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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?

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

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$A$ 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 $P$.

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