Talk:Netto's theorem

Completely unclear statement

One sentence reads:

"If one relaxes the requirement of continuity, then all smooth manifolds of bounded dimension have equal cardinality, the cardinality of the continuum."

It is entirely unclear what this means. If continuity is not being assumed, would it be too much trouble to state what *is* being assumed?

Apparently it is.

Also Since any two nonempty manifolds having positive dimension also have the cardinality of the continuum, what is the point of introducing this totally off-topic comment into the article?

I see no point whatsoever. — Preceding unsigned comment added by 2601:200:c082:2ea0:5134:6f4c:81da:368d (talk • contribs) 03:29, 27 June 2023 (UTC)

The next sentence begins "Therefore, ..." You might try reading it and seeing what having equal cardinalities implies about the existence of non-continuous mappings. —David Eppstein (talk) 04:31, 27 June 2023 (UTC)

Convex sets??

The article makes this claim (boldface mine): Osgood curves are continuous bijections from one-dimensional spaces to subsets of the plane that have nonzero area. They form Jordan curves in the plane. However, by Netto's theorem, they cannot cover the entire plane, unit square, or any other convex set.

with a reference to Sagan's book on space-filling curves. This (1) makes no sense (consider that (a) convexity is not a topological property and (b) a line segment in the plane is a convex set, and most certainly can be covered bijectively by a curve) and (2) does not appear to be supported by anything in Sagan's book. (I had a look at the three sections specifically cited, none of which mentions the word "convex" or the concept of convexity either explicitly or implicitly, and also searched for "convex" -- this is all using the version of the book available at archive.org -- and unsurprisingly found nothing resembling the claim in the article.)

There is a similar claim in the article on Osgood curves. I am about to change both to something that is (1) true and (2) actually supported by the reference to Sagan's book. — Preceding unsigned comment added by Gareth McCaughan (talk • contribs) 13:40, 27 June 2023 (UTC)

While you're at it, could you at least make an attempt to not replace simple phrasing like "convex set" with convoluted phrasing like "set that cannot be disconnected by removing any single point"? First of all, your version is even less true (a closed Osgood curve obviously covers itself, and because it is a closed curve it cannot be disconnected by removing any single point). And second, it's less understandable; see WP:TECHNICAL. Maybe you mean domain? The interior of a convex set (when it is nonempty) is a familiar example of a domain. —David Eppstein (talk) 17:52, 27 June 2023 (UTC)

Well, I was trying to replace it with something correct, which "convex set" demonstrably isn't. A curve is a continuous map from [0,1]. Netto's theorem is about the case where such a curve is a bijection, which in particular implies that its image is not a closed curve, so I do not believe your counterexample is a counterexample. The actual wording in Sagan's book (the thing that's cited as a reference for the false claim about convex sets, which so far as I can see is not supported by anything in the book) is: "Clearly, this proof applies not only to Q, W, and T but to any compact and connected set that contains at least two points and cannot be disconnected by the removal of one point." (Q, W, T are the unit square, the unit cube, and a particular triangular region.) Again, the claim with "convex set" is simply wrong; a unit line segment in the plane is a convex set and is literally the easiest thing in the world to cover bijectively with a curve. Gareth McCaughan (talk) 15:14, 1 July 2023 (UTC)

Convex set is correct, if you merely add a qualifier to avoid collinear sets. It is a special case of a more general claim that an Osgood curve cannot be a space-filling curve. It cannot fill all points of any region of the plane.

Your claim about "cannot be disconnected by the removal of one point" does NOT apply to Osgood curves in general. It is as wrong wrong wrong as the statement you are trying to replace. Osgood curves can be closed curves, requiring the removal of two points to disconnect. —David Eppstein (talk) 16:00, 1 July 2023 (UTC)

Let us stipulate for the moment that my application of the criterion in Sagan's book to Osgood curves is as incorrect as you say it is. (I think the incorrectness is purely superficial; if we switch from maps from [0,1] to maps from the unit circle, the condition just changes from "not disconnected by removing any one point" to "not disconnected by removing any two points".) It remains the case that convexity has nothing to do with anything here: Netto's theorem is about arbitrary smooth manifolds and for talk of convexity to mean anything one needs at least something like a metric. The current language about "two-dimensional regions" seems much better.

... But, actually, it's not quite clear to me that Netto's theorem _does_ imply the claim currently in the article (or my version, or the version with "convex"). Netto says: no continuous bijection from [0,1] to a 2-manifold. But what's currently being claimed is: no continuous bijection from [0,1] to a subset of a 2-manifold that contains (say) an open set. I expect that's _true_ but if you can get it just from Netto's theorem then I'm not seeing how. I think the proof of (the relevant special case of) Netto's theorem in Sagan's book does easily adapt to prove this stronger result but haven't checked carefully. Is there a way to get the stronger claim from Netto's theorem itself?

Gareth McCaughan (talk) 21:36, 1 July 2023 (UTC)

Re "I think the incorrectness is purely superficial": pot, meet kettle.

What Sagan's book actually states as Netto's theorem is that there is no continuous bijection from a 1-manifold to a 2-manifold. If you had a bijection b from a simple closed curve C to a set S containing an open subset O, then b would also be a bijection from the preimage of O to O. But the preimage of O is a 1-manifold (the preimage of an open set is open and an open subset of a manifold is a manifold) and O is a 2-manifold (again, an open subset of a manifold is a manifold). So b cannot exist. —David Eppstein (talk) 00:26, 2 July 2023 (UTC)

Yes, I thought you might have that reaction to what I said about superficiality! But consider: (1) convexity is simply not a topological notion; (2) in order to make "convex" work you have to add some condition like "of full dimension" or "containing an open set", and that condition itself does all the work. Convexity is simply an irrelevance here; having found that "convex" simpliciter is wrong, you can fix it by saying "convex and X" for various conditions X, but in each case X itself works too. Whereas "can't be disconnected by removing one/two points" is a condition you can state in the framework of smooth manifolds, and is how the proof actually works (at least in the case of mapping from an interval, which is e.g. the only case actually proved in the book cited as reference).

I like your argument for getting "can't cover an open set" from Netto's theorem, and am no longer concerned that it's not clear that Netto implies the claim in the article. (And, in view of that, I think that actually the Right Thing to say is something like "they cannot cover the entire plane, the unit square, or any other region containing a nonempty open set", because that avoids questions like "what about some more exotic set of Hausdorff dimension 2 but not containing a nonempty open set?" There's probably a proof that you can't cover the plane minus its rational points with the continuous image of an interval, but that seems harder to get from Netto. Though I might be missing something easy.) I think "open set" is better than "domain" because (1) "open set" is more widely familiar terminology and (2) "domain" is more specific by requiring connectedness, which is not in fact a requirement here.

Gareth McCaughan (talk) 15:56, 3 July 2023 (UTC)