$\{(x,y):x\in{U},y>f(x)\}$ is an open subset of $\mathbb{R}^{n+1}$ 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 $\mathbb{R}^{n+1}$
I known $f(x)$ is open by using theorem of equivalent between $f$ mapping $S$ is subset of $\mathbb{R}^n$ and $f$ is continuous on $S$. $f$ define on $U$ thus $f(x)$ is open of $\mathbb{R}^n$.
Now, I want to know how to prove $f(x)$ is open subset of $\mathbb{R}^{n+1}$.
Thanks.
 A: You are mixing up things a little, I think. When you write "theorem of equivalent between $f$ [...]" you probably mean 

Theorem. For a map $f \colon S \to T$ between metric spaces the following are equivalent 
  (1) $f$ is continuous(2) preimages of open sets in $T$ are open in $S$, that is for open $U \subseteq T$, $f^{-1}[U] \subseteq S$ is open.

It is a good idea to apply this here in showing that the epigraph 
$$ \def\epi{\mathop{\rm epi}}\def\R{\mathbb R}\epi(f) = \{(x,y) \in \R^{n+1} \mid y > f(x) \} $$
of a continuous $f \colon U \subseteq \R^n \to \R$ is open. You want $\epi(f)$ open, not $f$ ($f$ being an open map is something different and if we identify $f$ with its graph, then $f$ is closed in $U \times \R$, not open). To this end, consider $g \colon U \times R \to \R$, $(x,y) \mapsto y - f(x)$. Then $\epi(f) = g^{-1}[(0,\infty)]$ is open in $U \times \R$, as $U \times \R$ is open in $\R^{n+1}$, $\epi(f)$ is also.
A: Let's prove this from scratch, i.e., without appeal to "general theorems".
We have to prove that the set
$$\Omega:=\{({\bf x},y)\ |\ {\bf x}\in U, \ y>f({\bf x})\}\subset{\mathbb R}^{n+1}$$
is open. To this end consider a point $$({\bf x}_0,y_0)\in\Omega\ .$$
Then $y_0=f({\bf x}_0)+2\epsilon$ for some $\epsilon>0$. Since $U\subset{\mathbb R}^n$ is open and $f$ is continuous at ${\bf x}_0$ there is an open disk $D$ with center ${\bf x}_0$ such that $D\subset U$, and that
$$f({\bf x})<f({\bf x}_0)+\epsilon$$
for all ${\bf x}\in D$. It follows that
$$y-f({\bf x})>0\qquad\forall {\bf x}\in D,\quad \forall y>y_0-\epsilon\ ,$$
and this is implies that the  "hat box"
$$B:=D\times\ ]y_0-\epsilon,\ y_0+\epsilon[$$
with center $({\bf x}_0,y_0)$ is a subset of $\Omega$.
