proving a a set is closed I need help with proving that:
$$ \{ (x,y,z)\in \mathbb{R}^3 \ \vert \ x^2+y^2 <1, 0<z< x^2+ xy^2+y \} $$
(see also https://i.stack.imgur.com/V7VEq.png)
is a an open set.
it is easy to see that the circle and function are open sets but how can I prove the question as a whole?
 A: The intersection of two open sets is open. Your set is the intersection of the circle and the other shape, both of which are open as you've noted.
A: I'd solve it like this:
Your set might be written as
$$ \{ (x,y,z)\in \mathbb{R}^3 \ \vert \ x^2+y^2<1  \} \cap
\{ (x,y,z)\in \mathbb{R}^3 \ \vert \ x^2+xy^2+y -z>0  \} \cap 
\{ (x,y,z)\in \mathbb{R}^3 \ \vert \ z>0  \}.$$
To see that those sets are open, you can use the fact that the preimage of an open set under a continuous function is open. Take
\begin{align*}
f_1 :& \mathbb{R}^3 \rightarrow \mathbb{R}, \ f_1(x,y,z)=x^2 + y^2,\\
f_2 :& \mathbb{R}^3 \rightarrow \mathbb{R}, \ f_2(x,y,z)=x^2+xy^2+y-z,\\
f_3 : &  \mathbb{R}^3 \rightarrow \mathbb{R}, \ f_3(x,y,z)=z.
\end{align*}
Then we have
\begin{align*}
 \{ (x,y,z)\in \mathbb{R}^3 \ \vert \ x^2+y^2<1  \}&= f_1^{-1}((-\infty, 1)),\\
\{ (x,y,z)\in \mathbb{R}^3 \ \vert \ x^2+xy^2+y -z>0  \}&= f_2^{-1}((0,\infty)), \\
\{ (x,y,z)\in \mathbb{R}^3 \ \vert \ z>0  \}&= f_3^{-1}((0,\infty)).
\end{align*}
Thus, your set is the intersection of finitely many open sets and therefore open.
