Comment by yarg

Comment by yarg 3 days ago

This actually reminds me of another claim that I think is wrong for the opposite reason;

That that the Hilbert Curve covers the totality of the square; but the square contains all bound points of the form [real, real], and you can see from the rational construction of the recursive vertix generator that one of the two values for each co-ordinate pair must necessarily be a rational number (albeit one denominated by an infinite integer exponent of two).

Even if you covered all of [real, rational] + [rational, real] (which you don't), you'd still never reach all of [real, real].

Effectively 100% of the plane is not on the curve and 100% of the plane is within an infinitesimal distance of the curve.

Which I actually think is more interesting than saying that the whole damned thing is in there, which it isn't.

jepler 3 days ago

You're right that the hilbert curve only visits certain points in the unit square, and never a (non-rational,non-rational) point. While the Wikipedia article doesn't seem to mention it, other sources like [1] mention that the definition of a space-filling curve is one that comes arbitrarily close to any point within its space. I think you would be able to see that the iteration of the hilbert curve does get arbitrarily close to (say) the point (sqrt(2)/2, sqrt(2)/2).

[1]: https://people.csail.mit.edu/jaffer/Geometry/PSFC

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cooljoseph 2 days ago

The Hilbert curve does contain every point in the unit square. It is a limit of curves, and so can contain points even not in the intermediate constructions. This is similar to how the limit of 1/x as x -> infinity can be 0, even though 1/x never equals 0.

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- cooljoseph 2 days ago
  
  Also, a curve which gets arbitrarily close to every point in the unit square actually touches every point in the unit square. This is because (by definition) a curve is a continuous map from a compact space (the unit interval) to a Hausdorff space (R^2), and so its image is compact, and thus closed. A closed set contains every point that it is arbitrarily close to.
  
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  yarg 2 days ago
  
  If I travel one half of the distance from where I am to the finishing line an infinite number of times, I reach the finishing line but still never finish the race.
  With a Hilbert curve the entire plane becomes a limit.
  
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- yarg 2 days ago
  
  This doesn't seem to fly with the inductive fact that 1/2 of a power of two is always one over a power of two no matter how many times you perform the iteration.
  There are a countably infinite number of rationals between any two rationals, you can even keep splitting up those rational infinitesimal gaps into countably many rationals that are infinitesimal even relative to the earlier infinitesimals.
  And you still only end up with a countably infinite set of expressible locations and not the real continuum.
  Either x, y, or both are guaranteed to be a number of that form for all values on the curve.
  
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