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#1
Nov2108, 02:27 AM

P: 655

Are there surfaces that have a geodesic curve which completely covers the surface, or (if that's not possible) is dense in the surface?
In other words, if you were standing on the surface and started walking in a straight line, eventually you would walk over (or arbitrarily close to) every point on the surface. My current thinking is of a torus where you start walking at a wonky angle so that your path never repeats, like so, but I'm not sure if it quite works: 


#2
Nov2108, 02:30 AM

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P: 16,091

It's actually pretty easy to see that it does work. Pick any horizontal line: where does the geodesic pierce it?



#3
Nov2108, 02:37 AM

P: 655

Now actually, thinking about it a bit, this is a very roundabout way of showing that the cardinality of RxR is equal to the cardinality of R. This torus construction is directly analogous to the diagonal counting used when you show ZxZ = Z. EDIT: Scratch that, I can only show it is dense... 


#4
Nov2108, 04:00 AM

P: 534

Dense geodesic
While a geodesic on a torus might not do it, you can construct a continuous, surjective function from [0, 1] to [0, 1]^{2}, so that R = [0, 1] ≥ [0, 1]^{2} = R^{2}.



#5
Nov2108, 04:22 PM

P: 2,499




#6
Nov2108, 05:13 PM

P: 534

The idea of the spacefilling curve is that you can construct a sequence of continuous functions f_{n}: I → I^{2} (with I = [0, 1]) that converges uniformly to another, also continuous function f: I → I^{2} that is surjective (which can be proven by showing that every point in I^{2} is in the closure of f(I), which is just f(I) because I is compact). Munkres' Topology (§44 in the second edition) describes it better than I.
(If you're still not convinced that R = R^{2}, it's easy to find a surjection [0, 1) → [0, 1)^{2} using decimal expansions.) 


#7
Nov2108, 08:23 PM

P: 532

you can't have a geodesic completely covering a surface though (not if its a differentiable manifold), just dense. The geodesic would have zero area.



#8
Nov2108, 08:55 PM

P: 655




#9
Nov2108, 09:01 PM

P: 532

take a coordinate patch on the manifold and a segment of the geodesic lying in this patch. Using the coordinates, this gives a differentiable curve in R^{2}, which must have zero area. Adding up all the segments of the geodesic lying in this coord patch will give a set with zero area and can't cover the whole coordinate patch.



#10
Nov2108, 09:13 PM

P: 655




#12
Nov2208, 05:16 AM

P: 255




#13
Nov2208, 11:47 AM

P: 532

The point of the measure theoretic argument is that it applies to any manifold, not just the torus which was used as an example. You could also use the Baire category theorem.
and a geodesic will intersect a coordinate patch in countably many segments simply because any open subset of the real line is a union of countably many connected segments 


#14
Nov2208, 04:27 PM

P: 2,499

This quote is in response to my statement that the fractal dimension is always less than two in this particular example. Yes, the Peano curve passes through every point in the plane, but how do you get there? This is the limiting condition. How many iterations of the relevant (non intersecting) fractal generator does it take to get there from some arbitrary starting point? The fractal dimension is a real number (n), in this case on the interval D1 < n < D2 where D1 and D2 are whole number fractal dimensions. The upper bound of the interval is open. In what way can it be closed? The Peano curve (sometimes called a 'monster') is either a limit value or identical with all the points on a plane by definition. 


#15
Nov2208, 08:42 PM

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If not, then what exactly are you trying to say? The Peano curve, being spacefilling, is obviously of fractal dimension 2. Each of the chosen approximations to it, being a polygonal chain, are obviously of fractal dimension 1. (This is not a problem, because fractal dimension is not continuous) 


#16
Nov2308, 02:49 AM

P: 2,499

What is the cardinality of P itself? What is the cardinality of R^2? Why is the fractal dimension of approximations to P 'one' instead of a real number 'Dn' (actually a fraction) on the interval D1 < Dn < D2? Is P necessarily a polygonal chain?, 


#17
Nov2308, 04:48 AM

P: 534

I think you might getting a bit confused between the two curves discussed: the dense geodesic (which is not a spacefilling curve and does not cover the entire surface, but is only dense), and the Peano curve (which does cover the entire unit square). The geodesic, as gel said, has zero area, while the Peano curve has area 1.
The cardinality of the Peano curve P, being an image of [0, 1], has cardinality at most that of [0, 1], which is the same as that of R. But P is the entire unit square, so it has the cardinality of [0, 1]^{2}, which is that of R^{2}. Thus, P = R = R^{2}. P is the limit of a (uniformly converging) sequence of polygonal chains, but is not itself a polygonal chain. (Indeed, if it were, it would be the union of countably many lines, but each line has measure 0, so the entire curve must have measure 0. But P has measure 1.) Minor note: You say "The cardinality of all approximations to P is the cardinality of R (or aleph 0)", but [itex]\aleph_0[/itex] is the cardinality of the natural numbers, which is different from the cardinality of R, which is [itex]2^{\aleph_0}[/itex]. 


#18
Nov2308, 05:43 AM

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P: 16,091

[0,1] > [0,1] x [0,1](at least in the case where we consider curves that merely fill the unit square), thus proving RxR <= R 


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