Let $f:[0,1]\to [0,1]$ be a continuous function. Must it have a point $x$ that $f^{-1}(x)$ is at most countable?


Added: Must it have a point $x$ that $dim\_H(f^{-1}(x))=0$ ? ($dim\_H$ means the Hausdorff dimension)



Uncountable preimage of every point 
A simple modification of the ideas of André Henriques, Sergei Ivanov and others shows that it is possible that all fibers have Hausdorff dimension $1$. For completeness I write down a complete proof.


Form a Cantor set as follows: let $I\_0 = [0,1/3]$, $I\_1=[2/3,1]$. If $I\_{i\_1,\ldots, i\_n}$ has been defined, let $I\_{i\_1,\ldots,i\_n,0}, I\_{i\_1,\ldots,i\_n,1}$ be the two intervals obtained by removing from $I\_{i\_1,\ldots, i\_n}$ a central open interval of length $1/(n+2)$ times the length of $I\_{i\_1,\ldots, i\_n}$. The Cantor set $C$ is then
$$
C := \bigcap\_{n=1}^\infty C\_n :=\bigcap\_{n=1}^\infty \bigcup\_{i\_1\ldots i\_n} I\_{i\_1\ldots i\_n}.
$$
(This is just like the construction of the usual Cantor set, except that the relative lengths of the removed intervals tend to $0$ rather than staying constant).


We define a continuous function $f:C\to [0,1]$, and then extend $f$ to all of $[0,1]$ in an arbitrary way.


Let $x$ be a point of $C$. Then $x$ is coded by a unique sequence $i\_1 i\_2\ldots\in \{0,1\}^{\mathbb{N}}$, and we define
$$
f(x) = \sum\_{n=1}^\infty i\_{n^2} 2^{-n}.
$$
It is very easy to check that $f$ is indeed continuous.


Now let $t=[0,1]$, and let $t =\sum\_{n=1}^\infty a\_n 2^{-n}$ be a binary expansion of $t$. Then $f^{-1}(t)$ consists of those points in $C$ whose code $i\_1 i\_2\ldots$ satisfies $i\_{n^2} =a\_n$. In other words, except for places corresponding to perfect squares, the other elements of the sequence are completely arbitrary. 


It is a fairly easy exercise in the calculation of Hausdorff dimension to show that $f^{-1}(t)$ has dimension $1$. The easiest way to give a formal proof is perhaps to use the mass distribution principle (see e.g. Falconer's or Mattila's books): we define a measure $\mu$ on $f^{-1}(t)$ inductively as follows. We start by assigning a unit mass to $[0,1]$. After $k$ steps, we have assigned a mass to all intervals $I\_{i\_1\ldots i\_k}$, which is $0$ whenever $I\_{i\_1\ldots i\_k}\cap f^{-1}(t)=\varnothing$. Now, if $k+1=n^2$ for some $n$, then we specify that $\mu(I\_{i\_1\ldots i\_k a\_n}) = \mu(I\_{i\_1\ldots i\_k})$ and $\mu(I\_{i\_1\ldots i\_k (1-a\_n)})=0$. Otherwise, if $k+1$ is not a perfect square, then we specify that
$$
\mu(I\_{i\_1\ldots i\_k 0})=\mu(I\_{i\_1\ldots i\_k 1}) = \frac{1}{2}\mu(I\_{i\_1\ldots i\_k}).
$$
(In other words, at the stages where the next symbol isn't determined, split mass uniformly; otherwise, pass all the mass to the required interval of next level.)


In this way we have assigned a mass to each of the intervals in the construction of $C$, so that $\mu$ is a well defined measure, which is supported on $f^{-1}(t)$ by construction. Note that if $n^2\le k < (n+1)^2$, then $\mu(I\_{i\_1\ldots i\_k})=2^{k-n}$. Hence
$$
\lim\_{k\to\infty} \frac{\mu(I\_{i\_1\ldots i\_k})}{-\log(2^k)} = 1,
$$
for any sequence $i\_1 i\_2\ldots$, and since
$$
\lim\_{k\to\infty} \frac{\log|I\_{i\_1\ldots i\_k}|}{-\log(2^k)} = 1,
$$
it is easy to deduce that
$$
\lim\_{r\to 0} \frac{\log(\mu(B(x,r))}{\log r} = 1
$$
for $\mu$-almost every $x$ (indeed for all $x$ in the support of $\mu$). Since $\mu(f^{-1}(t))=1$, the mass distribution principle implies that $\dim\_H(f^{-1}(t))\ge 1$ (and hence it is exactly $1$).


Two concluding remarks. A straightforward modification of the argument shows that for any gauge function $\varphi(x)$ such that $\lim\_{x\to 0} \varphi(x)/x=+\infty$, there is a continuous function $f:[0,1]\to [0,1]$ such that every fiber $f^{-1}(t)$ has positive $\varphi$-dimensional Hausdorff measure.


Finally, these results are not really surprising, since continuity alone does not imply any Hausdorff dimension bounds. This is the same reason why space-filling curves exist - continuity does not prevent the dimension of the image from being as large as the ambient space, and likewise it does not prevent the dimension of the fibers from being as large as the ambient space.


