# Conceptual proof why a continuous surjective function $f: \mathbb R \setminus \mathbb Q \to \mathbb N$ can exist

I know that a continuous surjective function $f: \mathbb R \setminus \mathbb Q \to \mathbb N$ does in fact exist, but I do not know how I can construct it — I mean, how do I even attempt to do this? My only idea was something like this:
1. Choose any two integer values $i, j$ and find an irrational number between them
2. Set the value of the function of this irrational number to a natural number $n$.
3. Make the function map all irrational numbers between $i$ and $j$ map to $n$
4. Repeat this process infinitely many times, always choosing a different $n$, $i$ and $j$. This way we will cater for the entire set of natural numbers, without running out of the points in the domain ($continuum > \aleph_o$) I am making use of the fact that my function is not defined at $x \in \mathbb N$, therefore I can "jump" with its value without losing the continuity property.

What do you think of this? Is it a correct construction?

Essentially, your approach is correct. We can be a little bit more systematic (in your approach, what happens if we first choose $i=1$ and $j=3$, and then $i=0$ and $j=2$, so the two intervals $(i,j)$ overlap? What if we choose an irrational number whose value has already been set?) by using the function $\lfloor x \rfloor$, which, for any $i \in \mathbb{N}$, maps irrational numbers between $i$ and $i+1$ to $i$. This is equivalent to always choosing intervals $(i,i+1)$ in your approach.
We can now prove continuity a bit more rigorously than you have. The preimage of any $i \in \mathbb{N}$ is an open set because it is the intersection of $(i,i+1)$ with $\mathbb{R}\setminus\mathbb{Q}$. The preimage of any set $N \subset \mathbb{N}$ is a union of these open sets and is thus open. Therefore the preimage of an open set is open, so the function is continuous.
Your construction looks fine. Specifically, you can just use the floor or ceiling functions (and say that any negative number gets mapped to $0$). You do, of course, actually have to go through an $\epsilon$-$\delta$ thing or similar to prove that it is continuous, but it shouldn't be too much work.
The idea looks fine. You can make it work as follows: define $f(x)=1$ if $x<1$, $f(x)=2$ if $1<x<2$, $f(x)=3$ if $2<x<3$, $f(x)=4$ if $3<x<4$, and so on.
$\mathbb{R}\setminus \mathbb{Q}$ is homeomorphic to $\mathbb{N}^\omega$ in the product topology.
We can use any projection as a map as sought. You proposed map is essentially the $0$th projection, if you use the homeomorphism via continued fractions.