Metamath Proof Explorer

Theorem cjcn2

Description: The complex conjugate function is continuous. (Contributed by Mario Carneiro, 9-Feb-2014)

Ref Expression
Assertion cjcn2 ( ( 𝐴 ∈ ℂ ∧ 𝑥 ∈ ℝ+ ) → ∃ 𝑦 ∈ ℝ+ ∀ 𝑧 ∈ ℂ ( ( abs ‘ ( 𝑧 − 𝐴 ) ) < 𝑦 → ( abs ‘ ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ) < 𝑥 ) )

Proof

Step Hyp Ref Expression
1 cjf ⊢ ∗ : ℂ ⟶ ℂ
2 cjcl ⊢ ( 𝑧 ∈ ℂ → ( ∗ ‘ 𝑧 ) ∈ ℂ )
3 cjcl ⊢ ( 𝐴 ∈ ℂ → ( ∗ ‘ 𝐴 ) ∈ ℂ )
4 subcl ⊢ ( ( ( ∗ ‘ 𝑧 ) ∈ ℂ ∧ ( ∗ ‘ 𝐴 ) ∈ ℂ ) → ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ∈ ℂ )
5 2 3 4 syl2an ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ∈ ℂ )
6 5 abscld ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( abs ‘ ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ) ∈ ℝ )
7 cjsub ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ∗ ‘ ( 𝑧 − 𝐴 ) ) = ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) )
8 7 fveq2d ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( abs ‘ ( ∗ ‘ ( 𝑧 − 𝐴 ) ) ) = ( abs ‘ ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ) )
9 subcl ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝑧 − 𝐴 ) ∈ ℂ )
10 9 abscjd ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( abs ‘ ( ∗ ‘ ( 𝑧 − 𝐴 ) ) ) = ( abs ‘ ( 𝑧 − 𝐴 ) ) )
11 8 10 eqtr3d ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( abs ‘ ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ) = ( abs ‘ ( 𝑧 − 𝐴 ) ) )
12 6 11 eqled ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( abs ‘ ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ) ≤ ( abs ‘ ( 𝑧 − 𝐴 ) ) )
13 1 12 cn1lem ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝑥 ∈ ℝ+ ) → ∃ 𝑦 ∈ ℝ+ ∀ 𝑧 ∈ ℂ ( ( abs ‘ ( 𝑧 − 𝐴 ) ) < 𝑦 → ( abs ‘ ( ( ∗ ‘ 𝑧 ) − ( ∗ ‘ 𝐴 ) ) ) < 𝑥 ) )