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 ‘ ℂfld )
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 ‘ ℂfld )
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 )