and $\mathbb R\cup\{+\infty\}$, respectively. It is obvious that $U\subseteq\{(t,x)\in\mathbb R^2\mid t\in{\mathop{\rm pr}\nolimits}_tU,\,{\rm lb}_U(t)\leqslant x\leqslant{\rm ub}_U(t)\}$. See Figure \[fig:ObjectRelatedToXSimplicity\] that illustrates some objects related to $x$-simplicity.

\[lem:OnOpenSetLB&UB\] If a subset $\Omega$ of the $(t,x)$-plane is open, then its projection $I:={\mathop{\rm pr}\nolimits}_t\Omega$ is an open subset of $\mathbb R$ and the functions $a:={\rm lb}_\Omega$ and $b:={\rm ub}_\Omega$ are upper and lower semi-continuous, respectively.

Since for any ball contained in $\Omega$ its projection to the $t$-axis is an open interval contained in $I$, it is obvious that $I$ is an open set. Fix an arbitrary $t_0\in I$. If $a(t_0)\in\mathbb R$, then for any $\varepsilon>0$ with $a(t_0)+\varepsilon<b(t_0)$, the point $z_0=(t_0,a(t_0)+\varepsilon)$ belongs to $\Omega$ and thus there exists a $\delta>0$ such that the ball $B_\delta(z_0)$ is contained in $\Omega$. Therefore, for any $t\in(t_0-\delta,t_0+\delta)$ we have $t\in I$ and $a(t)<a(t_0)+\varepsilon$. Analogously, if $a(t_0)=-\infty$, then for an arbitrary $N>0$ with $-N<b(t_0)$, the point $z_0=(t_0,-N)$ belongs to $\Omega$ and again there exists a $\delta>0$ such that the ball $B_\delta(z_0)$ is contained in $\Omega$. Hence for any $t\in(t_0-\delta,t_0+\delta)$ we have $t\in I$ and $a(t)<-N$. In total, this means that the function $a$ is upper semi-continuous on $I$. The lower semi-continuity of $b$ is proved in a similar way.

In 