Metamath Proof Explorer


Theorem blsconn

Description: An open ball in the complex numbers is simply connected. (Contributed by Mario Carneiro, 12-Feb-2015)

Ref Expression
Hypotheses blsconn.j ⊢ J = TopOpen ⁡ ℂ fld
blsconn.s ⊢ S = P ball ⁡ abs ∘ − R
blsconn.k ⊢ K = J ↾ 𝑡 S
Assertion blsconn ⊢ P ∈ ℂ ∧ R ∈ ℝ * → K ∈ SConn

Proof

Step Hyp Ref Expression
1 blsconn.j ⊢ J = TopOpen ⁡ ℂ fld
2 blsconn.s ⊢ S = P ball ⁡ abs ∘ − R
3 blsconn.k ⊢ K = J ↾ 𝑡 S
4 cnxmet ⊢ abs ∘ − ∈ ∞Met ⁡ ℂ
5 blssm ⊢ abs ∘ − ∈ ∞Met ⁡ ℂ ∧ P ∈ ℂ ∧ R ∈ ℝ * → P ball ⁡ abs ∘ − R ⊆ ℂ
6 4 5 mp3an1 ⊢ P ∈ ℂ ∧ R ∈ ℝ * → P ball ⁡ abs ∘ − R ⊆ ℂ
7 2 6 eqsstrid ⊢ P ∈ ℂ ∧ R ∈ ℝ * → S ⊆ ℂ
8 2 blcvx ⊢ P ∈ ℂ ∧ R ∈ ℝ * ∧ x ∈ S ∧ y ∈ S ∧ t ∈ 0 1 → t ⁢ x + 1 − t ⁢ y ∈ S
9 7 8 1 3 cvxsconn ⊢ P ∈ ℂ ∧ R ∈ ℝ * → K ∈ SConn