I would like to know how is it possible to characterize all the open subsets of the real plane having the following property: for every straight line $\gamma$ in the plane there is a line $\gamma'$ which is parallel to $\gamma$ and whose intersection with the open subset is not connected (as a subspace of $\gamma'$).
Thanks!
We can characterize open subsets $U$ of the plane $\mathbb R^2$ without the property. Without loss of generality me may suppose that for each straight line $\gamma’$ ($x=const$) which is parallel to $y$-axis the intersection $U(x)=U\cap\gamma’$ is connected. Let $\pi$ be the projection of the plane onto the $x$-axis. Since the set $U$ is open then since the map $\pi$ is open the set $U_X=\pi(U)$ is open and the set $U(x)$ is an open segment (may be infinite in one or both sides) for each $x\in U_X$. Therefore we can define two maps $l(x)=\inf U(x)$ and $u(x)=\sup U(x)$ from the set $U_X$ into an extended real line $\bar{\mathbb R}=\mathbb R\cup\{-\infty\}\cup\{+\infty\}$. Since the set is $U$ open, the map $u$ is lower semicontinuous (that is for any $x\in U_X$ and any real $r$ such that $f(x)>r$ there exists a neighborhood $O(x)\subset U_X$ of the point $x$ such that $f(x’)>r$ for each point $x’\in O(x)$) and the map $l$ is upper semicontinuous (that is for any $x\in U_X$ and any real $r$ such that $f(x)<r$ there exists a neighborhood $O(x)\subset U_X$ of the point $x$ such that $f(x’)<r$ for each point $x’\in O(x)$). Conversely, for any open set $U_X\subset\mathbb R$ and any maps $u$ and $l$ from the set $U_X$ into $\bar{\mathbb R}$ such that the map $u$ is lower semicontinuous, the map $l$ is upper semicontinuous, and $l(x)<u(x)$ for each $x\in U_X$, the set $U=\{(x,y)\in\mathbb R^2:x\in U_X, l(x)<y<u(x)\}$ is open. Remark that by Example 5.5.20 from Engelking’s “General topology” in such a case there should exist a continuous selection $f$, that is a map $f:U_X\to U$ such that $\pi f(x)\equiv x$.