Show that there does not exist a strictly increasing function $f : \mathbb Q \to \mathbb R$ such that $f(\mathbb Q) = \mathbb R$. 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!         
 A: Here is a proof that does not use the uncountability of $\mathbb{R}$.  Suppose $f:\mathbb{Q}\to\mathbb{R}$ is strictly increasing, and let $\alpha\in\mathbb{R}$ be any irrational number.  Let $A=(-\infty,\alpha)\cap\mathbb{Q}$ and $B=(\alpha,\infty)\cap\mathbb{Q}$.  Then every element of $f(A)$ is less than every element of $f(B)$, so $\sup f(A)\leq \inf f(B)$.  Moreover, if $x\in A$, then $x+\epsilon\in A$ for sufficiently small rational $\epsilon>0$, so $f(x)<f(x+\epsilon)\leq \sup f(A)$.  Similarly, if $x\in B$, then $f(x)>\inf f(B)$. Thus $$f(\mathbb{Q})=f(A)\cup f(B)\subseteq (-\infty,\sup f(A))\cup (\inf f(B),\infty),$$ so any $y\in [\sup f(A),\inf f(B)]$ is not in the image of $f$.
A: 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.
A: 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.
A: We don't need the fact that $f$ is increasing. Suppose that $f(\Bbb Q)=\Bbb R$, then for any $x\in\Bbb R$ there is a $r_x\in\Bbb Q$ such that $f(r_x)=x$. 
Since there are uncountable many $x\in \Bbb R$, the set of all $r_x$ must also be uncountable. However, $\Bbb Q$ is countable, a contradiction. 
