Metamath Proof Explorer


Theorem sqcn

Description: The square function on complex numbers is continuous. (Contributed by NM, 13-Jun-2007) (Proof shortened by Mario Carneiro, 5-May-2014)

Ref Expression
Hypothesis sqcn.j ⊢ J = TopOpen ⁡ ℂ fld
Assertion sqcn ⊢ x ∈ ℂ ⟼ x 2 ∈ J Cn J

Proof

Step Hyp Ref Expression
1 sqcn.j ⊢ J = TopOpen ⁡ ℂ fld
2 2nn0 ⊢ 2 ∈ ℕ 0
3 1 expcn ⊢ 2 ∈ ℕ 0 → x ∈ ℂ ⟼ x 2 ∈ J Cn J
4 2 3 ax-mp ⊢ x ∈ ℂ ⟼ x 2 ∈ J Cn J