real-analysis
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.
First, let me give some terminology that I will use here:
Suppose that $E\subseteq\Bbb R,$ $x_0\in\Bbb R,$ and $f:E\to\Bbb R$ are given.
We will say that $x_0$ is $E$-supremal if $$x_0=\sup\{x\in E:x<x_0\},$$ and $E$-infimal if $$x_0=\inf\{x\in E:x>x_0\}.$$ Equivalently, $x_0$ is $E$-supremal if $E\cap(x_0-\delta,x_0)$ is non-empty for all $\delta>0,$ and $E$-infimal if $E\cap(x_0,x_0+\delta)$ is non-empty for all $\delta>0$. Note that $x_0$ is a limit point of $E$ iff $x_0$ is $E$-supremal and/or $E$-infimal. If $x_0\in E$, then $x_0$ is isolated in $E$ iff $x_0$ is neither $E$-supremal nor $E$-infimal.
If $x_0$ is a limit point of $E$ and $L\in\Bbb R$, we say that $L$ is the limit of $f(x)$ as $x$ approaches $x_0$--written $L=\lim_{x\to x_0}f(x)$--if for all $\epsilon>0$ there is some $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $x\in E$ and $0<|x-x_0|<\delta$. We say that $\lim_{x\to x_0}f(x)$ fails to exist if there is no such $L$, or if $x_0$ is not a limit point of $E$.
If $x_0$ is $E$-supremal and $L\in\Bbb R$, we say that $L$ is the limit of $f(x)$ as $x$ approaches $x_0$ from below--written $L=\lim_{x\to x_0^-}f(x)$--if for all $\epsilon>0$ there is some $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $x\in E\cap(x_0-\delta_\epsilon,x_0).$ We say that $\lim_{x\to x_0^-}f(x)$ fails to exist if there is no such $L$, or if $x_0$ is not $E$-supremal.
If $x_0$ is $E$-infimal and $L\in\Bbb R$, we say that $L$ is the limit of $f(x)$ as $x$ approaches $x_0$ from above--written $L=\lim_{x\to x_0^+}f(x)$--if for all $\epsilon>0$ there is some $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $x\in E\cap(x_0,x_0+\delta_\epsilon).$ We say that $\lim_{x\to x_0^+}f(x)$ fails to exist if there is no such $L$, or if $x_0$ is not $E$-infimal.
Now, suppose $x_0\in E$. Recall that $f$ is said to be continuous at $x_0$ iff for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta.$ It can be shown that $f$ is continuous at $x_0$ iff either $x_0$ is isolated in $E,$ or $x_0$ is a limit point of $E$ and $f(x_0)=\lim_{x\to x_0}f(x).$ $f$ is said to be discontinuous at $x_0$ iff it is not continuous there--that is, iff there is some $\epsilon_0>0$ such that for all $\delta>0$ we have $|f(x)-f(x_0)|\ge\epsilon_0$ for some $x\in E$ with $0<|x-x_0|<\delta.$ Note that if $f$ is discontinuous at $x_0$, then $x_0$ is $E$-supremal and/or $E$-infimal (why?).
You should be able to prove the following:
Lemma: Suppose $E\subseteq\Bbb R,$ $x_0\in E,$ and $f:E\to\Bbb R$ are given, with $f$ discontinous at $x_0.$ Then exactly one of the following $5$ situations occurs:
(i) $x_0$ is $E$-supremal and $E$-infimal. There is some $L\in\Bbb R$ with $L\ne f(x_0)$ and $L=\lim_{x\to x_0}f(x).$
(ii) $x_0$ is $E$-supremal and not $E$-infimal. There is some $L\in\Bbb R$ with $L\ne f(x_0)$ and $L=\lim_{x\to x_0^-}f(x).$
(iii) $x_0$ is $E$-infimal and not $E$-supremal. There is some $L\in\Bbb R$ with $L\ne f(x_0)$ and $L=\lim_{x\to x_0^+}f(x).$
(iv) $x_0$ is $E$-supremal and $E$-infimal. There are some $L_1,L_2\in\Bbb R$ with $L_1\ne L_2$, $L_1=\lim_{x\to x_0^-}f(x),$ and $L_2=\lim_{x\to x_0^+}f(x).$
(v) $x_0$ is $E$-supremal but $\lim_{x\to x_0^-}f(x)$ fails to exist, and/or $x_0$ is $E$-infimal but $\lim_{x\to x_0^+}f(x)$ fails to exist.
We say that $f$ has a removable discontinuity at $x_0$ if (i) holds; that $f$ has an essential discontinuity at $x_0$ if (v) holds; that $f$ has a jump discontinuity at $x_0$ in the other three situations.
Now for the kicker: real-valued monotone functions on subsets of the reals can have only jump discontinuities, and can only have at most countably many of those.
Let me give you an idea of how to prove this. (Toward the end, I will use the fact that the rationals are countable and dense in the reals.)
