Show that if \(U\) is an open connected subspace of \(\mathbb{R}^2\), then \(U\) is path connected. [Hint: Show that given \(x_0 \in U\), the set of points that can be joined to \(x_0\) by a path in \(U\) is both open and closed in \(U\).]
Solution
If \(U=\varnothing\) there is nothing to join. Otherwise fix \(x_0\in U\) and let \(C\) be the set of \(x\in U\) joined to \(x_0\) by a path in \(U\); the constant path shows \(x_0\in C\).
1. If \(x\in U\) and \(B(x,\varepsilon)\subseteq U\), then either \(B(x,\varepsilon)\subseteq C\) or \(B(x,\varepsilon)\cap C=\varnothing\).
The ball is convex, so any two of its points are joined by a line segment in \(B(x,\varepsilon)\subseteq U\). If some \(w\in B(x,\varepsilon)\) lies in \(C\), concatenating a path in \(U\) from \(x_0\) to \(w\) with the segment from \(w\) to any \(y\in B(x,\varepsilon)\) gives a path in \(U\) from \(x_0\) to \(y\), so \(B(x,\varepsilon)\subseteq C\).
2. \(C\) is open and closed in \(U\).
Every \(x\in U\) has a ball \(B(x,\varepsilon)\subseteq U\) because \(U\) is open. By step 1, if \(x\in C\) the ball lies in \(C\), and if \(x\notin C\) the ball lies in \(U-C\).
3. Q.E.D.
\(C\) is nonempty, and by step 2 it is open and closed in the connected space \(U\), so \(C=U\). Hence every point of \(U\) is joined to \(x_0\), and any two points of \(U\) are joined through \(x_0\).