“…Again, we must have points from U 1 (say u 1 2 ) and from U 2 (say u 2 2 ) inside the other arc determined by ±u 3 1 . Again, for the same reason, U i ∩ relint(B(±u 3 1 , π/3)) = ∅, i = 1, 2. Assuming that u 3 1 = e 1 , Proposition 1.1(4) implies that U i ∩ {x ∈ S 1 : x 2 ≤ 0} = ∅, i = 1, 2.…”