Connectedness of a set defined analytically How can I prove the following set: $$S=\left\{(x,y,z)\in\mathbb{R^3}:x^2+y^2+z^2+\arctan^2(xyz^2)=7\right\}$$ is connected? I thought that I might somehow use the implicit function theorem to prove that $S$ is the union of graphs of continuous functions from connected sets, but practically I'm quite confused on how to proceed.
Thanks
 A: Write $S_\pm = \{ (x, y, z)\in S: \pm z \ge 0\}$. Then $S = S_+ \cup S_-$ and $S_+ \cap S_- = \{ (x, y, 0): x^2 + y^2 = 7\}$. Thus it suffices to show that each $S_\pm$ is connected. Since $S_- = P S_+$, where $P(x, y, z) = (x, y, -z)$, it suffices to show that $S_+$ is connected.
First, for each $(x, y)$ so that $x^2 + y^2 \le 7$, there is exactly one $z\ge 0$ so that $(x, y, z) \in S$.
To see this, for each such $(x, y)$, the function
$$f(z) = z^2 + \arctan^2 (xyz^2)$$
is strictly increasing on $(0,\infty)$ since
$$ f'(z) = 2z + 4\arctan (xyz^2) \frac{xyz}{1+ (xyz^2)^2} >0.$$
Thus there is exactly one point $z(x, y)\ge 0$ so that
$$ f(z(x, y)) = 7-x^2-y^2\Leftrightarrow (x, y, z(x, y))\in S.$$
Think of $z = z(x, y)$ as a surjective mapping
$$z: \{ (x, y): x^2 + y^2 \le 7\} \to S_+.$$
Then $z$ is continuous: indeed one can show that $z$ is smooth: write
$$ F: \mathbb R^3_+ \to \mathbb R^3, \ \ \ F(x, y, z) = x^2 + y^2 + f(z).$$
(note $f$ depends also on $x, y$). Then
$$ \nabla F = \begin{pmatrix} 1 & 0 & * \\ 0& 1 & * \\ 0 & 0 & f' \end{pmatrix}$$
Since $f' \neq 0$, $F$ is a local diffeomorphism by the inverse function theorem. Thus
$$ (x, y, z(x, y)) = F^{-1} (x, y, 7),$$
implying that $z$ is smooth.
Since $\{x^2 + y^2 \le 7\}$ is connected and $z$ is continuous and surjective, $S_+$ is also connected and we are done.
