as the supremum (resp. infinum) of a nonempty set that is bounded from above (resp. below), and $\tilde x_1\leqslant\tilde x_2$. The line segment $\{\tilde t_0\}\times(\hat x_1,\hat x_2)$ does not intersect the curve $C$ and contains the point $(\tilde t_0,\tilde x_0)$, which belongs to the interior $\Gamma$. Therefore, this segment is contained in $\Gamma$. The value of $\varepsilon$ is defined as in the previous case.

The chosen values of $\tilde t_0$, $\varepsilon$, $\tilde x_1$ and $\tilde x_2$ then satisfy the claimed properties.

\[def:XSimplePiece\] If there exists an open interval $I$ of the $t$-axis such that the intersection of a subset $U$ of the $(t,x)$-plane by the strip $I\times\mathbb R$ has an $x$-simple connected component, then we call this component an *$x$-simple piece* of $U$.

\[rem:OnSpecificOpenSetsWithNoXSimplePieces\] Suppose that for an open set $\Omega$ of the $(t,x)$-plane the subset $J$ of $t$’s from ${\mathop{\rm pr}\nolimits}_t\Omega$ with connected $\Omega_t$’s is dense in ${\mathop{\rm pr}\nolimits}_t\Omega$. Then the set $\Omega$ contains no $x$-simple pieces if an only if the complement of $J$ in ${\mathop{\rm pr}\nolimits}_t\Omega$ is also dense in ${\mathop{\rm pr}\nolimits}_t\Omega$.

![“Infinitely punctured open square” without $x$-simple pieces[]{data-label="fig:InfinitelyPuncturedOpenSquare"}](Ris2){width=".5\linewidth"}

\[ex:InfinitelyPuncturedOpenSquare\] The set $\Omega:=\big((0,1)\times(0,1)\big)\setminus\big\{(2^{-k}l,1-2^{-k}),\,l=1,\dots,2^k-1,\,k\in\mathbb N\big\}$ is open, connected, and contains no $x$-simple pieces. See Figure \[fig:InfinitelyPuncturedOpenSquare\].

Open $x$-simple 