# Question about moving planes method.

In the paper Inequalities for second-order elliptic equations with applications to unbounded domains I, by Berestycki, Caffarelli and Nirenberg (page $486$), they considered a set $\Omega=\mathbb{R}^{n-j}\times\omega$, where $\omega\subset\mathbb{R}^{j}$ is a bounded smooth domain, convex in the direction $y_1$ ($z\in\Omega$ then $z=(x,y)=(x_1,...,x_{n-j},y_1,...,y_j)\in\mathbb{R}^{n-j}\times\omega$). For a $\mu>0$ they defined the sets: $$\Gamma=\{y\in\partial\omega;y_1\geq\mu\},$$ $$\Gamma^\mu=\{y\in\overline\omega;y^\mu\in\Gamma\},$$ this set is the reflection of $\Gamma$ in the plane $\{y_1=\mu\}$, and $$\Gamma^\mu_+=\Gamma^\mu\cap\omega.$$ They said that is easy to see, if $j\geq2$ then the closure of $\Gamma^\mu_+$ meets the boundary of $\omega$, but this is not true when $j=1$. Someone know why? I can't see a counter example.

This is better done by drawing pictures, but basically what you are doing is you are taking a set, and reflecting its boundary over the plane $y_1 = \mu$.
Now, when the spatial dimension $j \geq 2$, $\partial \omega \cap \{y_1 = \mu\}$ is nonempty since $\omega$ is a bounded domain, so $\Gamma^\mu_+$ must meet $\partial \omega$ at least there.
When $j=1$, consider $\omega = (-5,1)$ and $\mu = 0$. Then $\Gamma^\mu \cap \partial \omega$ is empty.