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 can be joined to $x_0$ by a path in $U$ is both open and closed in $U$.)
This should not be too difficult, but I am stuck at some point.
Let $x_0 \in U$
Let $A = \{ x \in U | \text{there is a path connecting $x$ and $x_0$} \}$ $\subset U$.
I want to show that $A$ is open, so let $x \in A$. By openness of $U$, we can find a basic open set $\prod_{i=1}^2 (a_i,b_i)$ such that $\prod_{i=1}^2 (a_i,b_i) \subset U$. But I am stuck in showing that $\prod_{i=1}^2 (a_i,b_i)$ lies entirely in $A$.
Best Answer
Let $x\in A$. Then take an open ball $B$ containing $x$ small enough such that $B\cap U = B$. We know that open balls are path connected (you can easily construct a path), so $B\subseteq A$, and $A$ is open. To show that $A$ is closed, consider $x\in U\setminus A$ and show that there exists an open ball $B$ containing $x$ such that $B\cap A = \emptyset$ (hint: what would happen if $p\in B\cap A$?)