Continuous function from $[0,1]$ to $[0,1]$ whose fibers are infinite

Question

Show that there exists a continuous function $f:[0,1]\rightarrow [0,1]$ such that $\forall y\in[0,1]$, all the fibers $f^{-1}(\{y\})$ have infinite cardinality.

Answer 1

You can take a continuous surjective path $\phi : [0,1] \to [0,1]^2$ and project $\pi$ to one argument to get $f = \pi \circ \phi$. Then each fiber of $f$ is the preimage of an uncountable line segment. See Space-Filling Curves.

Answer 2

The Peano-curve (see http://en.wikipedia.org/wiki/Space-filling_curve) is a uniformly continuous space filling curve $\phi:[0,1]\to [0,1]^2$. Compose it with a projection $\pi$ of the first coordinate. Projections are continuous and the composition of continuous functions is continuous, so $\pi \circ \phi$ is continuous. But for every $y \in [0,1]$ there were uncountably many $x \in[0,1]$ with $(x,y) \in [0,1]^2$ so $\pi\circ \phi$ will have uncountable fibers for every $y \in [0,1]$.

Edit: To clarify on @JasonDeVito's comment.

Certainly not every space filling curve has this property. For instance, the projection of a space filling curve that fills the unit circle in $\mathbb{R}^2$ might not have infinite fibers at $x=\pm 1$.

The reason the fibers must be infinite in this case is because for any given $x$, there are infinitely many $y$ such that $(x,y) \in [0,1]^2$. Just for concreteness, take $x=0$. Then $(0,1),(0,1/2),(0,1/3),\ldots,(0,1/n)$ are all elements of $[0,1]^2$. Since the Peano curve fills the square, this gives distinct $t_1,t_2,t_3\ldots \in [0,1]$ with $\phi(t_i) = (0,1/i)$. But the projection $\pi$ doesn't look at the second components, it only sees the first, so $(\pi \circ \phi)(t_i) = 0$ for all $i$. This gives infinitely many elements in the fiber of $x=0$. The exact same argument can be used for all $x \in [0,1]$ and it may be slightly modified to show that all the fibers are uncountable, not just infinite.