Does the graph of a continuous function have an empty interior? Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous function and consider the set $\{ (x,f(x)) : x \in \mathbb{R} \}$ = G. Then, we obviously know that this set $G$ is closed. Now, I am having some difficulty trying to find its interior. My guess is that $\operatorname{Int} G = \varnothing $. But, how can I prove this? Any help would be greatly appreciated.
thanks
 A: The projection $\pi:G\to\Bbb R$ is a homeomorphism. Can you show $\Bbb R\times\{0\}\subset \Bbb R^2$ has empty interior?
A: I think you need the function (EDIT: curve; please see below) to be injective; otherwise you have cases like that of a space-filling curve :http://en.wikipedia.org/wiki/Space-filling_curve  . If you want to turn it into a function  $f: \mathbb R \rightarrow \mathbb R$ , then you can extend continuously left and right. The existence of the continuous extension is guaranteed, e.g., by Tietze extension theorem.
EDIT: like was pointed out, this is not actually a function, but a curve (and I wrongly assumed you meant a curve, not a function). For a function,  consider a ball about the pair $(x,f(x))$. Since f is a function, any ball $B((x,f(x));r)$ for any $r>0$ cannot contain any horizontal strip above $f(x)$, so the image must contain an empty interior. Say there is a closed ball $B$ in the graph. This ball will then contain a small vertical strip  , which implies that some point in the domain has two images, which cannot happen for a function. 
A: This set has measure $0$ in $\mathbb{R}^2$, so its interior must be empty.
