Continuous function with an open set I have a tough question regarding sets and continuity. It is as follows: 
"For each function $f: \mathbb{R^2} \rightarrow \mathbb{R}$, we define the set $E(f) \subset \mathbb{R^3}$ by \begin{equation} E(f) := \{(x,y,z) \in \mathbb{R^3} | z > f(x,y)\}. \end{equation}
Show that if $f$ is continuous,t hen $E(f)$ is open."
Help is greatly appreciated!
 A: Hint:


*

*Try to construct a continuous function $F:\mathbb R^3\to\mathbb R$ such that $E(f) = F^{-1}(A)$ for some open set $A$

*$z> f(x,y) \iff f(x,y)-z<0$

*$(-\infty, 0)$ is an open set.

A: All convergent sequences in the complement of $E(f)$
\begin{equation}
\{(x_i, y_i, z_i)\}\textrm{, }\ (x_i, y_i, z_i) \in \mathbb{R^3}\setminus E(f) \textrm{ (that is, with }\ z_i \le f(x_i, y_i)\textrm{)}\
\end{equation}
have limit $(x, y, z) \in \mathbb{R^3}\setminus E(f)$ because $z \le f(x, y)$ for the continuity of $f$ and for the property of limit operation on inequalities.
So the complement of $E(f)$ is closed, because it contains the limit of all its convergent sequences. 
So $E(f)$ is open.
A: Hint:
Let function $g:\mathbb R^3\to\mathbb R$ be prescribed by $\langle x,y,z\rangle\mapsto f(x,y)$.
Let function $h:\mathbb R^3\to\mathbb R$ be prescribed by $\langle x,y,z\rangle\mapsto z$.
Let function $[g,h]:\mathbb R^3\to\mathbb R^2$ be prescribed by $\langle x,y,z\rangle\mapsto\langle g(x,y,z),h(x,y,z)\rangle$.
Let $k:\mathbb R^2\to\mathbb R$ be prescribed by $\langle u,v\rangle\mapsto u-v$.
Then $E(f)$ can be recognized as the preimage of open set $(-\infty,0)$ w.r.t. composition $k\circ[g,h]$.
