**Question.** Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ whose restriction $f|\mathbb Q^\omega$ is injective?



Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ with injective restriction $f|\mathbb Q^\omega$?
It looks like no. 


Assume the contrary. We may start with two distinct rationals $q\_1,p\_1$ such that the sets $f(q\_1\times \mathbb{R}^{\omega-1})=f(\{(q\_1,\cdot,\cdot,\dots)\})$ and $f(p\_1\times \mathbb{R}^{\omega-1})$ intersect. Indeed, for any $p\_1$ the set $f(p\_1\times \mathbb{R}^{\omega-1})$ has non-empty interior, choose reals such that $f(p\_1,x\_2,x\_3,\dots)$ is interior point of this set. After that any $q\_1$ close enough to $p\_1$ works, since $f(q\_1,x\_2,\dots)$ is close to $f(p\_1,x\_2,\dots)$.


After that choose small intervals $\Delta\_2,\Delta\_3,\dots$ and $\delta\_2,\delta\_3,\dots$ such that the sets $f(q\_1\times \Delta\_2\times \Delta\_3\times \dots)$ and $f(p\_1\times \delta\_2\times \delta\_3\times \dots)$ intersect and have diameter less than 1. After that choose $q\_2\in \Delta\_2$ and $p\_2\in \delta\_2$ such that $f(q\_1\times q\_2\times \Delta\_3\times \dots)$ and $f(p\_1\times p\_2\times \delta\_3\times \dots)$ intersect. Proceeding this way, we get two rational sequences with the same value of $f$.


