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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?

flag	asked Mar 18 2010 at 21:59 Cristos A. Ruiz 661●6●10

Tom Leinster · Accepted Answer · 2010-03-18 22:04:24Z UTC

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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.

answered Mar 18 2010 at 22:04

Tom Leinster
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Great I didn't thought of that one, can I ask differentiability? – Cristos A. Ruiz Mar 18 2010 at 22:14

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Yes and en.wikipedia.org/wiki/Volterra's_function is an example. – Johannes Hahn Mar 18 2010 at 22:20

Is it obvious that these curve are continuous? – Tran Chieu Minh Mar 19 2010 at 7:52

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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 2010 at 14:03

Petya · Answer 2 · 2010-03-18 22:04:35Z UTC

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Take a projection of a well-known Peano curve, which is a surjective continuous mapping $[0,1] \to [0,1]^2$ to a factor.

answered Mar 18 2010 at 22:04

Petya
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Snap ! – Tom Leinster Mar 18 2010 at 22:07

You are few seconds faster, Tom. – Petya Mar 18 2010 at 22:08

Continuous function from $[0,1]$ to $[0,1]$

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