15

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.

nullUser
  • 27,877
  • 2
    I suppose you mean a nonconstant monotone function. – Srivatsan Oct 26 '11 at 01:06
  • 1
    "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? – hmakholm left over Monica Oct 26 '11 at 01:11
  • 6
    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 01:26
  • 3
    This thread on mathoverflow is relevant. – joriki Oct 26 '11 at 01:38
  • 2
    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 $\mathbb{R} \setminus \mathbb{Q}$. – nullUser Oct 26 '11 at 01:39
  • 2
    @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 01:43
  • @Henning: Regarding the answer you just deleted, see this answer by JDH in the MO thread I linked to above. – joriki Oct 26 '11 at 01:59
  • @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... – hmakholm left over Monica Oct 26 '11 at 02:02
  • @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. – Brian M. Scott Oct 26 '11 at 02:04
  • 1
    @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." :-) – joriki Oct 26 '11 at 02:07
  • 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. – N. S. Oct 26 '11 at 21:00

2 Answers2

21

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.

Brian M. Scott
  • 616,228
  • 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 07:32
  • @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 07:46
12

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.

Community
  • 1
Pete L. Clark
  • 97,892