Show that Set in $M:=\{x\in \Bbb R^3 : x_1^2\ge2(x_2^3+x_3^3) \}$ is closed I have to show this regarding the Euclidean metric.
I've already shown that it isn't bounded by showing that the $d(x,y)\:\forall x,y \in M$ isn't bounded.
I know that in order to show the closedness i have to show that the complement is open, but how do I show that $$M:=\{x\in \Bbb R^3 : x_1^2\lt 2(x_2^3+x_3^3) \}$$ is open? (I know that a set is open if the $\epsilon$-ball around every element is also in the set but how do I show this in this particular case)
I have to determine whether the set is closed and whether it's bounded for the following set too, but I don't have any ideas how to show either:
$$M:=\{x\in \Bbb R^3 : x_1 \ge -1, \:x_2\lt2, ||x||_2\le\sqrt5\}$$
Thanks in advance

Let $f(x_1,x_2,x_3)= x_1^2-2(x_2^3+x_3^3)$, which is a polynomial, hence continuous. Then the set in question is $f^{-1}([0,\infty))$ which is closed, hence by the definition of continuity, the inverse image of this closed set is closed.
