I am considering here convex sets and star domains, as subsets of some fixed $\mathbb{R}^n$.
I use the notation $(S,x_0)$ for a star domain $S$, and by that I mean : $\forall x \in S$, the line segment from $x_0$ to $x$ is in $S$. I also use the notation $B(x,r)$ for an open ball of center $x$ and radius $r$.
If $(S,x_0)$ is a star domain and $C$ a convex set containing $x_0$, it's easy to see that $(S \cap C, x_0)$ is a star domain.
But if $(S,x_0)$ is star domain and $C$ a convex set not containing $x_0$, $S \cap C$ is not necessarily a star domain, as it can fail to be connected, as in the case of $S = (\{0 \} \times [0,1]) \cup ([0,1] \times \{ 0 \})$ and $C = $ the closed line segment between $u = (0,\frac{1}{2})$ and $v = (\frac{1}{2},0)$.
I'm looking for examples of one of the following (note that an open ball is a convex set) :
(1) $(S,x_0)$ is a star domain such that : $\exists x \neq x_0 : \exists r > 0 : S \cap B(x,r)$ is not a star domain.
(2) $(S,x_0)$ is a star domain such that : $\forall x \neq x_0 : \exists r > 0 : S \cap B(x,r)$ is not a star domain.
(3) $(S,x_0)$ is a star domain such that : $\forall x \neq x_0 : \forall r > 0 : S \cap B(x,r)$ is not a star domain.
(4) $(S,x_0)$ is a star domain such that : $\exists x \neq x_0 : \forall r > 0 : S \cap B(x,r)$ is not a star domain.
Thanks.
Edit :
For (3) and (4), I actually meant :
(3) $(S,x_0)$ is a star domain such that : $\forall x \neq x_0 : \forall r \in ]0,\lVert x-x_0 \rVert] : S \cap B(x,r)$ is not a star domain.
(4) $(S,x_0)$ is a star domain such that : $\exists x \neq x_0 : \forall r \in ]0,\lVert x-x_0 \rVert] : S \cap B(x,r)$ is not a star domain.
As they are written, (3) and (4) cannot really happen, since for $r> \lVert x_0-x\rVert>0$ you'll get $x_0\in S\cap B(x,r)$.
So I'll henceforth assume you meant $$[\forall /\exists]x\ne x_0,\forall r\in(0,\lVert x-x_0\rVert), \text{et cetera}\ldots$$
See (3)
See (3)
Consider $\Bbb R^2\supseteq (S,0)=\{(x,\alpha x)\,:\,\alpha\in\Bbb Q,\ x\in\Bbb R\}$
See (3)