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Metamath Proof Explorer

Theorem addccncf

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

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

Proof

Step Hyp Ref Expression
1 addccncf.1 ⊢ 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝑥 + 𝐴 ) )
2 eqid ⊢ ( TopOpen ‘ ℂfld ) = ( TopOpen ‘ ℂfld )
3 2 addcn ⊢ + ∈ ( ( ( 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 cncfmptid ⊢ ( ( ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ ) )
7 5 5 6 mp2an ⊢ ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ )
8 7 a1i ⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ ) )
9 cncfmptc ⊢ ( ( 𝐴 ∈ ℂ ∧ ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
10 5 5 9 mp3an23 ⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
11 2 4 8 10 cncfmpt2f ⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ ( 𝑥 + 𝐴 ) ) ∈ ( ℂ –cn→ ℂ ) )
12 1 11 eqeltrid ⊢ ( 𝐴 ∈ ℂ → 𝐹 ∈ ( ℂ –cn→ ℂ ) )