Metamath Proof Explorer


Theorem cncfcn1

Description: Relate complex function continuity to topological continuity. (Contributed by Paul Chapman, 28-Nov-2007) (Revised by Mario Carneiro, 7-Sep-2015)

Ref Expression
Hypothesis cncfcn1.1 𝐽 = ( TopOpen ‘ ℂfld )
Assertion cncfcn1 ( ℂ –cn→ ℂ ) = ( 𝐽 Cn 𝐽 )

Proof

Step Hyp Ref Expression
1 cncfcn1.1 𝐽 = ( TopOpen ‘ ℂfld )
2 ssid ℂ ⊆ ℂ
3 1 cnfldtopon 𝐽 ∈ ( TopOn ‘ ℂ )
4 3 toponrestid 𝐽 = ( 𝐽t ℂ )
5 1 4 4 cncfcn ( ( ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( ℂ –cn→ ℂ ) = ( 𝐽 Cn 𝐽 ) )
6 2 2 5 mp2an ( ℂ –cn→ ℂ ) = ( 𝐽 Cn 𝐽 )