Problem. A set is said to be pathwise connected if any two points in can be joined by (piecewise-smooth) curve entirely contained in . The purpose of this exercise is to prove that an open set is pathwise connected if and only if is connected. a) Suppose first that is open and pathwise connected, and that it can be written as =12 where 1 and 2 are disjoint non-empty open sets. Choose two points w1 to w2. Consider the parametrization z : [0,1] of this curve with z(0) = w1 and z(1) = w2, and let t = sup 0t1 {t : z(s) 1 for all 0 s < t} Arrive at a contradiction by considering the point z(t). b) Conversely, suppose that is open and connected. Fix a point w and let 1 denote the set of all points that can be joined to w by a curve contained in . Also, let 2 denote the set of all points that cannot be joined to w by a curve in . Prove that both 1 and 2 are open, disjoint and their union is . Finally, since 1 is non-empty conclude that = 1 as desired