Metamath Proof Explorer


Theorem abscn2

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

Ref Expression
Assertion abscn2 ( ( 𝐴 ∈ ℂ ∧ 𝑥 ∈ ℝ+ ) → ∃ 𝑦 ∈ ℝ+𝑧 ∈ ℂ ( ( abs ‘ ( 𝑧𝐴 ) ) < 𝑦 → ( abs ‘ ( ( abs ‘ 𝑧 ) − ( abs ‘ 𝐴 ) ) ) < 𝑥 ) )

Proof

Step Hyp Ref Expression
1 absf abs : ℂ ⟶ ℝ
2 ax-resscn ℝ ⊆ ℂ
3 fss ( ( abs : ℂ ⟶ ℝ ∧ ℝ ⊆ ℂ ) → abs : ℂ ⟶ ℂ )
4 1 2 3 mp2an abs : ℂ ⟶ ℂ
5 abs2difabs ( ( 𝑧 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( abs ‘ ( ( abs ‘ 𝑧 ) − ( abs ‘ 𝐴 ) ) ) ≤ ( abs ‘ ( 𝑧𝐴 ) ) )
6 4 5 cn1lem ( ( 𝐴 ∈ ℂ ∧ 𝑥 ∈ ℝ+ ) → ∃ 𝑦 ∈ ℝ+𝑧 ∈ ℂ ( ( abs ‘ ( 𝑧𝐴 ) ) < 𝑦 → ( abs ‘ ( ( abs ‘ 𝑧 ) − ( abs ‘ 𝐴 ) ) ) < 𝑥 ) )