Talk:Space-filling curve

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A curve filling a countably dimensional hypercube

You should describe such a curve. Also, you should make a link to a description of the discontinuous mapping of the unit interval onto the unit square, discovered by Georg Cantor. -- Leocat 14:13, 25 October 2006 (UTC)[reply]

Bijection, homeomorphism and self-intersection

So this space-filling curve is a continuous bijection, right? Why isn't it a homemomorphism (it certainly can't be). Because its inverse isn't continuous? That must be it. Lethe | Talk 07:42, Mar 3, 2005 (UTC)

Space filling curves aren't one to one, so aren't bijections.(Balthamos 23:35, 11 June 2006 (UTC))[reply]

Any space-filling curve must be densely self-intersecting

Wasn't one of the attributes of the Peano curve the property of non self-intersection? WLD 13:14, 23 Apr 2005 (UTC)

There cannot be any non-self-intersecting (i.e. injective) continuous curve filling up the unit square, because that will make the curve a homeomorphism from the unit interval onto the unit square (using the fact that any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism), but the unit-square (which has no cut-point) is not homeomorphic to the unit interval (all points of which, except the endpoints, are cut-points).

A.D.

Self-avoiding property of finite approximations

Under the Properties heading, the article says

Peano's original construction is self-contacting at all finite approximations, but Hilbert's construction is self-avoiding for all finite approximations.

But this is not true: First, Peano's original construction does not even use finite approximations. Instead, it is defined directly in terms of the trinary digits of the argument (!). Second, his construction could be described in terms of self-avoiding finite approximations, as the figure in the article clearly shows. And third, properties of finite approximations are not properties of the curve, so they don't belong under the Properties heading anyway. Accordingly, I am going to delete the statement. Hanche 16:37, 2 September 2007 (UTC)[reply]

Proof that a square and its side contain the same number of points

The formal proof kindly provided by JRSpriggs (around which I built this section) involves the definition of a right-inverse which seems not to be a ("proper") function, because it yields nonunique results... This is because the original space-filling curve is self-intersecting. Was I clear enough? Am I right? Is this right inverse compatible with Cantor–Bernstein–Schroeder theorem? Paolo.dL 15:19, 7 September 2007 (UTC)[reply]

Using the axiom of choice, we can get a right inverse to any surjective function. In fact, asserting the existence of an inverse to every surjective function is AC. I sense there might be away to get around AC by using directly some sort of dual to C-B-S, but I doubt there's actually a constructive proof. Depends what you call "proper" I guess. --192.75.48.150 17:40, 7 September 2007 (UTC)[reply]