Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a continous function. I want to prove the set $E(f)$ given by
$E(f) = \{(x,y,z)\in\mathbb{R}^3 \ | \ z > f(x,y) \}$
is open.
What I have tried so far:
- Suppose $E(f) = \emptyset$, then obviously $E(f)$ is open.
If $E(f) \neq \emptyset$, let $(x_1,x_2,x_3) = x \in E(f)$ be arbitrary. Now we need to show $\exists_{R>0}: B(x,R) \subset E(f)$, where $B(x,R)$ denotes the open sphere around the point $x$ with radius $R$, that is; $B(x,R) = \{ y \in \mathbb{R}^3 \ | \ |y - x| < R \}$.
We have that $f$ is continuous, thus $\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{y \in D(f)}: |x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$.
Somehow I should be able to pick a $R>0$ (depending on some other variables) such that for any $p \in B(x,R)$, I can show that $p_z > f(p_x, p_y)$ using the definition above. I have had a few attempts at this, without any success unfortunately.
I also tried creating a closed rectangle around the point $(x_1,x_2)$ and using the fact that $f$ will have a maximum and minimum in this rectangle, given $f$ is continuous. However; this did not seem to work out as well.
The region is the inverse image of the open set $(0,\infty) \subset \mathbb{R}$ under the continuous function $g(x,y,z) := z-f(x,y)$.