proving a a set is closed

Question

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?

Answer 1

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.

Answer 2

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.