The question is in the title, actually for $n=2$, it's okay, for $n=3$, I thought that :
$\{(x_1,...,x_n) \; | \; x_1^2+x_2^2-{x_{3}}^2 < 0 \}, \{(x_1,...,x_n) \; | \; x_1^2+x_2^2-{x_{3}}^2 > 0 \}$ are the connected component.
Actually, for $n \geq 4$, I have the impression that $\{(x_1,...,x_n) \; | \; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 \}, \{(x_1,...,x_n) \; | \; x_1^2+...+{x_{n-1}}^2-x_n^2 > 0 \}$ are the connected component, cause each are open and closed, but I need to prove that it's connected.
Anyone could help me, please ?
Thank you !
Let’s note $S_n= \{(x_1,...,x_n) \in \mathbb R^n \; | \; x_1^2+...+{x_{n-1}}^2-x_n^2 \neq 0\}$
Case $n=2$
$S_2=\{(x,y) \in \mathbb R^2 \mid x^2 \neq y^2\} =\{(x,y) \in \mathbb R^2 \mid x \neq y \wedge x \neq -y\}$ has 4 connected components.
Case $n \ge 3$
Let’s prove that in that case, $S_n$ has 3 connected components, namely:
$$\begin{aligned} S_n^+ &= \left\{(x_1,...,x_n) \in \mathbb R^n \mid \sqrt{x_1^2+...+{x_{n-1}}^2}< x_n\right\}\\ S_n^0 &= \left\{(x_1,...,x_n) \in \mathbb R^n \mid -\sqrt{x_1^2+...+{x_{n-1}}^2}< x_n < \sqrt{x_1^2+...+{x_{n-1}}^2}\right\}\\ S_n^- &= \left\{(x_1,...,x_n) \in \mathbb R^n \mid x_n < -\sqrt{x_1^2+...+{x_{n-1}}^2}\right\} \end{aligned}$$
Those 3 opens are in fact path connected. $S_n^+$ and $S_n^-$ are convex. Following question: when is the epigraph a convex cone? is a good basis for the proof.
$S_n^0$ is path connected and therefore connected. Use « vertical » lines going to the hyperplane $x_n=0$ and one or two segment lines in the hyperplane $x_n=0$, to avoid passing through the origin.