# Can an analytic curve fill the x,y plane?

Can a curve, connected and differentiable everywhere, fill the x,y plane? My view is that a curve can only fill the x, y plane if it has singular points.

quasar987
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Try googling "differentiable space-filling curve".

The Wiki article ( near the end) states that a differentiable space filling curve cannot exist but offered no reference for a proof. I proved (at least to my own satisfaction) that a spiral originating at the origin can cross any circle centered on the origin at only one point, no matter how tightly wound it is. Since all the remaining points on the circle are not included in the spiral, the spiral cannot fill the plane.

However, this is not a general proof. One might imagine that a self-intersecting otherwise differentiable curve could fill a space (although I didn't believe so). Can you point me to a proof? Thanks

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quasar987
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The first link for me was http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2005;task=show_msg;msg=2554.0001.0001 and there is a proof there based on Sard's lemma. The proof is for a square-filling curve but as far as I can see, the proof holds without change for a space-filling curve.

The notion of a "negligible set" is used in that proof. In case you don't know what that means, by definition, a subset of R is said to be negligible if it has measure 0. The measure of a subset of the real line is a measure of its "length". Indeed, the measure of [a,b] is b-a and in particular, [0,1] has measure 1 and so it is not negligible.

The first link for me was http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2005;task=show_msg;msg=2554.0001.0001 and there is a proof there based on Sard's lemma. The proof is for a square-filling curve but as far as I can see, the proof holds without change for a space-filling curve.

The notion of a "negligible set" is used in that proof. In case you don't know what that means, by definition, a subset of R is said to be negligible if it has measure 0. The measure of a subset of the real line is a measure of its "length". Indeed, the measure of [a,b] is b-a and in particular, [0,1] has measure 1 and so it is not negligible.

The proof states that the mapping [0,1] --> [0,1] x [0,1] cannot be everywhere differentiable, but to the extent that it depends on Sard's lemma, it is not clear to me. Sard's lemma deals with critical points whereas I was thinking in terms of singular points. I'm suggesting that a space filling curve must contain singular points (such as Peano's 'monster' curve) .

A related question is: if we ignore the singularities at self intersections, can an otherwise analytic curve be space filling?

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You can have a space filling curve that's differentiable almost everywhere, but that's about it.

You can have a space filling curve that's differentiable almost everywhere, but that's about it.

Thanks Tibarn. When you say 'almost' everywhere, I assume you mean after you have excluded self-intersecting points. Is that correct?

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Thanks Tibarn. When you say 'almost' everywhere, I assume you mean after you have excluded self-intersecting points. Is that correct?

No, I assume he means differentiable almost everywhere (i.e. except on a set of measure 0). Any curve can be reparameterized so that it is locally constant - and therefore zero derivative - except on a set of measure zero. Just compose it with a http://en.wikipedia.org/wiki/Singular_function" [Broken].

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