real analysis - Function whose image of every open interval is $(-infty,infty)$

How to find a function from reals to reals such that the image of every open interval is the whole of R?

Is there one which maps rationals to rationals?

Answer

Though this can be done explicitly with enough cleverness (for example with the Conway base 13 function), I rather like the following choice-based argument, which requires almost no thought once you're familiar with transfinite recursion.

Consider the set consisting of all ordered triples of reals $(a,b,c)$ with $a

Now we build our function $f$ recursively along this well-order, fixing the value of $f$ at one new point at each step. At the step corresponding to $(a,b,c)$, we want to ensure that there exists a point $x$ in the open interval $(a,b)$ such that $f(x)=c$. Since we've only fixed $f$ at less-than-continuum-many points, and $(a,b)$ has cardinality continuum, we can choose an $x$ in $(a,b)$ such that $f(x)$ is not yet fixed, and fix $f(x)$ to be $c$.

This recursion gives us a partial function from $\mathbb{R}$ to $\mathbb{R}$ that already satisfies our requirements. We can make it total by just setting $f(x)$ to be $0$ (say) wherever $f(x)$ is not yet defined.

If we additionally want $f$ to map rationals to rationals, we can simply set $f(x)=0$ for every rational $x$ before commencing the recursion.