Let $\mathcal{C} \subset [0,1]$ denote the usual Cantor set (i.e. the usual definition based on removed middle third open intervals succesively). Define $S = \mathcal{C} - E$ where $E$ is the set $\{0,1 \} \cup E'$, and $E'$ is the set of all endpoints of all the middle third open intervals successively removed from $[0,1]$ to construct the usual cantor set $\mathcal{C}$.
My question is, does there exist a function $f : S \rightarrow \mathbb{R}_{\geq 0}$ with the following properties:
(i) If $\{s_n \}_{n \geq 1} \subset S$ satisfies $\lim_{n \rightarrow \infty} s_n = s$, for $s \in S$, and $\lim_{n} f(s_n) \in \mathbb{R}_{\geq 0}$ exists, then $\lim_{n} f(s_n) = f(s)$. In other words $\mathcal{G}(f)$ is closed (since it is sequentially closed and both $S$ and $\mathbb{R}_{\geq 0}$ are first countable).
(ii) If $\{s_n \}_{n \geq 1} \subset S$ satisfies $\lim_{n} s_n = e \in E$, then $\lim_{n} f(s_n) = +\infty$
Edit (25th August) : As pointed out by mihaild, $\liminf_{n} f(s_n), \limsup_{n} f(s_n)$ must both be $\infty$ in (ii). I have updated (ii) accordingly