Is $\mathbb{R}^{d_1}\times (0,\infty)^{d_2}$ a $(\varepsilon, \delta)$-locally uniform domain, for integer $d_1 \geq 0, d_2 \geq 0$? I am trying to get a grasp on these domains and am having a hard time proving this.
The definition:
A domain $\Omega \subseteq \mathbb{R}^n$ is $(\varepsilon, \delta)$-locally uniform for $\varepsilon, \delta>0$ if any $x,y \in \Omega$ with $\| x-y \| \leq \delta$ can be joined by a rectifiable arc $\gamma \subset \Omega$ satisfying for any point $z \in \gamma$
$l(\gamma) \leq \varepsilon \| x-y \|$, and
$\min\{l(\gamma'), l(\gamma'')\} \leq \mathrm{dist}(z, \partial \Omega)$,
where $l(\gamma)$ is the Euclidean length of the arc $\gamma$, and $\gamma'$ and $\gamma''$ are the two components of $\gamma\setminus \{z\}$. $\partial \Omega$ is the boundary of $\Omega$.
My first instinct was that, for a simple domain like that, a line segment should work as said arc. Simple counter-example is, suppose $\Omega=(0, \infty) \times (0, \infty)$, and put $x=(x_0, \beta_0)$ and $y=(\beta_1, y_1)$ such that $\beta_0$ and $\beta_1$ are arbitrarily small, the distance between $x$ and $y$ are lower than $\delta$, and $y_1=C x_0$ for a (very) large $C$. Then there is $z$ on the line segment between $x$ and $y$ that is closer to the boundary than to $x$ or $y$. I spent a while on this example and couldn't come up with an arc that would satisfy the second condition..