real analysis - Let $f:(mathbb{R}setminusmathbb{Q})cap [0,1]to mathbb{Q}cap [0,1]$. Prove there exists a continuous$f$.

I'm working on the following problem from N.L. Carother's Real Analysis:

Let $I=(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]$ with its usual metric. Prove that there is a continuous function $g$ mapping $I$ onto $\mathbb{Q}\cap[0,1]$.

My thoughts:

I feel the preimage of open sets definition of continuity will be the easiest way to prove this. If I could show $V\subset \mathbb{Q}\cap [0,1]$ is open for all open sets $V$, and I could show that $f^{-1}(V)$ is open as well, then that would mean $f$ is continuous. I've considered trying to prove that $(\mathbb{Q}\cap [0,1])^c$ is closed, but that doesn't seem much easier. I know $\mathbb{Q}$ is dense in $\mathbb{R}$, and so maybe I can use that to say that $B_{\epsilon}(x)\setminus\{x\}\cap(\mathbb{Q}\cap[0,1])\neq\emptyset$, which would mean every $x\in\mathbb{Q}\cap[0,1]$ is a limit point of $\mathbb{Q}\cap[0,1]$, but I still don't see how this could be helpful.

Any hints on how to proceed would be appreciated. Thanks.

Answer

Hint: let $a_1=0, a_2=1/2, a_3=2/3, \ldots, a_i = 1-1/i$ and define $f$ to be constant on $(\mathbb R \setminus \mathbb Q) \cap [a_i, a_{i+1}]$.