Main Question or Discussion Point
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.
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) .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.
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].Thanks Tibarn. When you say 'almost' everywhere, I assume you mean after you have excluded self-intersecting points. Is that correct?