Prove $x^2-y^2I'm trying to prove $x^2-y^2<z^2 $ is open. I believe this is a cone shape, can I simply assert that $∀ x ∈ P$ there exists $r > 0$ such that $B_r(x) ⊆ P$, where $P$ is my subset? Or is there an easier way to explicitly prove why it is open?
 A: 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. 
A: For any space $V$, if $f$ and $g$ are continuous from $V$ to $\Bbb R$ then for any $a,b\in \Bbb R$ the functions  $(af+ bg)(v)=a\cdot f(v)+ b\cdot g(v)$ and    $(afg)(v)=a\cdot f(v)\cdot g(v)$ are continuous from $V$ to $\Bbb R.$ 
It follows that if $f,g,h$ are continuous from $V$ to $\Bbb R$ then $(f^2+g^2-h^2)(v)=f(v)^2+g(v)^2-h(v)^2$ is continuous from $V$ to $\Bbb R.$ 
For $v=(x,y,z)\in \Bbb R^3$ let $f(v)=z$ and $g(v)=y$ and $ h(v)=x.$ Then $f,g,h$ are continuous from $\Bbb R^3$ to $\Bbb R$. So $i(v)=(f^2+g^2-h^2)(v)=z^2+y^2-x^2$ is continuous from $\Bbb R^3$ to $\Bbb R$.   
$\Bbb R^+$ is open in $\Bbb R$. So $\{(x,y,z)\in \Bbb R^3: x^2-y^2<z^2\}=i^{-1}\Bbb R^+$ is open in $\Bbb R^3.$ 
A: 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.
