The following exercise in an analysis text and I am trying to solve it without concepts of general topology but fail.
Show that there does not exist a strictly increasing function $f : \mathbb Q \to \mathbb R$ such that $f(\mathbb Q) = \mathbb R$.
Attempt 1. Suppose that the function $f(D) = \mathbb R$ is monotone. If its image $f(D)$ is an interval, then the function $f$ is continuous. So, if we suppose by contradiction that a strictly increasing function $f : \mathbb Q \to \mathbb R$ exists such that $f(\mathbb Q) = \mathbb R$ it must be continuous.
Attempt 2. Since intersection of the set of irrational numbers of the domain is empty so by a convergence of a sequence in the domain $\mathbb Q$ in either case of converging to a rational or irrational number there is nothing to reach a contradiction.
Attempt 3. The function $f$ is injective and it's not surjective so the inverse function is not defined such that I can use theorems about an inverse of a function.
Please help!
Best Answer
A strictly increasing function is necessarily injective. If $f(\mathbb{Q})=\mathbb{R}$, then the function is also surjective (by definition). This would imply that there exists a bijection between $\mathbb{Q}$ and $\mathbb{R}$.
EDIT: For the sake of completion, the statement is still true if $f$ is not assumed to be strictly increasing.
To see this, note that there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}$. Therefore, if there were a surjection from $\mathbb{Q}$ to $\mathbb{R}$, there would be one from $\mathbb{N}$ to $\mathbb{R}$. But note that Cantor's argument actually shows that there is no surjection from $\mathbb{N}$ to $\mathbb{R}$. qed.
As a curiosity, note that this argument does not use AC.