Is $S=\{(x,y,z)\in\Bbb R^3\mid x^2+y^2\ge z\}$ path-connected?
My thoughts:
I think it is, but I can't construct a path from arbitrary points $p,q\in S$ explicitly.
If we imagine the curve $f(x,y)=x^2+y^2$ as the trace of a parabola $y=x^2$ rotated around the $y$ axis, for every $p\in S$, there is a parabola $y=x^2-\varepsilon_p$ rotated around the $y$ axis for some angle $\vartheta_p$ passing through $p$.
Then, for $p,q\in S,$ we could look at the circular annulus on the level $L=\max\{\pi_3(p),\pi_3(q)\}$ bounded by the concentric circles on level $L$, which are the result of rotations of the parabolas $y=x^2-\varepsilon_p$ and $y=x^2-\varepsilon_q$ around the $y$ axis.
So, for $p, q\in S,$ we could travel along their parabolas all the way to their intersections (let's call them $r$ and $s,$ respectively) with the respective circles on the level $L$. Then, since every circlular annulus in $\Bbb R^2$ is path connected, we could find a path from $r$ to $s$.
So, for example, we travel from $p$ to $r$ along the parabola $y=x^2-\varepsilon_p$ rotated for some angle $\vartheta_p$ around the $y$ axis, then, we travel from $r$ to $s$ in the circular annulus and, from $s,$ we travel to $q$ along the parabola $y=x^2-\varepsilon_q$ rotated for some angle $\vartheta_q$ around the $y$ axis.
However, I'm not sure if my conjecture is true. If it is, can we describe the path from $p$ to $q$ in $S$ explicitly?
Take a point $(x,y,z)$ in your set. First note that every point $(x',y',z)$ with $|x'|\le |x|, |y'|\le |y|$ is also in your set. Hence you can connect $(x,y,z)$ to $(0,0,z)$ by a line $x=t, y=t, z=z$. The point $(0,0,z)$ can be connected to $(0,0,0)$ by the line $x=0,y=0, z=t$. The intervals from these lines that you need, are inside your set. Hence every point in your set can be connected to $(0,0,0)$ by a path which is contained in the set. Thus your set is path-connected.