Problem: Let $g$ be harmonic on $U$ where $U \subseteq \mathbb{R}^2$ is open and connected. Show that if $g$ is non-constant on $U$, then $g(U)$ is open in $\mathbb{R}$.
I noticed that we could show that $g$ does not have a minimum in $U$ by taking the contrapositive of the Minimum Principle:
Let g be harmonic on $U$ where $U$ is open and connected. If $g$ has a minimum in $U$, then $g$ is constant.
I'm currently stuck here and not too sure how to proceed. Any suggestion?
On
$f(U)$ will be an interval (or a ray, or $\mathbb{R}$), because $U$ is connected and $f$ is continuous.
If $x\in U$, $f(U)$ must contain a nhood of $f(x)$. Because if not, $x$ would be an extreme point of the interval, that is, there would be a maximum of a minimum at $x$, which can´t happen. Thus $f(U)$ is open.
Since $U$ is connected and $g$ is continuous, the image $g(U)$ must be connected. The connected subsets of $\Bbb R$ are precisely the intervals. (Possibly unbounded on one or both ends.)
Now, let $y \in g(U)$. Then, $y = g(x)$ for some $x \in U$. As you noted: $g$ cannot have a minimum (and the same reasoning applied to $-g$ shows that it cannot have a maximum either).
Thus, there exist $x_1, x_2 \in U$ such that $g(x_1) < g(x) < g(x_2)$. Since $g(U)$ is an interval, we see that $$g(x) \in (g(x_1), g(x_2)) \subset g(U),$$ as desired.