Prove that the set $ A= \{(x_{1},x_{2},x_{3},x_{4}) \in \mathbb R^{4} | x_{1} ^2 + x_{2}^2+ x_{3}^2< x_{4}^2\}$ is linearly connected. I have completely run out of ideas how this can be solved.
$\cdot$Set is linearly connected if every two points in the set can be connected by a sequence of rational curves in the set.
The set is not linearly connected because it can be split to two non-empty sets by the plane $x_4=0$ which doesn't contain any points from the set. There are no points for which $x_4=0$ because a sum of squares can't be negative in $\Bbb{R}$. The points $(0,0,0,1)$ and $(0,0,0,-1)$ are both in the set and each of them is on a different side of the dividing plane.