Why can't there be a monotone function with domain $\mathbb{R}$ and range $\mathbb{R} \setminus \mathbb{Q}$?

Why can't there be an increasing function with domain $\mathbb{R}$ and range $\mathbb{R} \setminus \mathbb{Q}$?

Edit: By range I mean the image of the function's domain, i.e. the function admits every irrational value.

I feel like there should be a bijection between every irrational value it takes and the number of discontinuities it has, and I know monotone functions have at most countably many discontinuities, so this would be a contradiction.

But I don't know how to show it.

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I suppose you mean a nonconstant monotone function. – Srivatsan Oct 26 '11 at 1:06
"Range" is a somewhat ambiguous word. By "range $\mathbb R\setminus\mathbb Q$" do you mean that the image of the function must be this entire set, or just that every value of the function must be irrational? – Henning Makholm Oct 26 '11 at 1:11
Mapping into $\mathbb R \setminus \mathbb Q$? Easily done. Mapping onto $\mathbb R \setminus \mathbb Q$? Not possible. Indeed, think of the discontinuities. In fact, think of just one discontinuity. – GEdgar Oct 26 '11 at 1:26
This thread on mathoverflow is relevant. – joriki Oct 26 '11 at 1:38
@Kb100: Wikipedia has heard of such a definition. The concluding sentence of the introduction of that article says: "Because of this ambiguity, it is a good idea to specify whether it is the image or the codomain being discussed." – joriki Oct 26 '11 at 1:43

The basic reason that there can be no monotone mapping of $\mathbb{R}$ onto $\mathbb{R}\setminus\mathbb{Q}$ is the completeness of the linear order on $\mathbb{R}$.

Suppose that $f:\mathbb{R}\to\mathbb{R}\setminus\mathbb{Q}$ is an increasing surjection. Let $L=\{x\in\mathbb{R}:f(x)<0\}$; clearly $L$ is bounded above (e.g., by the real number $y$ such that $f(y)=\sqrt2$), so $L$ has a least upper bound $u$. Now what can $f(u)$ be?

• If $u\notin L$, then $f(u)>0$, and $f(x)\ge f(u) > 0$ for every $x\ge u$, so $f[\mathbb{R}]\cap(0,f(u))=\varnothing$, and $f$ isn’t a surjection.

• If $u\in L$, then $f(u)<0$, but $f(x)>0$ for every $x>u$, so $f[\mathbb{R}]\cap (f(u),0)=\varnothing$, and again $f$ is not a surjection.

In either case we have a contradiction, so $f$ cannot be a surjection.

If $f$ were a decreasing surjection, $-f$ would be an increasing surjection, so a decreasing function from $\mathbb{R}$ to $\mathbb{R}\setminus\mathbb{Q}$ can’t be a surjection either.

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Just because the function is increasing, doesn't mean that it will reach 0, or 1. Also I don't think you can assume there is a real number y with f(y)=1 if you're also assuming f:R->R\Q. I'm sure you can fix both these issues by a linear transformation of your increasing function (e.g. all increasing functions are a linear transform of an increasing function that has both positive and negative values). – tttppp Oct 26 '11 at 7:32
@tttppp: You’re right about $1$; that was supposed to be a positive irrational, and I’ve fixed it. The other part isn’t a problem: if $f$ doesn’t stretch that far, it certainly isn’t surjective. But I’ve fixed that the easy way by making it a proof by contradiction. – Brian M. Scott Oct 26 '11 at 19:41
Ah yes - I missed that the question was about hitting all irrationals. Very easy to visualise, thanks! :-) – tttppp Oct 27 '11 at 7:46

Note first that the OP is probably using "range" in the sense of "image", not in the sense of "codomain".

I suggest you use Theorem 28 of these notes on discontinuities of weakly monotone functions. Here is the basic idea (I will assume $f$ is weakly increasing).

Case 1: The function is everywhere continuous. Then its image $f(\mathbb{R})$ must be an interval, by the Intermediate Value Theorem, so it certainly can't be $\mathbb{R} \setminus \mathbb{Q}$.

Case 2: The function is discontinuous at at least one point $a \in \mathbb{R}$. Here's where Theorem 28 comes in: then we have either $f(a^-) = \lim_{x \rightarrow a^-} f(x) < f(a)$ and the image contains no values in $(f(a^-),f(a))$ or $f(a) < f(a^+) = \lim_{x \rightarrow a^+} f(x)$ and the image contains no values in $(f(a),f(a^+))$. But every nonempty open interval contains irrational numbers.

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+1 for a nice clean answer (as usual!). – lhf Oct 26 '11 at 20:38