Let $S$ be the set of self-avoiding sequences in $\mathbb{R}$: $$S = \{s: \mathbb{N} \rightarrow \mathbb{R}: s(m) \neq s(n) \text{ if }m \neq n\}.$$ Consider $S$ with the topology of pointwise convergence, and $C(S,S)$ the associated continuous functions on $S$. For any sequence $s$ in $S$, let $\text{ran}(s)$ be the corresponding set of reals.

Is there $f \in C(S,S)$ such that $\text{ran}(f(s))$ is always dense in $\mathbb{R}$ and disjoint from $\text{ran}(s)$?

Without continuity, this would be as simple as listing the intervals with rational endpoints, and choosing one point in each interval minus $\text{ran}(s)$. With continuity, I don't know.

Background: I'd like to show that for any sequence of reals, we can find a dense sequence avoiding it, constructively and without using countable choice. I'd be happy to see an answer on that too. I think the question above, phrased without constructvity, gets at much the same issue.

Let $S$ be the set of injective sequences in $\mathbb{R}$: $$S = \{s: \mathbb{N} \rightarrow \mathbb{R}: s(m) \neq s(n) \text{ if }m \neq n\}.$$ Consider $S$ with the topology of pointwise convergence, and $C(S,S)$ the associated continuous functions on $S$. For any sequence $s$ in $S$, let $\text{ran}(s)$ be the corresponding set of reals.

Is there $f \in C(S,S)$ such that $\text{ran}(f(s))$ is always dense in $\mathbb{R}$ and disjoint from $\text{ran}(s)$?

Without continuity, this would be as simple as listing the intervals with rational endpoints, and choosing one point in each interval minus $\text{ran}(s)$. With continuity, I don't know.

Background: I'd like to show that for any sequence of reals, we can find a dense sequence avoiding it, constructively and without using countable choice. I'd be happy to see an answer on that too. I think the question above, phrased without constructvity, gets at much the same issue.