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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.

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Do you know how to accept answer? –  Paul Oct 2 '12 at 13:35
    
I did not understand your question... –  Tomás Oct 2 '12 at 13:41
    
Oh!! Ok thanks, i will do that. –  Tomás Oct 2 '12 at 14:01

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up vote 4 down vote accepted

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.

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If $x=0.d_1d_2\ldots d_n\ldots$ is the decimal expansion of $x$, define $f(x)=0.d_2d_4\ldots d_{2n}\ldots$ Clearly f is continuous and every fiber has infinite (actually continuum) cardinality.

Sorry for the writing. I'm not momentarily familiar with LaTex.

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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.

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To make sure I'm undestanding, is it a fact that every space filling curve, when projected, has uncountable fibers? Or is it just the special example Peano found? –  Jason DeVito Oct 2 '12 at 17:36
    
@JasonDeVito See my edit. –  nullUser Oct 2 '12 at 18:10
    
It would be interesting to see how the plot of such a function would look like. –  Marco Oct 2 '12 at 20:36
    
@nullUser: Thanks for that, I was just being extremely dense. –  Jason DeVito Oct 2 '12 at 21:32
    
@Marco probably something like this: en.wikipedia.org/wiki/File:WeierstrassFunction.svg not to scale obviously. –  nullUser Oct 2 '12 at 22:53

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