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Are there continuous functions $f:I\to I$ such that $f^{-1}(\{x\})$ is countably infinite for every $x$? Here, $I=[0,1]$.

The question "Infinity-to-one function" answers is similar but without the condition that it be countable. (The range was $S^2$, not $I$, but the accepted answer also worked for $I$.)

I doubt one exists, since I haven't been able to come up with one, but I'm not sure. There's probably some topological reason why this is impossible.

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Akiva Weinberger
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  • Doesn't $[0,1] \to [0,1]$, $x \mapsto \sin(\frac{1}{x})$ satisfy this? – AnalysisStudent0414 Jan 04 '16 at 18:09
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    @AnalysisStudent0414 Almost, but what's $f(0)$? – Akiva Weinberger Jan 04 '16 at 18:10
  • The reason I posted this as a comment and not as an answer – AnalysisStudent0414 Jan 04 '16 at 18:15
  • One way of disproving existence of such function would be to prove that $I$ is not a disjoint union of countably infinite closed sets (to be honest I don't think this is true, but I don't see immediately such partition). – Wojowu Jan 04 '16 at 18:15
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    @Wojowu : But I think it is, you can construct such a disjoint union by transfinite recurrence : take a $x$, construct a sequence $(U^{(x)}n){n \in \Bbb N}$ that converge to $x$, take another $x'$, construct a sequence $(U^{(x')}n){n \in \Bbb N}$ that have no common terms with $U^{(x)}$ and that converge to $x'$, rinse-repeat and Zorn-Lemma. No, I have a gut feeling that compactness is involved here – Tryss Jan 04 '16 at 18:21
  • @AkivaWeinberger : each set is $U^{(x)}$, the elements of a sequence, hence countable – Tryss Jan 04 '16 at 18:23
  • @Tryss Yes, that seems to work. So my idea fails. – Wojowu Jan 04 '16 at 18:24
  • do you have any examples which come close ? the function $x\cos(x)$ with $x\neq0$ and $f(0)=0$ get us at least one point with countable preimage and everything elsse with finite preimage. Using this we can extend it to have any finite subset with finite preimage and everything else with finite preimage. Do you have examples that come closer (for example, on which gives an infinite set with uncountably infinite preimages)? – Asinomás Jan 04 '16 at 19:29
  • @dREaM Do you mean $x\cos\frac1x$? Also, I think I can get countably many points to have a countable preimage by modifying the Cantor function. It has a countably infinite amount of flat lines, so I think I could replace each of them with shifted, shrunk versions of $x\cos(\frac1x)$. – Akiva Weinberger Jan 04 '16 at 21:11
  • @Wojowu it is indeed true that $[0,1]$ is not a countable union of pairwise disjoint non-empty closed sets, this is true by Sierpinski's theorem for continua. How does this imply the fact that no such $f$ exists? Then we have an uncountable partition of countable sets, quite different. – Henno Brandsma Jan 04 '16 at 21:40
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    @HennoBrandsma I am talking about union of (uncountably many) countable closed sets, not countably many closed sets. – Wojowu Jan 04 '16 at 21:43
  • If the answer to the Q is "no",the compactness of $I$ ought to matter. Because replacing $I$ with $(0,1)$ in the Q , it becomes trivial : Take $ f=g h$ where $ h:(0,1)\to R$ is a homeomorphism and $g(x)=\sin x.$ – DanielWainfleet Jan 04 '16 at 23:25

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Here is an example. Start with the Cantor function $g:[0,1]\to[0,1]$, i.e. the function that sends $x=\sum a_n/3^n$ to $\sum a_n/2^{n+1}$ if every $a_n$ is $0$ or $2$ and is locally constant off of the Cantor set $K$. Note that for each dyadic rational $q\in(0,1)$, there is a unique (nondegenerate) interval $[a_q,b_q]$ with $a_q,b_q\in K$ such that $g(x)=q$ for all $x\in[a_q,b_q]$, and $[0,1]\setminus K$ is the disjoint union of the intervals $(a_q,b_q)$. Define a function $h:[0,1]\to[0,1]$ by saying $h=g$ on $K$, and on each interval $[a_q,b_q]$, $h$ is a finite-to-one continuous surjection $[a_q,b_q]\to[q,q+1/2^n]$, where $2^n$ is the denominator of $q$ (in lowest terms), and $h(a_q)=h(b_q)=q$.

The function $h$ is clearly continuous when restricted to each interval $[a_q,b_q]$, and in particular is continuous off of $K$. As you approach a point of $K$ without staying in a single interval $[a_q,b_q]$, you pass through infinitely many such intervals $[a_q,b_q]$, with the denominators of the numbers $q$ getting larger and larger, and so $h$ remains continuous because $g$ is continuous. Thus $h$ is continuous on all of $[0,1]$.

I claim every point of $[0,1]$ except $0$ has countably infinitely many preimages under $h$ (since $h(x)\geq g(x)$ for all $x$, $0$ is the only preimage of $0$). It is clear that every point has countably many preimages: $h$ agrees with $g$ on $K$ and $g$ is finite-to-one on $K$, and off of $K$, $h$ is finite-to-one on each of the countably many intervals $[a_q,b_q]$. Now if $x\in[0,1]$ and $q\in(0,1)$ is any dyadic rational obtained by truncating a binary expansion of $x$ at some point, then by construction, $h$ takes the value $x$ somewhere on the interval $[a_q,b_q]$ (since $q\leq x\leq q+1/2^n$). If $x\neq0$, then there are infinitely many different dyadic rationals $q\in(0,1)$ that can be obtained by truncating a binary expansion of $x$ (if $x$ is a dyadic rational, use the binary expansion of it that ends in $1$s). So we find that $x$ must have infinitely many preimages unless $x=0$.

Finally, it is easy to modify $h$ to give $0$ infinitely many preimages. For instance, define $f(x)=i(x)$ if $x\in[0,1/2]$ and $f(x)=h(2x-1)$ if $x\in[1/2,1]$, where $i:[0,1/2]\to[0,1]$ is any countable-to-one continuous function that achieves the value $0$ infinitely many times and satisfies $i(1/2)=0$ (it is easy to construct such a function by appropriately modifying the function $x\mapsto x\sin^2(1/x)$). Then $f$ is a continuous function $[0,1]\to[0,1]$ with countably infinitely many preimages for each point.

Eric Wofsey
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