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 2


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 space-filling curves.) Then the composite $f$ of $g$ with first projection $[0, 1]^2 \to [0, 1]$ has the required property.

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    $\begingroup$ Great I didn't thought of that one, can I ask differentiability? $\endgroup$ Mar 18, 2010 at 22:14
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    $\begingroup$ 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? $\endgroup$
    – abcdxyz
    Mar 19, 2010 at 7:52
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    $\begingroup$ Space-filling curves, such as the Peano curve, are indeed continuous. The existence a not-necessarily-continuous 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 space-filling 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 well-known Peano curve, which is a surjective continuous mapping $[0,1] \to [0,1]^2$ to a factor.

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    $\begingroup$ Snap ! $\endgroup$ Mar 18, 2010 at 22:07
  • $\begingroup$ You are few seconds faster, Tom. $\endgroup$
    – Petya
    Mar 18, 2010 at 22:08

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