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. 



Given a sequence of reals, we can find a dense sequence avoiding it, but can we find one continuously?
I think there is not such an $f:S\to S$. Consider the sequence $x^t\in S$ continuously depending on $t\in[0,1]$, such that $x\_0^t=-t$ and $x\_n^t=1/n $ for all $n\ge1$. Since $f(x^1)$ is dense, for some index, say $17$, we have $-1 <f\_{17}(x^1)<0$. Therefore, for $t=1$, we have $$-t=x^t\_0<f\_{17}(x^t)<x^t\_n=1/n$$ for all $n\ge1$. By continuity this must hold for all $0\le t\le1$, but it is impossible for $t=0$.


