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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?

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2 Answers 2

up vote 24 down vote accepted

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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2  
Great I didn't thought of that one, can I ask differentiability? –  Cristos A. Ruiz Mar 18 '10 at 22:14
6  
Yes and en.wikipedia.org/wiki/Volterra's_function is an example. –  Johannes Hahn Mar 18 '10 at 22:20
    
Is it obvious that these curve are continuous? –  Tran Chieu Minh Mar 19 '10 at 7:52
1  
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. –  Tom Leinster 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.

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2  
Snap ! –  Tom Leinster Mar 18 '10 at 22:07
    
You are few seconds faster, Tom. –  Petya Mar 18 '10 at 22:08

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