I'm trying to show there does not exist a strictly increasing function:
$f: \mathbb{Q} \rightarrow \mathbb{R}$ that is surjective.
I started by assuming that such a function exists. Then, this implies the function is continuous because it is strictly monotone with an interval as its image. Now, I think I should somehow use that to contradict that there are enough values in $\mathbb{R}$ for each value in $\mathbb{Q}$, but I'm stuck. Hints?
[Math] Strictly Monotone Surjective Function
calculusreal-analysis
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Best Answer
The proof that follows avoids using the fact that the cardinality of $\mathbb R$ is larger than the one of $\mathbb Q$.
Let $x_0\in \mathbb R\smallsetminus\mathbb Q$. Then as $f$ is (strictly) increasing, then the limit $$ a=\lim_{x\to x_0^-}f(x), $$ exists and it is a real number. In fact for every $x,y\in\mathbb Q$, with $x<x_0<y$, $$ f(x)<a<f(y), $$ as $f$ is strictly increasing.
This means that $a\not\in\mathrm{Ran}(f)$, and hence $f$ is not surjective.
Note. Let me explain better why for every $x,y\in\mathbb Q$, with $x<x_0<y$, we have that $f(x)<a<f(y)$. As $x<x_0<y$, there are $x_1,y_1\in\mathbb Q$, such that $x<x_1<x_0<y_1<y$, and since $f$ is strictly increasing, we have that $$ f(x)<f(x_1)\le \lim_{z\to x_0^-}f(z)\le f(y_1)<f(y). $$