# Prove that $A = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : e^{x_1^2 + x_2^2} < 1 + x_3^2 \} \subset \mathbb{R}^3$ is open and connected

I need to prove that set $$A = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : e^{x_1^2 + x_2^2} < 1 + x_3^2 \} \subset \mathbb{R}^3$$ is open and connected in $$\mathbb{R^3}$$

So, I did a change of variable $$x = x_1^2 + x_2^2$$ and $$y = x_3^2$$ and then it's easy to show, that set $$\tilde{A} = \{ (x, y) \in \mathbb{R^2}_+: e^x < 1 +y \} \subset \mathbb{R}^2$$ is open and connected in $$\mathbb{R^2}$$. Does it follow that the set $$A \subset \mathbb{R^3}$$ is also open and connected?

• Wouldn't it be easier and better to prove a more general result: for any continuous functions $f$ and $g$, $\{ f < g \}$ is open? Dec 17, 2020 at 20:49

$$A$$ is open because $$f : \mathbb R^3 \to \mathbb R, f(x_1,x_2,x_3) = e^{x_1^2 + x_2^2} - x_3^2$$ is continuous and $$A = f^{-1}((-\infty,1))$$. It is not connected because the plane $$P = \{ (x_1,x_2,x_3) \in \mathbb R^3 \mid x_3 = 0\}$$ has empty intersection with $$A$$ (note that $$e^{x_1^2 + x_2^2} \ge e^0 = 1$$). But the points $$(0,0,\pm1) \in A$$ lie on different sides of $$P$$.