If \(X=\RR^n\) with the usual Euclidean topology and \(U\subseteq X\) is an open, connected subspace of \(X\), show that \(U\) is also path connected.
Solution
If \(U=\varnothing\) there is nothing to prove. Otherwise fix \(x_0\in U\) and let \(P=\{x\in U:\text{there is a path in }U\text{ from }x_0\text{ to }x\}\).
1. \(P\) is open in \(U\).
Let \(x\in P\) and choose \(r>0\) with \(B(x,r)\subseteq U\). Each \(y\in B(x,r)\) is joined to \(x\) by the segment \(t\mapsto(1-t)x+ty\), which lies in the convex ball. Concatenating a path from \(x_0\) to \(x\) with this segment shows \(y\in P\), so \(B(x,r)\subseteq P\).
2. \(U\setminus P\) is open in \(U\).
Let \(x\in U\setminus P\) and choose \(r>0\) with \(B(x,r)\subseteq U\). If some \(y\in B(x,r)\) were in \(P\), a path from \(x_0\) to \(y\) followed by the segment from \(y\) to \(x\) would put \(x\) in \(P\). So \(B(x,r)\subseteq U\setminus P\).
3. Q.E.D.
By steps 1 and 2, \(P\) is clopen in the connected space \(U\), and \(x_0\in P\), so \(P=U\). Given \(x,y\in U\), the reverse of a path from \(x_0\) to \(x\) followed by a path from \(x_0\) to \(y\) is a path from \(x\) to \(y\) in \(U\).