Problem: Prove that there is no strictly increasing function $f: Q \rightarrow R$ such that $f(Q)=R$. You may not use cardinality.
Where I am: Since $f$ is strictly increasing, $f$ has an inverse (I will denote $f^{-1}$ by $g$). I know $g(R)=Q$. I know how to get a contradiction using $g$ by using the Intermediate Value Theorem. However, I need to have $g$ be continuous in order to apply the IVT. I do know that g is also strictly increasing and 1 to 1. I just need help on showing $g$ is continuous, and I can take it from there.
Update: Using the solution provided by T.A.E., I think I may have come up with something. Please give me feedback on this new attempt. I will be using the sequence definition of continuity as well.
Suppose such an $f$ exists. Since $f(Q)$ is an interval and $f$ is strictly increasing, $f$ is continuous (by a Theorem we were taught). Let $\{l_n\}$ in Q be such that $\{l_n\}$ is strictly increasing and converges to $\sqrt{2}$. Then $\{f(l_n)\}$ is an increasing sequence that is bounded above by some real number. By the Monotone Convergence Theorem, $\{f(l_n)\}$ converges to some $L$. There is a $q$ in $Q$ such that $f(q)=L$ by the definition of $f$. But this contradicts the continuity of $f$ because $\{l_n\}$ converges to $\sqrt{2}$ which is irrational so $\sqrt{2}$ is not equal to $q$.
Best Answer
Assume there is such a function $f$. I'll show that this leads to a contradiction.
Let $\{ l_{n}\}$ be an increasing sequence of rationals converging to $\sqrt{2}$, and let $\{ r_{n}\}$ be a decreasing sequence of rationals converging to $\sqrt{2}$. Then $\{f(l_{n})\}_{n=1}^{\infty}$ is an increasing sequence which is bounded above and, so, converges to some $L$. Similarly $\{f(r_{n})\}_{n=1}^{\infty}$ converges to some $R$. Furthermore, $L=R$ because, if not, $f$ cannot assume any value in $(L,R)$. By assumption, there is some rational $q$ such that $f(q)=L=R$. However, $l_{n} < q < r_{n}$ must hold because $f(l_{n}) < f(q) < f(r_{n})$ and because $f$ is strictly increasing. But there is no rational $q$ such that $l_{n} < q < r_{n}$ for all $n$. This contradiction proves the result.