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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 :)

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Here is a relevant reference: 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 1:41
    
@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 Answer 1

up vote 8 down vote accepted

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.

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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 0:01
    
@mitch just make one up or take a homeomorphism from the square to the disk –  yoyo Mar 29 '11 at 1:13
    
@yoyo: OK. Is the obvious 'interpret the cartesian coords as polar coords' homeomorphic? –  Mitch Mar 29 '11 at 14:35
    
Thanks for the answer. –  hattenn Mar 29 '11 at 18:26

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