TITLE: Show that there does not exist a strictly increasing function $f : \mathbb Q \to \mathbb R$ such that $f(\mathbb Q) = \mathbb R$.
QUESTION [7 upvotes]: 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!

REPLY [3 votes]: I want to give another proof which actually uses the assumptions:
Let $x$ be some irrational number and $(x_n)$ a rational monotone increasing sequence converging to $x$. Then $f(x_n)$ is a mononte increasing sequence and bounded (take some rational $y > x$, then $f(x_n)$ is bounded from above by $f(y)$). Hence the sequence $f(x_n)$ converges to $y$ in $\mathbb{R}$. By surjectivity there is some rational $z$ with $f(z) = y$. By assumption $z > x_n$ for all $n$ and thus $z \geq x$. Since $x$ is irrational we have $z > x$. Now there exists a rational number $x'$ with $z > x' > x > x_n$. We obtain $f(x_n) < f(x') < f(z)$, but then $f(x')$ would already be an upper bound of $f(x_n)$ and thus $f(z) = y \leq f(x') < f(z)$ which is a contradiction.