# Existence of Function Taking Every Value Uncountably Many Times

I am stuck on the following problem coming from a plane geometry theorem I'm working on. This is a generalization of that problem which might actually prove easier to handle. All the techniques that I have tried have failed, and it wasn't in Counterexamples in Analysis, which is the only collection of results on this sort of thing I'm familiar with. Here, by a function $f$ we mean any function, not necessarily continuous or measurable.

Is it possible to find a subset $A \subset \mathbb{R}$ and a function $f: A \rightarrow [0,1]$ such that $f$ takes every value uncountably many times, $f^{-1}(x)$ is totally disconnected for every $x \in [0,1]$ (equivalently, $f$ is not constant on any interval), and satisfying the following local monotonicity condition:

If $a_n \rightarrow a$ are points of $A$ and $a_n \leq a$, then $\limsup f(a_n) \leq f(a)$, and

if $a_n \rightarrow a$ are points of $A$ and $a \leq a_n$, then $f(a) \leq \liminf f(a_n)$?

Thanks!

• – Rahul Aug 8 '18 at 10:26
• Is there any function that satisfies your "local monotonicity condition" other than $f(x)=x$? Did you mean to compare the limsup and liminf with $f(a)$ rather than $a$? – Henning Makholm Aug 8 '18 at 10:29
• Henning, yes. Fixed! – John Samples Aug 8 '18 at 10:33
• What about some modified form of the Weisterstrass function that could rapidly oscillate around each value in $[0,1]$? But I'm not sure whether this would satisfy the monotonicity criterion. – Jam Aug 8 '18 at 11:01

Let $A=[0,\frac23]$. To compute $f(x)$, write $x$ as a decimal fraction, and follow these rules:

• If $x$ can be written with only even digits, convert each of its digits to a binary digit by mapping $0, 4, 8$ to binary $0$ and $2, 6$ to binary $1$.

• Otherwise, let $x_1$ and $x_2$ be the numbers closest to $x$ on each side that do consist of only eve digits and interpolate linearly between $f(x_1)$ and $f(x_2)$. Since $x_1$ will end in repeating $8$s and $x_2$ will end in repeating $0$s, both map to repeating binary $0$s, so $f(x_1)\ne f(x_2)$ and this linear interpolation will not make $f$ constant on an interval.

This function is continuous so certainly satisfies your "monotonicity" criterion, and hits every number in $[0,1]$ continuum many times.

(If you don't require $A$ to be an interval, you could also just let it consist of the Cantor set of numbers that can be written with even digits, and skip the linear interpolation).

• I will draw that out and check, hopefully it works! Or maybe not, I was expecting it to be impossible haha. My particular problem requires that all the sets $f^{-1}(x)$ be a Cantor Set, but it seems like I can just use your function and then select a Cantor Set in each preimage, then restrict to their union, yes? This is the problem my question stems from: math.stackexchange.com/questions/2872483/… – John Samples Aug 8 '18 at 11:02
• That's fine if it is a Cantor Set. The Cantor Set partitions itself so everything can work the same. – John Samples Aug 8 '18 at 11:08
• @JohnSamples: Yes, see my recent final paragraph. If you take $A$ to be all numbers that can be written with the digits $0, 2, 4, 6$ and don't end in repeating $2$s or $6$s, then every preimage is a middle-$\frac{8}{10}$s Cantor set and all preimages are translations of each other! – Henning Makholm Aug 8 '18 at 11:08
• It seems like a fun kind of math. I wish I could study it all haha – John Samples Aug 8 '18 at 11:11

Take any space filling curve, i.e., any continuous surjective function $g : [0, 1] \to [0, 1] \times [0, 1]$ and arrange for $g$ to be non-constant on any open interval (either by adjusting an arbitrary space filling $g$ to shrink any open interval where it is constant to a point or by picking a specific $g$ like the Peano curve that doesn't suffer from this problem). Then define $f(x) = \pi_1(g(x))$, where $\pi_1 : [0, 1]\times[0, 1]\to [0, 1]$ is projection onto the first factor of the product ($\pi_1(x, y) = x)$. Then $f : [0, 1] \to [0, 1]$ is continuous and so certainly meets your monotonicity conditions and for any $x \in [0, 1]$, $f^{-1}(x)$ comprises the set $\{y : [0, 1] \mid g(y) \in \{x\} \times [0, 1]\}$, which is uncountable because $g$ is surjective and $\{x\} \times [0, 1]$ is uncountable.