Proof that $\mathbb{R}^3$ minus $z$-axis is a star domain How can I prove that $\mathbb{R}^3$ minus $z$-axis is a star domain?
Due to conditions for star domain, I think I have to prove that the set is simply connected or convex. 
I have found a family of closed curves that surround $z$-axis.
How can I conclude? 
 A: $\Bbb R^3$ without $z$-axis is not a star domain. 
Given any $(x,y,z)\in\Bbb R^3$ with $x\neq 0$ or $y\neq 0$, consider $(-x,-y,z)$. This is a different point, but the line from $(x,y,z)$ to $(-x,-y,z)$ goes through $(0,0,z)$, which lies on the $z$-axis.
A: This is false. $\Bbb{R}^3$ minus the whole $z$-axis is not a star domain:
Proof: Let $v = (x, y, z)$ be any point not on the $z$-axis. Then either $x \neq 0$ or $y \neq 0$. Hence either $-x \neq x$ or $-y \neq y$. And so $-v = (-x, -y, -z)$ is different from $v$ but still not on the $z$-axis. However, now the line segment $\ell(t) = tv + (1 - t)(-v), 0 \leq t \leq 1$ from $v$ to $-v$ includes the origin (which is on the $z$-axis) as you can see by plugging $t = 0.5$ in $\ell(t)$. This is contrary to what is required for a star domain. 
A: The set you are considering is not even simply connected, and there is a very simple way of proving this. Consider the cylindrical coordinates 
$$
x=r\cos \phi, \quad y=r\sin \phi, \quad z=z.$$ 
The differential form 
$$\omega=\frac{xdy-ydx}{x^2+y^2},$$ 
which can be written in cylindrical coordinates as $\omega=d\phi$, is defined on the domain in consideration, that is $\mathbb R^3\setminus\{z\text{ axis }\}$, and is manifestly closed, BUT it is not exact; this means that there are closed curves $\gamma$ such that 
$$
\oint_\gamma \omega \ne 0.$$ 
One such curve is the circle $C=\{(\cos \phi, \sin \phi, 0)\}$; indeed, using the representation in cylindrical coordinates, we immediately see that 
$$
\oint_C \omega = \int_0^{2\pi} d\phi = 2\pi\ne 0.$$ 
Thus, the domain cannot be simply connected, let alone star shaped.
