Does there exist a continuous function $f:[0,1]\rightarrow [0,1]$ such that $f$ takes every value in $[0,1]$ an infinite number of times?
2 Answers
Yes. In fact, there exists such an $f$ taking every value uncountably many times.
Take a continuous surjection $g: [0, 1] \to [0, 1]^2$. (Such things exist: they're spacefilling curves.) Then the composite $f$ of $g$ with first projection $[0, 1]^2 \to [0, 1]$ has the required property.

2$\begingroup$ Great I didn't thought of that one, can I ask differentiability? $\endgroup$ Mar 18, 2010 at 22:14

6$\begingroup$ Yes and en.wikipedia.org/wiki/Volterra's_function is an example. $\endgroup$ Mar 18, 2010 at 22:20

$\begingroup$ Is it obvious that these curve are continuous? $\endgroup$– abcdxyzMar 19, 2010 at 7:52

1$\begingroup$ Spacefilling curves, such as the Peano curve, are indeed continuous. The existence a notnecessarilycontinuous surjection from [0, 1] to [0, 1]^2 is much easier to prove  it's just an argument about cardinality of sets. Often one defines a spacefilling curve as a uniform limit of a sequence of curves, and uses the fact that a uniform limit of continuous maps is continuous. $\endgroup$ Mar 19, 2010 at 14:03
Take a projection of a wellknown Peano curve, which is a surjective continuous mapping $[0,1] \to [0,1]^2$ to a factor.

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