- A set Ω is said to be pathwise connected if any two points in Ω can be joined by a (piecewise-smooth) curve entirely contained in Ω. The purpose of this exercise is to prove that an open set Ω is pathwise connected if and only if Ω is connected.
- Suppose first that Ω is open and pathwise connected, and that it can be written as \(\Omega = \Omega _ { 1 } \cup \Omega _ { 2 }\) where \(\Omega _ { 1 }\) and \(\Omega _ { 2 }\) are disjoint non-empty open sets. Choose two points \(w _ { 1 } \in \Omega _ { 1 }\) and \(w _ { 2 } \in \Omega _ { 2 }\) and let \(\gamma\) denote a curve in Ω joining \(w _ { 1 }\) to \(w _ { 2 }\) . Consider a parametrization \(z : [ 0 , 1 ] \to \Omega\) of this curve with \(z ( 0 ) = w _ { 1 }\) and \(z ( 1 ) = w _ { 2 }\) , and let
\[ t ^ {*} = \sup _ {0 \leq t \leq 1} \{t: z (s) \in \Omega_ {1} \text { for all } 0 \leq s < t \}. \]
Arrive at a contradiction by considering the point \(z ( t ^ { * } )\)
- Conversely, suppose that Ω is open and connected. Fix a point \(w \in \Omega\) and let \(\Omega _ { 1 } \subset \Omega\) denote the set of all points that can be joined to w by a curve contained in Ω. Also, let \(\Omega _ { 2 } \subset \Omega\) denote the set of all points that cannot be joined to w by a curve in Ω. Prove that both \(\Omega _ { 1 }\) and \(\Omega _ { 2 }\) are open, disjoint and their union is Ω. Finally, since \(\Omega _ { 1 }\) is non-empty (why?) conclude that \(\Omega = \Omega _ { 1 }\) as desired.
The proof actually shows that the regularity and type of curves we used to define pathwise connectedness can be relaxed without changing the equivalence between the two definitions when Ω is open. For instance, we may take all curves to be continuous, or simply polygonal lines.
Solution
1. (a) An open pathwise connected set \(\Omega\) is connected.
1.1. With \(\Omega=\Omega_1\sqcup\Omega_2\), \(w_1,w_2\), \(z\), and \(t^*\) as in the statement, \(z(t^*) \notin \Omega_1\).
Suppose \(z(t^*) \in \Omega_1\). Since \(z(1)=w_2\in\Omega_2\) and \(\Omega_1\cap\Omega_2=\varnothing\), \(t^*<1\). By continuity of \(z\) and openness of \(\Omega_1\) there is \(\varepsilon > 0\) with \(z(t) \in \Omega_1\) for all \(t \in (t^* - \varepsilon, t^* + \varepsilon)\cap[0,1]\). Together with the definition of \(t^*\), this gives \(z(s)\in\Omega_1\) for all \(0\le s<\min(t^*+\varepsilon,1)\), so \(t^*\) is not the supremum.
1.2. \(z(t^*) \notin \Omega_2\).
Since \(z(0)=w_1\in\Omega_1\) and \(\Omega_1\) is open, \(t^*>0\), and by definition of \(t^*\) there are \(s<t^*\) arbitrarily close to \(t^*\) with \(z(s) \in \Omega_1\). If \(z(t^*) \in \Omega_2\), then openness of \(\Omega_2\) and continuity give \(\varepsilon>0\) with \(z(s)\in\Omega_2\) for \(\abs{s-t^*}<\varepsilon\), contradicting \(\Omega_1 \cap \Omega_2 = \varnothing\).
1.3. Q.E.D.
By steps 1 and 2, \(z(t^*)\in\Omega\) lies in neither \(\Omega_1\) nor \(\Omega_2\), which contradicts \(\Omega=\Omega_1\cup\Omega_2\). Hence \(\Omega\) is not a disjoint union of two nonempty open sets, so \(\Omega\) is connected.
2. (b) An open connected set \(\Omega\) is pathwise connected.
2.1. With \(w\), \(\Omega_1\), \(\Omega_2\) as in the statement, \(\Omega_1\) is open.
Let \(p \in \Omega_1\) be joined to \(w\) by a curve \(\gamma\) in \(\Omega\), and let \(D \subset \Omega\) be a disc centered at \(p\). Every \(q \in D\) is joined to \(w\) by \(\gamma\) followed by the segment from \(p\) to \(q\), which lies in \(D\). Hence \(D \subset \Omega_1\).
2.2. \(\Omega_2\) is open.
Let \(p \in \Omega_2\) and let \(D \subset \Omega\) be a disc centered at \(p\). If some \(q \in D\) lay in \(\Omega_1\), then the segment from \(p\) to \(q\) followed by a curve from \(q\) to \(w\) would join \(p\) to \(w\) in \(\Omega\), contradicting \(p\in\Omega_2\). Hence \(D \subset \Omega_2\).
2.3. Q.E.D.
By definition \(\Omega_1\) and \(\Omega_2\) are disjoint with union \(\Omega\), and \(w \in \Omega_1\) via the constant curve. By steps 1 and 2 both sets are open. Since \(\Omega\) is connected and \(\Omega_1\ne\varnothing\), \(\Omega_2 = \varnothing\), so \(\Omega = \Omega_1\): every point of \(\Omega\) is joined to \(w\), and any two points are joined through \(w\).