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Is it possible to construct a continuous function from $[0,1] \to \mathbb{R}$ whose length is infinite?

TorsionSquid
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2 Answers2

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Yes. Try $f(x)=\sqrt{x}\sin(1/x)$. This curve has unbounded variation, so in particular it is not rectifiable.

Fedorius Neckbeard
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NOTE: Because 1. is a discrete function, the below does not actually describe a continuous function.

Yes. Note that:

  1. A bijection exists between $[0, 1]$ and $(0, 1)$

Then:

  1. A bijection exists between $(0, 1)$ and $ℝ$   —   $f(x) = \ln(\frac{1}{x}-1)$

Or, to use a more classic example:

  1. A bijection exists between $(0, 1)$ and $(-π,π)$   —   $f(x) = 2πx-π$
  2. A bijection exists between $(-π,π)$ and $ℝ$   —   $f(x) = \tan(x)$

∴  A bijection exists between $[0,1]$ and $ℝ$

Community
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Zaz
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  • This isn't continuous. – Wojowu Oct 25 '15 at 18:03
  • Oh, of course, my bad. I'll leave the answer here though for the sake of anyone coming here from Google who might be looking for this. Unless anyone has any objections or knows of a better question to move this answer to. – Zaz Oct 25 '15 at 18:59