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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.2011-10-26
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    "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?2011-10-26
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    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.2011-10-26
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    [This thread on mathoverflow](http://mathoverflow.net/questions/279/questions-about-ordering-of-reals-and-irrationals) is relevant.2011-10-26
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    I have never heard of a definition of "range" other than the set of values which are outputs of the function, i.e $\mathrm{Ran}f = \{ f(x) : x \in \mathrm{dom}f \}$. Nevertheless, I mean that the function is a surjection onto $\mathbb{R} \setminus \mathbb{Q}$.2011-10-26
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    @Kb100: [Wikipedia](http://en.wikipedia.org/wiki/Range_(mathematics)) 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."2011-10-26
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    @Henning: Regarding the answer you just deleted, see [this answer](http://mathoverflow.net/questions/279/questions-about-ordering-of-reals-and-irrationals/7458#7458) by JDH in the MO thread I linked to above.2011-10-26
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    @joriki: Yes, I had the same insight right after posting, which is why I deleted the answer. I could have rewritten it, but there would be almost nothing left...2011-10-26
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    @joriki: Possibly I’m just old-fashioned or of a very strongly set-theoretic orientation, but I take *image* to be the default interpretation of *range* in such contexts. Until I ran into a couple of these discussions here, it had never occurred to me that this was in any way unusual.2011-10-26
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    @Brian: I don't think it indicates that you're old-fashioned -- the Wikipedia article says that "This usage [image] is more common in *modern* mathematics." :-)2011-10-26
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    One comment: there exists functions from $\mathbb{R}$ to $\mathbb{R} \setminus \mathbb{Q}$ which are onto and have only countably many discontinuities. It is really the monotony which matters in this problem.2011-10-26

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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).2011-10-26
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    @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.2011-10-26
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    Ah yes - I missed that the question was about hitting all irrationals. Very easy to visualise, thanks! :-)2011-10-27
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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 119 in $\S$6.3 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 119 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.