Open subset of $\mathbb{R}^{n+1}$ proof 
Let $G\subset \mathbb{R}^n$ be open and $f:G\to \mathbb{R}$ be
  continuous on $G$. Show that the set $$\{(a,b)\in\mathbb{R}^{n+1}:a\in
 G,b>f(a)\}$$ is an open subset of $\mathbb{R}^{n+1}$.

How can I prove this? I don't even know where to start. 
 A: Step 1: the function $g:(a,b)\longmapsto b-f(a)$ is continuous on $G\times \mathbb{R}$. So your set $S=g^{-1}((0,+\infty))$ is open in $G\times \mathbb{R}$.
Step 2: since $G$ is open, $G\times\mathbb{R}$ is open in $\mathbb{R}^{n+1}$.
Step 3: if $U$ is open in a topological space $X$, then $V\subseteq U$ is open in $U$ for the induced topology (that is $V=U\cap W$ for some open set $W$ in $X$) if and only if $V$ is open in $X$.
Conclusion: since $S$ in open in $G\times \mathbb{R}$ (step 1) which is open in $\mathbb{R}^{n+1}$ (step 2), it is open in $\mathbb{R}^{n+1}$ (step 3). The short argument once we have clarified all this is:
$$
S=g^{-1}((0,+\infty))=(G\times\mathbb{R})\cap W
$$
for some $W$ open in $\mathbb{R}^{n+1}$.
Note: It might be interesting to see that, although tempting given the strict inequality defining $S$, the proof that the complement is closed is not necessarily easier in this situation. So I added it below.
Complement argument: consider the complement of your set
$$
S^c=G^c\times\mathbb{R}\cup\{(a,b)\;;\;a\in G\; b\leq g(a)\}.
$$
Since $G^c$ is closed the lhs is clearly closed. Now let $(a,b)$ be a limit point of $S^c$. It is either a limit point of the lhs or the rhs , or both. If it is a limit point of the lhs, it belong to it as we have just seen it is closed. Now if it is a limit point of the rhs, there are two cases to consider. First $a\not\in G$, then $(a,b)$ belongs to the lhs. Second, $a\in G$. Then $b\leq g(a)$ by continuity of $g$ at $a$ (as we have $(a_n,b_n)$ tending to $(a,b)$ with $a_n$ converging to $a$ in $G$ and $b_n\leq f(a_n)$ for all $n$, which yields $b\leq f(a)$ at the limit). So again $(a,b)$ belongs to $S^c$. So $S^c$ contains all its limit points. Hence it is closed and $S$ is open.
