Prove that there cannot be a continuous onto function $f :\Bbb R \to \Bbb R \setminus \Bbb Q$ Prove that there cannot be a continuous onto function $f :\Bbb  R \to \Bbb R \setminus \Bbb Q$
Please let me know if you agree with my proof:
Suppose, toward a contradiction, that there exists such a continuous onto function $f :\Bbb  R \to \Bbb R \setminus \Bbb Q$ . Since $\Bbb R$ is connected, $f(\Bbb R) \subset \Bbb R\setminus \Bbb Q$ must also be connected. Pick any $q \in \Bbb Q$ and let $U=(\infty, q)$ and $V=(q, \infty)$. Note that $U$ and $V$ are non-empty, disjoint and open. Then $U\cup V$ is a disconnection of $\Bbb R\setminus \Bbb Q$, a contradiction. Furthermore, ${\Bbb R}=f^{-1}(U \cup V)$ is also a contradiction since $\Bbb R$ is connected.
 A: Your proof just needs a cleaning up; set, for simplicity, $\mathbb{I}=\mathbb{R}\setminus\mathbb{Q}$. If $q\in\mathbb{Q}$, then
$$
(-\infty,q)\cap\mathbb{I}
\qquad\text{and}\qquad
(q,\infty)\cap\mathbb{I}
$$
are nonempty, disjoint and open subsets of $\mathbb{I}$, whose union is $\mathbb{I}$. Thus $\mathbb{I}$ is not connected and therefore it cannot be the image of a continuous function from a connected space.
The last sentence should be removed as it's inessential and badly written.

Actually, much more can be said: any continuous function $f\colon \mathbb{R}\to\mathbb{I}$ is constant, because every subset of $\mathbb{I}$ with at least two points is disconnected. Indeed, if $a\ne b$ are elements of $X\subset\mathbb{I}$, take a rational $q$ such that $a<q<b$ and repeat the argument above with $(-\infty,q)\cap X$ and $(q,\infty)\cap X$.
A: If two irrational numbers were in the image, then by intermediate value theorem any number between them would be in the image as well. Thus, some rational will have to get covered. So, at most one irrational may be in the image. So, $f$ must be a constant function.
