Metamath Proof Explorer


Theorem sub2cncf

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

Ref Expression
Hypothesis sub2cncf.1 ⊢ 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝐴 − 𝑥 ) )
Assertion sub2cncf ( 𝐴 ∈ ℂ → 𝐹 ∈ ( ℂ –cn→ ℂ ) )

Proof

Step Hyp Ref Expression
1 sub2cncf.1 ⊢ 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝐴 − 𝑥 ) )
2 eqid ⊢ ( TopOpen ‘ ℂfld ) = ( TopOpen ‘ ℂfld )
3 2 subcn ⊢ − ∈ ( ( ( TopOpen ‘ ℂfld ) ×t ( TopOpen ‘ ℂfld ) ) Cn ( TopOpen ‘ ℂfld ) )
4 3 a1i ⊢ ( 𝐴 ∈ ℂ → − ∈ ( ( ( TopOpen ‘ ℂfld ) ×t ( TopOpen ‘ ℂfld ) ) Cn ( TopOpen ‘ ℂfld ) ) )
5 ssid ⊢ ℂ ⊆ ℂ
6 cncfmptc ⊢ ( ( 𝐴 ∈ ℂ ∧ ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
7 5 5 6 mp3an23 ⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
8 eqid ⊢ ( 𝑥 ∈ ℂ ↦ 𝑥 ) = ( 𝑥 ∈ ℂ ↦ 𝑥 )
9 8 idcncf ⊢ ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ )
10 9 a1i ⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ ) )
11 2 4 7 10 cncfmpt2f ⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ ( 𝐴 − 𝑥 ) ) ∈ ( ℂ –cn→ ℂ ) )
12 1 11 eqeltrid ⊢ ( 𝐴 ∈ ℂ → 𝐹 ∈ ( ℂ –cn→ ℂ ) )