Q:

Prove there is no strictly increasing function $f$ from irrationals to reals.

Prove there does not exist a strictly increasing function $f\colon \mathbb{R} - \mathbb{Q} \rightarrow \mathbb{R}$ such that $f(\mathbb{R} - \mathbb{Q}) = \mathbb{R}$.
I imagine the best way to go about this is by contradiction. So suppose there does exist a strictly increasing function. Then I think I have to use the fact that since $f$ is strictly increasing, it is one-to-one. However, I am not sure.
Any help is appreciated.

A:

Consider the sets $f(\Bbb I^-)$ and $f(\Bbb I^+)$, where $\Bbb I=\Bbb R\setminus\Bbb Q$. For any $x\in\Bbb I^-, y\in\Bbb I^+$, we have $x<0<y\implies f(x)<f(y)$, so each element of $f(\Bbb I^-)$ is less than each element of $f(\Bbb I^+)$. Now both sets are nonempty, so they partition $\Bbb R$ into two pieces, and there is a point $\alpha$ in the middle which is equal to the supremum of $f(\Bbb I^-)$ and the infimum of $f(\Bbb I^+)$. But since $f(\Bbb I^-)\cup f(\Bbb I^+)=\Bbb R$, $\alpha$ is in one of the two sets, and so  $\alpha=f(\beta)$ for some $\beta\in\Bbb I^+$ or $\Bbb I^-$. In the first case, $f(\beta/2)$ is a number less than the infimum; in the second case it is greater than the supremum - a contradiction.
Note: this is a direct application of the following simple topological property: $f$ is an order isomorphism, so under the order topologies, $\Bbb R$ and $\Bbb I$ must be homeomorphic. But $\Bbb I$ is disconnected (in particular, at 0) and $\Bbb R$ is not.

