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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
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If you click your name at the top of the page, you can find the questions you have asked in this website. Then you can find the answers to the questions you have asked before. On the right of each answer, you can find a check mark. By clicking it, you accept the answer. Accepting answer is the way to appreciate others' effort of answering your question. Without accepting answer, you will have "0% accept rate", as you have now. Then others will not be willing to answer questions you have asked in future. –  Paul Oct 2 '12 at 13:54
    
Oh!! Ok thanks, i will do that. –  Tomás Oct 2 '12 at 14:01
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up vote 3 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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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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