Let $S= (\mathbb{R}^{2} - S^{1}) \cup \{(1,0)\} $. Show that, for every straight line $r$ in $\mathbb{R}^{2}$ it has to be $r\cap S$ open in $r$ but S not is open in $\mathbb{R}^{2}$.
I can see a little with S set sketch but I can not solve it indeed
The set $S$ is not open in $\Bbb R^2$ because each open ball $B$ centered at a point $p=(1,0)\in S$ is not contained in $S$ (because $B$ contains an arc of the circle $S^1$). From the other hand, the set $S’\setminus\{p\}=\Bbb R^2\setminus S^1$ is open in $\Bbb R^2$ as a complement to a compact set $S^1$. Now let $r$ be an arbitrary straight line. If $r$ does not contain the point $p$ then the set $r\cap S=r\cap S’$ is open in $r$. If $r$ contains the point $p$ then the set $r\cap S’$ is $r$ without at most one point, so it is open in $r$.