Apostol's Calculus - Prove the set is open (inequality) This is from Apostol's Calculus, Vol. II Section 8.3 #3(a).
Prove that $S=\{(x,y,z) \mid z^2 - x^2 - y^2 -1>0\}$ is open.
The only way to prove that a set is open which has been covered so far is to prove that for an arbitrary point $\mathop a\limits^{\small \to}   \in S$ there is an open ball $B(\mathop a\limits^{\small \to}  , r) \subset S$ by finding an explicit $r$ and showing that $\mathop x\limits^{\small \to}   \in B(\mathop a\limits^{\small \to}  , r) \implies \mathop x\limits^{\small \to}   \in S$. Nothing topological (compactness, completeness, etc.) has been discussed. For some reason I just am stuck trying to figure out an $r$ which would work.
Hints > Complete answers
 A: Note that $|x-x'| \leq \|(x,y,z)-(x',y',z')\|$. Furthermore, we have $|x^2-(x')^2| = |x-x'| |x+x'|$. Hence if $|x|, |x'| < L$, then $|x^2-(x')^2| \leq 2 L \|(x,y,z)-(x',y',z')\|$. The same applies to $y,y', z, z'$, of course.
Pick $(x,y,z) \in S$, let $M = \max\{|x|,|y|,|z|\}$ and $r = \min \{M, \frac{1}{6M}(z^2-(1+x^2+y^2))\}$ (which is $ >0$, of course).
Suppose $\|(x,y,z)-(x',y',z')\| < r$. Then we have $|x^2-(x')^2| < 2M r$, and similarly for the other coordinates.
Finally, we have $(z')^2-(1+(x')^2+(y')^2) > z^2-2Mr - (1 +x^2+y^2) - 4Mr$. By choice of $r$, we have $z^2 - (1 +x^2+y^2) \geq 6Mr$, from which we obtain $(z')^2-(1+(x')^2+(y')^2) > 0$, hence $(x',y',z') \in S$. It follows that $S$ is open since $B((x,y,z), r) \subset S$.
A: Choose arbitrary $(a,b,c)\in S$. Assume without loss of generality that $a,b,c\geq 0$. Put $k=c^2-a^2-b^2-1$.
Then, for any $(a',b',c')\in B((a,b,c),r)$ (with $0<r<c$, $r$ to be specified later -- notice that $c>0$) we have that
$$c'^2-a'^2-b'^2-1\geq (c-r)^2-(a+r)^2-(b+r)^2-1=k-2r(a+b+c)-r^2$$
Thus, we can see that if we choose $r>0$ small enough to have $r^2+2r(a+b+c)<k$ (and $r<c$), $B((a,b,c),r)$ will be contained in $S$.
A: Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be the function $f(x, y, z) = z^2 - x^2 - y^2 - 1$.
Since $f$ is continuous and $(0, \infty)$ is open in $\mathbb{R}$, $S = f^{-1}((0, \infty))$ is open.
