(Exercise 10 of the previous section is a particular case of this general result.) In an open subset $U$ of a locally path connected space $X$ each path component and their unions are open in $X$ and, therefore, open and close in $U$ . Since $U$ is connected, there is only one path component. Another way to show this is as follows. We prove that an open subspace of a locally path connected space is locally path connected (this implies that the components and the path components of the subspace are the same). Indeed, if $U$ is an open subset of a locally path connected space $X$ , $x\in U$ and $V\subseteq U$ is an neighborhood of $x$ in $U$ then $V$ is open in $X$ as well (since $U$ is open) and contains a path connected neighborhood $W\subseteq V\subseteq U$ of $x$ which is also open in $U$ .