Prove $x^2-y^2<z^2$ is open I'm trying to prove $x^2-y^2&lt;z^2 $ is open. I believe this is a cone shape, can I simply assert that $∀ x ∈ P$ there exists $r &gt; 0$ such that $B_r(x) ⊆ P$, where $P$ is my subset? Or is there an easier way to explicitly prove why it is open?
Let $(x_0,y_0,z_0)$ be a point of your set. Because the inequality is strict, you have some margin around the first two coordinates: if you add, say, $\epsilon$, to the first two coordinates the point you get is still in the set. This shows a small (2D) circle around the point is still in the set .
Then, for each of the points in the set, if you add the same $\epsilon$ (or even more!) to $z_0$ you get a looser inequality and thus you get another point of your set.
In this way we get a whole (3D) box around our point. Inside this box we can inscribe a ball, thus the openness.
HINT

Try to find an obviously open set that is mapped to your set by a continuous function. Then it will be open by definition of a continuous function, since it must map open sets to open sets.