Characterization of an open set in the punctured plane by its intersection with lines through the origin

Question

Let $U$ be an open subset of the punctured plane $ℝ^2 \setminus \{0\}$ such that for every line $L$ through the origin, the subspace $U∩L$ has precisely two connected components. Is it necessarily true that $U$ is homeomorphic to the open annulus? Or, failing that, necessarily not simply connected? Are there higher dimensional analogues to this?

Answer 1

Here's a counterexample to the first question. Let $$A=\{(x,y): 0<x^2+y^2<1,y>0\}\setminus\{(0,1/2)\}$$ and $$B=\{(x,y):1<x^2+y^2<4,x>0\}\setminus\{(3/2,0)\}.$$ Notice that $A\cap L$ has $1$ component for every line through the origin except the horizontal and vertical lines, for which it has $0$ and $2$ components, respectively. Similarly $B\cap L$ has $0$ components if $L$ is horizontal, $2$ components if $L$ is vertical, and $1$ otherwise. So if $U=A\cup B$, then $U\cap L$ has exactly $2$ components for each $L$. But $U$ is disconnected, so it is not homeomorphic to an annulus (in fact it is homeomorphic to a disjoint union of two annuli).