Prove that Set is open $\{(x,y) \in R^2: 0Prove that $\{(x,y)  \in R^2: 0<y<f(x) \}$ is an open set, where $f$ denotes a continuous function.
Is there an easy way to show this? I feel like all my current attempts feel a bit over the top, since I tried to prove the definition of an open set directly. May  there be any theorem I could use? Thanks!!
 A: Let me give you a hint to get you started. Do you know how to show that $g: (x,y) \mapsto f(x) - y$ is continuous? Also remember that the intersection of two open sets is open.

Your set is then $g^{-1}( (0, +\infty)) \cap \{ (x,y) \, | \, y > 0 \}$. Since $g$ is continuous, $g^{-1}( (0, \infty) )$ is open. The set $\{ (x,y) \, | \, y > 0 \} = \mathbb{R} \times (0, +\infty)$ is open by definition of the product topology on $\mathbb{R}^2$. 
In order to show that $g$ is continuous, you can proceed as follows: The projection $(x,y) \mapsto x$ is continuous, so the map $(x,y) \mapsto f(x)$ is the composition of continuous functions and is thus continuous. Linear combinations of continuous functions are continuous, the projection $(x,y) \mapsto y$ is continuous and thus $g$ is continuous. 
A: Hint : Try drawing a picture, take a general continuous  function, I am sure that will help you. After drawing a graph, take any general point on the shaded area, try drawing a ball around any point. Please keep in mind $y<f(x)$. 
A: Let me present another proof that does not rely on the continuity
of the map: $(x,y)\mapsto f(x)-y$.
Denote $A=\{(x,y)\mid0<y\}$ and $B=\{(x,y)\mid y<f(x)\}$. Note that
$A=\mathbb{R}\times(0,\infty)$ which is open in $\mathbb{R}^{2}$.
Moreover,
\begin{eqnarray*}
B & = & \cup_{a\in\mathbb{R}}\{(x,y)\mid y<a\}\cap\{(x,y)\mid a<f(x)\}\\
 & = & \cup_{a\in\mathbb{R}}(\mathbb{R}\times(-\infty,a))\cap (f^{-1}((a,\infty))\times\mathbb{R}).
\end{eqnarray*}
For each $a\in\mathbb{R}$, the sets $\mathbb{R}\times(-\infty,a)$
and $f^{-1}((a,\infty))\times\mathbb{R}$ are open subsets of $\mathbb{\mathbb{R}}^{2}$.
Therefore, their intersection $(\mathbb{R}\times(-\infty,a))\cap (f^{-1}((a,\infty))\times\mathbb{R})$
is open. Finally, $B$ is union of open sets, so $B$ is open. Note
that $\{(x,y)\mid0<y<f(x)\}=A\cap B$, which is also open.
