Closure of the graph of the Conway base 13 function Let $f$ be the Conway base $13$ function. Is the closure of the graph of $f$ equal to $\mathbb{R}^2$? Or symbolically $$\overline{G_f} = \mathbb{R}^2, \text{ where } G_f = \{ (x, \ f(x)) \mid x \in \mathbb{R} \} $$
As $f$ takes all the values in $\mathbb{R}$ on any interval $I \subset \mathbb{R}$ it would imply that $$\forall a \in \mathbb{R^2}, \forall \epsilon \in \mathbb{R}^*_+, \exists x \in \mathbb{R} : d\big(a, (x, f(x)\big) < \epsilon \\ \implies \forall a \in \mathbb{R}^2, d(a, G_f) = 0 \implies \overline{G_f} = \mathbb{R}^2.$$
Is my sketch of a proof correct?

Yes. Another way to see it intuitively is to observe that there can be no open ball in the complement of the graph, since such a ball would overlap with a vertical strip over an interval, revealing that the function on that interval was not surjective.
