = ( 𝑥 · ( 1 / 𝐴 ) ) )
5 4 mpteq2dva ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 ∈ ℂ ↦ ( 𝑥 / 𝐴 ) ) = ( 𝑥 ∈ ℂ ↦ ( 𝑥 · ( 1 / 𝐴 ) ) ) )
6 1 cnfldtopon 𝐽 ∈ ( TopOn ‘ ℂ )
7 6 a1i ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → 𝐽 ∈ ( TopOn ‘ ℂ ) )
8 7 cnmptid ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( 𝐽 Cn 𝐽 ) )
9 reccl ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℂ )
10 7 7 9 cnmptc ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 ∈ ℂ ↦ ( 1 / 𝐴 ) ) ∈ ( 𝐽 Cn 𝐽 ) )
11 1 mulcn · ∈ ( ( 𝐽 ×t 𝐽 ) Cn 𝐽 )
12 11 a1i ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → · ∈ ( ( 𝐽 ×t 𝐽 ) Cn 𝐽 ) )
13 7 8 10 12 cnmpt12f ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 ∈ ℂ ↦ ( 𝑥 · ( 1 / 𝐴 ) ) ) ∈ ( 𝐽 Cn 𝐽 ) )
14 5 13 eqeltrd ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝑥 ∈ ℂ ↦ ( 𝑥 / 𝐴 ) ) ∈ ( 𝐽 Cn 𝐽 ) )