Prove that $\{(x,y,z) \in \mathbb{R}^3 \mid z^2-x^2-y^2-1>0 \}$ is an open set in $3$-space. Prove that $\{(x,y,z) \in \mathbb{R}^3 \mid z^2-x^2-y^2-1>0 \}$ is an open set in $3$-space.
I'm getting no clue as to how to proceed in order to prove this formally.
 A: $f(x,y,z) = z^2-x^2-y^2-1$ is continuous from $R^3 \to R$, and $(0,\infty)$ is open , thus $f^{-1}(0,\infty)$ is open, and you are done.
A: In case you are not familiar with the topological concept of continuity:
Let $A:=\{(x,y,z)\in\mathbb R^3:z^2-x^2-y^2>1\}$ and let $(x,y,z)\in A$ be arbitrary. Then there is some natural number $n$ such that $1+\dfrac{1}{n}<z^2-x^2-y^2.$ Now define $\varepsilon:=\min\left\{1,\dfrac{1}{3n(|x|+|y|+|z|+1)}\right\}.$ It follows that if $(u,v,w)$ is any point of the open ball $B$ of radius $\varepsilon$ and center $(x,y,z)$ then 
$$
\begin{aligned}
z^2-x^2-y^2-w^2+u^2+v^2&\leqslant|z^2-x^2-y^2-w^2+u^2+v^2|\\\\&=|(u+x)(u-x)+(v+y)(v-y)+(z+w)(z-w)|\\\\&\leqslant|u+x||u-x|+|v+y||v-y|+|z+w||z-w|\\\\&\leqslant\|(x-u,y-v,z-w)\|(|x|+|y|+|z|+|u|+|v|+|w|)\\\\&<\varepsilon(2(|x|+|y|+|z|)+3\varepsilon)\\\\&\leqslant3\varepsilon(|x|+|y|+|z|+1)\\\\&\leqslant\dfrac{1}{n}
\end{aligned}
$$
and hence $z^2-x^2-y^2<\dfrac{1}{n}+w^2-u^2-v^2$ and since $1+\dfrac{1}{n}<z^2-x^2-y^2$ then $1<w^2-u^2-v^2,$ which means that $(u,v,w)\in A$ and thus, since this point is an arbitrary point of $B,$ every point of the open ball $B$ is a point of $A$ and since $(x,y,z)$ is an arbitrary point of $A,$ we conclude that every point of $A$ is an interior point of $A,$ which means that $A$ is open.
