Exercise: show that if $U$ is a connected open subspace of $ R^2$, then $U$ is path connected.

Question

Show that if $U$ is a connected open subspace of $ R^2$, then $U$ is path connected.

The idea was to show that given $x_{0} \in U$, the set of points that can be joined to $x_{0}$ by a path in $U$ is open and closed in $U$, however I have not been able to do this. Could you give me any suggestion? Another idea to do this, is assuming that $U$ is not connected by paths, that is, assuming that $a, b \in U$ exist with $a < b$ such that the interval $[a, b]$ of points $X$ is not entirely contained in $U$. However I have not been able to conclude anything. Is this reasoning correct?

Definition: Given two points $x$ y $y$ from the space $X$, a path in $X$ that joins $x$ with $y$ is a continuous application $f \colon [a, b] \rightarrow X$ of some closed interval of the real line at $X$, so that $f(a) = x$ and $f(b) = y$. A space $X$ is said to be path connected if each pair of $X$ points can be joined by a path at $X$.

Answer 1

This is a general mechanism used for connected sets. You have to get used to it. If you want to show that some property holds on all of the connected set, you define an agreement set $A$, where the property you want to verify holds. Showing that $A$ is non-empty (usually given as an assumption), open, and closed, implies that $A = U$. And usually some sort of continuity is used in proving that $A$ is open or closed.

Here, we fix $x_0\in U$ and define $$A = \{x \in U \mid x \text{ may be connected to }x_0 \text{ with a path in }U\}.$$

• $A \neq \varnothing$. Because $x_0 \in U$. Take the constant path on $x_0$.
• $A$ is open in $U$. If $x$ can be joined to $x_0$, $U$ being open gives $r>0$ such that $B(x,r) \subseteq U$. And every point $y$ in this ball can be joined to $x_0$ by concatenating the path from $x_0$ to $x$ with a line segment joining $x$ to $y$. This means that $B(x,r) \subseteq A$. Since $x$ was arbitrary, $A$ is open.
• $A$ is closed in $U$. Take $x \in U$ that cannot be joined to $x_0$. For every $r>0$ such that $B(x,r) \subseteq U$, no $y \in B(x,r)$ can be joined to $x_0$. Else, if there is a "bad" $y$, you get the contradiction of joining $x_0$ to $x$ by concatenating the path joining $x_0$ to $y$ with a line segment joining $y$ to $x$.

So $U = A$ is path-connected.