Metamath Proof Explorer


Theorem subcncf

Description: The subtraction of two continuous complex functions is continuous. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses subcncf.a ⊢ ( 𝜑 → ( 𝑥 ∈ 𝑋 ↦ 𝐴 ) ∈ ( 𝑋 –cn→ ℂ ) )
subcncf.b ⊢ ( 𝜑 → ( 𝑥 ∈ 𝑋 ↦ 𝐵 ) ∈ ( 𝑋 –cn→ ℂ ) )
Assertion subcncf ( 𝜑 → ( 𝑥 ∈ 𝑋 ↦ ( 𝐴 − 𝐵 ) ) ∈ ( 𝑋 –cn→ ℂ ) )

Proof

Step Hyp Ref Expression
1 subcncf.a ⊢ ( 𝜑 → ( 𝑥 ∈ 𝑋 ↦ 𝐴 ) ∈ ( 𝑋 –cn→ ℂ ) )
2 subcncf.b ⊢ ( 𝜑 → ( 𝑥 ∈ 𝑋 ↦ 𝐵 ) ∈ ( 𝑋 –cn→ ℂ ) )
3 eqid ⊢ ( TopOpen ‘ ℂfld ) = ( TopOpen ‘ ℂfld )
4 3 subcn ⊢ − ∈ ( ( ( TopOpen ‘ ℂfld ) ×t ( TopOpen ‘ ℂfld ) ) Cn ( TopOpen ‘ ℂfld ) )
5 4 a1i ⊢ ( 𝜑 → − ∈ ( ( ( TopOpen ‘ ℂfld ) ×t ( TopOpen ‘ ℂfld ) ) Cn ( TopOpen ‘ ℂfld ) ) )
6 3 5 1 2 cncfmpt2f ⊢ ( 𝜑 → ( 𝑥 ∈ 𝑋 ↦ ( 𝐴 − 𝐵 ) ) ∈ ( 𝑋 –cn→ ℂ ) )