Is this an open set? Let $f : \mathbb{R}\rightarrow\mathbb{R}$ be a continuous function and define $\mathbf{A}=\{\, (x,y) \in \mathbb{R}^2 \mid y < f(x) \,\}$. I need to express $\mathbf{A}$ as unions of cartesian product of two open sets. 
Let $$\mathbf{A}=\bigcup_{y \in \mathbb{R}}\Big(\{x \in \mathbb{R} \mid f(x)  > y\} \times \{y' \in \mathbb{R} \mid y' < y < \sup{f(\mathbb{R})}\}\Big)\;.$$ Does that work?
 A: Hint: Define $F:\mathbb{R}^2\to\mathbb{R}$ by $F(x,y)=f(x)-y$. I'm pretty sure you can see this is continuous if $f$ is. What is $F^{-1}(0,\infty)$? And, how does this help us?
PS, to be open, you don't need to be the product of open sets. Those just happen to form a basis for the topology. 
A: HINT: In general you cannot express $\mathbf{A}$ as a product of open sets, and you do not need to do so in order to show that it’s open; you need only express it as a union of products of open sets. You seem to have tried to do this, but in general a set 
$$\{x\in\Bbb R:f(x)>y\}\times\{y'\in\Bbb R:y'<y\}$$
is not disjoint from $\mathbf{A}$.
The most straightforward way to show that $\mathbf{A}$ is open is to pick a point $\langle a,b\rangle\in\mathbf{A}$ and show that there are open sets $U$ and $V$ in $\Bbb R$ such that $\langle a,b\rangle\in U\times V\subseteq\mathbf{A}$. (This is not the shortest or most elegant approach, but it is the most straightforward, from the definitions.)
Since $\langle a,b\rangle\in\mathbf{A}$, you know that $b<f(a)$. Let $\epsilon=\frac12\big(f(a)-b\big)$; $f$ is continuous, so there is a $\delta>0$ such that $|f(x)-f(b)|<\epsilon$ whenever $|x-a|<\delta$. Use $\delta$ and $\epsilon$ to construct $U$ and $V$.
A: If we just need to show that this set is open, let $g(x,y) = f(x)-y$ and notice that the set in question is $g^{-1}(0,\infty)$.  By continuity, preimages of open sets are open.  
