Given a sequence of reals, we can find a dense sequence avoiding it, but can we find one continuously? 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.  
 A: 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 $t=0$.
A: If you replace the reals $\mathbb{R}$ with Cantor space $2^{\mathbb{N}}$ or with Baire space $\mathbb{N}^{\mathbb{N}}$ (homeomorphic to the space of irrationals), then the answer is yes. Indeed, one can have the function defined on the whole space of sequences, not just the injective ones. 
To see this, define $f(x_0,x_1,...)=(y_0,y_1,...)$, where $y_k$ extends the $k^{th}$ finite sequence $u_k$, and diagonalizes the $x_n's$ in a canonical way beyond the length of $u_k$, so that the $|u_k|+j^{th}$ digit of $y_k$ is different from the $|u_k|+j^{th}$ digit of $x_j$. This is a continuous function, since any finitely many digits for the output are determined by finitely many digits of the input. The $y_k$'s are not among the $x_n$, since they diagonalize against this list, and the $y_k$'s are dense, because $y_k$ extends $u_k$. (By changing the diagonalization procedure slightly, it is easy to arrange that the $y_k$ are all distinct.)