Proof Idea: Suppose $E\subseteq\Bbb R,$ that $f:E\to\Bbb R$ is monotone function, and that $f$ is discontinuous at some point $x_0\in E$. (WLOG, suppose $f$ is nondecreasing. The proof when $f$ is nonincreasing will be similar.) By definition, then, there is some $\epsilon_0>0$ such that for all $\delta>0,$ there exists $x\in E$ where $0<|x-x_0|<\delta$ and $|f(x)-f(x_0)|\ge\epsilon_0.$ We must address three cases.
(Case 1): If $x_0$ is not $E$-supremal, then $x_0$ must be $E$-infimal since $f$ is discontinuous at $x_0$. Moreover, since $f$ is non-decreasing, we must have that $f(x)\ge f(x_0)+\epsilon_0$ for all $x\in E\cap(x_0,\infty),$ for otherwise, this would violate our choice of $\epsilon_0$. (Why?) Then $\{f(x):x\in E,x>x_0\}$ is a non-empty set bounded below by $f(x_0)+\epsilon_0,$ so since $f$ is nondecreasing then we can conclude (why?) that $$\lim_{x\to x_0^+}f(x)=\inf\{f(x):x\in E,x>x_0\}\ge f(x_0)+\epsilon_0>f(x_0).$$ Thus, we have a type (iii) discontinuity.
(Case 2): If $x_0$ is not $E$-infimal, then in a similar fashion to Case 1, we conclude that $x_0$ is $E$-supremal, that $\lim_{x\to x_0^-}f(x)$ exists and is at most $f(x_0)-\epsilon_0,$ meaning we have a type (ii) discontinuity.
(Case 3): If $x_0$ is both $E$-supremal and $E$-infimal, then since $f$ is nondecreasing, we can conclude that $$\lim_{x\to x_0^+}f(x)=\inf\{f(x):x\in E,x>x_0\}\ge f(x_0)$$ and likewise that $\lim_{x\to x_0^-}f(x)\le f(x_0).$ By our choice of $\epsilon_0$, we can show that these two limits cannot be equal, so we have a type (iv) discontinuity.
Thus, $f$ must have a jump discontinuity at $x_0.$
Now, let $Y$ be the union of the "gaps" in $\text{range}(f)$--more precisely: $$Y=\{y\in\Bbb R:\exists x_1,x_2\in E\text{ s.t. }f(x_1)<y<f(x_2),\text{ and }\exists a<y<b\text{ s.t. }(a,b)\cap\text{range}(f)=\emptyset\}.$$ Note that $Y$ is a union of pairwise disjoint open intervals, and is disjoint from $\text{range}(f)$. (Why?) Note further that if $f$ has a jump discontinuity at any point $x_0\in E$, then at least one (and at most two) of the component intervals of $Y$ has an endpoint at $f(x_0)$. (Why?) Thus, we readily have a one-to-one function from the set of discontinuities of $f$ into the set of component intervals of $Y$--in particular, say that for any given point of discontinuity $x_0$, we take $x_0$ to the component interval of $Y$ with $f(x_0)$ for an endpoint, and if we have a choice, take $x_0$ to the component interval of $Y$ whose lower endpoint is $f(x_0)$.
Now, let $\{q_n\}_{n\in\Bbb N}$ be any enumeration of the rationals, and note that for each component interval $J$ of $Y,$ there is some least $n$ such that $q_n\in J$--call this $n_J$ for each such $J$. Different components of $Y$ correspond to different $n_J$ (why?), so the function $J\mapsto n_J$ is a one-to-one function from the set of component intervals of $Y$ into the naturals, meaning that $Y$ has at most countably-many component intervals. Since we also have a one-to-one function from the set of discontinuities of $f$ into the set of component intervals of $Y$, then $f$ has at most countably-many discontinuities.
For continuity, fix $\varepsilon>0$. Consider the sets $A=(f(x_0)-\varepsilon/2,f(x_0))$ and $B=(f(x_0),f(x_0)+\varepsilon/2)$. If $x_0\neq a$,there must be a point $x_1\in[a,b]$ such that $f(x_1)\in A$. If $x_0\neq b$, there must be a point $x_2$ such that $f(x_2)\in B$. Since $f$ is strictly increasing, $x_1<x_0<x_2$. Let $\delta=\min\{x_0-x_1,x_2-x_1\}$. Then for all $x$ such that $x_1<x<x_2$, $f(x_1)<f(x)<f(x_2)$, so $$|f(x)-f(x_0)|\leq|f(x)-f(x_1)|+|f(x_1)-f(x_0)|<\varepsilon.$$
If $x_0=a$, we let $\delta=x_2-x_0$, where $x_2$ is as described above, and then the result follows in the same manner.
If $x_0=b$, we let $\delta=x_0-x_1$, where $x_1$ is as described above, and the result follows in the same manner.
If $f:[a,b]\rightarrow [c,d]$ is bijective, monotonic and continuous, $f^{-1}:[c,d]\rightarrow [a,b]$ is bijective and monotonic, so it is continuous for the same reason as $f$.
Best Answer
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.