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?

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    $\begingroup$ $\Bbb Q$ is countable, $\Bbb R$ is not? $\endgroup$ – David Mitra Mar 2 '14 at 20:43
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    $\begingroup$ What do you understand "continuous" to mean when $\mathbb Q$ is involved? Topology with the rationals is weird. $\endgroup$ – Ben Millwood Mar 2 '14 at 20:43
  • $\begingroup$ I'm in a intro to real analysis class, so I'm only allowed to use basic definitions of continuity. I cannot use the cardinality of $\mathbb{Q}$ or $\mathbb{R}$ $\endgroup$ – Paul Malinowski Mar 2 '14 at 20:45
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    $\begingroup$ Note that the strictly increasing condition implies that the function is also injective $\endgroup$ – Mark Bennet Mar 2 '14 at 20:45
  • $\begingroup$ For the sake of the problem, is there a way to do it only using the sequential or episolon-delta definition of continuity? $\endgroup$ – Paul Malinowski Mar 2 '14 at 20:47

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 $$ f(x)<f(x_1)\le \lim_{z\to x_0^-}f(z)\le f(y_1)<f(y). $$

  • $\begingroup$ The step where you note that $x_0 > x \in \mathbb Q$ implies $a > f(x)$ might need some unpacking, since it relies on some non-trivial properties of $\mathbb Q$ (specifically, that it is densely ordered); in particular, it would not hold if we replaced $\mathbb Q$ with $\mathbb Z$ (although the conclusion would obviously still be true). $\endgroup$ – Ilmari Karonen Mar 2 '14 at 21:44
  • $\begingroup$ is there a way to do this without using left/right limits? $\endgroup$ – Paul Malinowski Mar 7 '14 at 21:32

Let $p$ be an irrational number. Show that the supremum of $f(x)$ for $x \in \mathbb Q$ with $x < p$ and the infimum of $f(x)$ for $x \in \mathbb Q$ with $x > p$ are be equal. If this is $y$, then there is no rational $x$ with $f(x) = y$.


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