Theorem sub1cncf

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

	Ref	Expression
Hypothesis	sub1cncf.1	⊢ 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝑥 − 𝐴 ) )
Assertion	sub1cncf	⊢ ( 𝐴 ∈ ℂ → 𝐹 ∈ ( ℂ –cn→ ℂ ) )

Proof

Step	Hyp	Ref	Expression
1		sub1cncf.1	⊢ 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝑥 − 𝐴 ) )
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		eqid	⊢ ( 𝑥 ∈ ℂ ↦ 𝑥 ) = ( 𝑥 ∈ ℂ ↦ 𝑥 )
6	5	idcncf	⊢ ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ )
7	6	a1i	⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ ) )
8		ssid	⊢ ℂ ⊆ ℂ
9		cncfmptc	⊢ ( ( 𝐴 ∈ ℂ ∧ ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
10	8 8 9	mp3an23	⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
11	2 4 7 10	cncfmpt2f	⊢ ( 𝐴 ∈ ℂ → ( 𝑥 ∈ ℂ ↦ ( 𝑥 − 𝐴 ) ) ∈ ( ℂ –cn→ ℂ ) )
12	1 11	eqeltrid	⊢ ( 𝐴 ∈ ℂ → 𝐹 ∈ ( ℂ –cn→ ℂ ) )