Let $X$ and $Y$ be toplogical spaces, and call a function $f:X\to Y$ wild if the preimage $f^{-1}(\{y\})$ is dense in $X$ for every $y\in Y$ -- or, equivalently, if the image of every nonempty open subset of $X$ is all of $Y$.
Can every function $X\to Y$ be approximated pointwise by wild functions?
For $\mathbb Q\to\mathbb Q$ the answer is yes -- if we want to approximate $f$, then select a surjection $h:\mathbb Z\to \mathbb Q$, and approximate $f$ by $$ f_n(x) = \begin{cases} h(\lfloor \tan q \rfloor) & \text{if }x=(2p+1)2^{-q}\text{ for }p,q\in\mathbb Z, q>n \\ f(x) & \text{otherwise} \end{cases} $$
For $\mathbb Q\to\mathbb R$ the answer is no for the trivial reason that there are no wild functions.
For $\mathbb R\to\mathbb R$, we can extend the construction for $\mathbb Q\to\mathbb Q$, if only a Vitali set $A$ exists:
$$ f_n(x) = \begin{cases} a+h(\lfloor \tan q \rfloor) & \text{if }x=a+(2p+1)2^{-q}\text{ for }a\in A, p,q\in\mathbb Z, q>n \\ f(x) & \text{otherwise} \end{cases} $$
Question: Is the answer still yes for $\mathbb R\to\mathbb R$ if we don't have enough choice to construct a Vitali set?
Wild functions $\mathbb R\to\mathbb R$ exist no matter whether we have choice or not -- such as Conway's base-13 function -- but it is not obvious that there will be enough of them to be dense in $\mathbb R^{\mathbb R}$ under pointwise convergence.
(Inspired by this question.)
Sleeping on it helps. :-)
A sufficient conditions for wild functions to be dense is that there exists one wild function $X\to\mathbb N\times Y$. Calling this function $g$, we can approximate any $f$ by
$$ f_n(x) = \begin{cases} y &\text{when }g(x)=\langle k,y\rangle\text{ with }k>n \\ f(x) &\text{otherwise} \end{cases} $$
In the $\mathbb R\to\mathbb R$ case we can manufacture a suitable $g$ given a wild function $h:\mathbb R\to\mathbb R$, such as Conway's base-13 function:
$$ g(x) = \begin{cases} \langle 0,0\rangle &\text{if }h(x)\text{ is an odd multiple of }\pi/2 \\ \langle \lfloor h(x)/\pi \rfloor, \tan(h(x))\rangle &\text{otherwise} \end{cases}$$
This same construction works whenever we have a wild $X\to Y$ and a surjection $Y\to\mathbb N\times Y$.
We see that the topology of $Y$ doesn't really matter. The only case where the topology on $Y$ can matter is if $Y$ and $\mathbb N\times Y$ are not equipotent, which requires $Y$ to be finite (boring case) or the Axiom of Choice to fail.