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As you already know, the interior of a circle is represented by an inequality. For example,

$$x^2+y^2\leq1$$

for the unit circle. Today I was thinking by myself and I wondered if there is a curve that could represent every point inside of a circle. Maybe with a spiral like this,

If you can't represent it perfectly with a curve, what would be the closest way to represent it?

This question is asked merely out of curiosity, it may be completely irrelevant or meaningless :)

Community
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hattenn
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    Here is a relevant reference: http://en.wikipedia.org/wiki/Space-filling_curve. The curve can't be one-to-one, but it can be done. – Jonas Meyer Mar 28 '11 at 20:34
  • Why can't it be one-to-one? Isn't the Peano curve one-to-one? – Mitch Mar 28 '11 at 23:59
  • @Mitch: A continuous bijective map from a compact space to a Hausdorff space is a homeomorphism, but $[0,1]$ is not homeomorphic to $[0,1]^2$. So a space-filling curve such as the Peano curve, which is continuous and surjective, cannot be one-to-one. – Noah Stein Mar 29 '11 at 01:41
  • @Jonas, @Noah: thanks, I'll ask in what way the Peano curve is not 1-to-1 separately. – Mitch Mar 29 '11 at 14:14

1 Answers1

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Since topologically a disc and a square are the same, most of what you might want to know about this falls under the heading of Space-filling curves. To summarize, the answer to your main question is that the disc $D^2$ is the image of the interval $[0,1]$ under a continuous map, but not a one-to-one (non-intersecting) continuous map. So it depends on exactly what you mean by curve.

Noah Stein
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  • Is there an explicit version of the Peano curve directly for a disk (rather than the usual $[0,1]^2$ – Mitch Mar 29 '11 at 00:01
  • @mitch just make one up or take a homeomorphism from the square to the disk – yoyo Mar 29 '11 at 01:13
  • @yoyo: OK. Is the obvious 'interpret the cartesian coords as polar coords' homeomorphic? – Mitch Mar 29 '11 at 14:35