Continuous real-valued function and open subset Let $f$ be a continuous real-valued function defined on an open subset $U$ of $\mathbb{R}^n$. 
Show that $\{(x,y):x\in{U},y>f(x)\}$ is an open subset of $\mathbb{R}^{n+1}$

Let $\forall{x}\in{X}, X\subset{U}$
Using the theorem, for a function $f$ mapping $S\subset{\mathbb{R}^n}$ into $\mathbb{R}^m$, it is equivalent to $f$ is continuous in $S$
so we can say $f(x)$ is continuous on $U$. Also, by following $U$, $f(x)$ is also open which is one of what I want to prove.
But how does it so sure about it maps to $\mathbb{R}^{n+1}$ but not $\mathbb{R}^{n+2}$, $\mathbb{R}^{n+3}$, ...

Hint: the function
$$
g:(x,y)\longmapsto y-f(x)
$$
is defined and continuous on $U\times \mathbb{R}$, which is open in $\mathbb{R}^{n+1}$. Now try to express your set with this function.
