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?


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.

  • 2
    $\begingroup$ Great I didn't thought of that one, can I ask differentiability? $\endgroup$ Mar 18 '10 at 22:14
  • 6
    $\begingroup$ Yes and en.wikipedia.org/wiki/Volterra's_function is an example. $\endgroup$ Mar 18 '10 at 22:20
  • $\begingroup$ Is it obvious that these curve are continuous? $\endgroup$
    – abcdxyz
    Mar 19 '10 at 7:52
  • 1
    $\begingroup$ 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 '10 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.

  • 2
    $\begingroup$ Snap ! $\endgroup$ Mar 18 '10 at 22:07
  • $\begingroup$ You are few seconds faster, Tom. $\endgroup$
    – Petya
    Mar 18 '10 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.