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Is there a necessary and sufficient condition for the boundary of a planar region to be a finite union of Jordan curves?

flag	asked Nov 9 2009 at 11:22 Jaikrishnan 485●7

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Facetious comment: a necessary and sufficient condition is that the boundary is a finite union of Jordan curves. Less facetious comment: I've never understood this sort of question. Implicit in it is some sort of metamathematical principle: you want the equivalent condition to be "a long way from the condition you state" in some metric, but I've never understood the metric. – Kevin Buzzard Nov 9 2009 at 11:28

I am sorry for being imprecise. I just want some reformulations of the conditions which are "easier" to work with. For instance, I think one necessary and sufficient condition is that the interior of the closure of the region is is the region itself. – Jaikrishnan Nov 9 2009 at 12:34

I don't think the "is the interior of its closure" condition does what you want. The interior of the closure of the open upper half plane is itself, for instance. Or even for a bounded region, take ${(x,y): 0 < x < 1, 0 < y < \sin (1/x)}$ – Martin M. W. Nov 9 2009 at 14:21

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@Martin: I think we need to be clear on what a “region” is. To me, a region is by definition open and connected, which rules out your last example. (Some authors even define a region to be bounded, I think.) Still, a variant of your example is illuminating: Just replace $0<y$ by $-2<y$. – Harald Hanche-Olsen Nov 9 2009 at 15:39

@Harald: By region I mean a open and connected set. Thank you Martin and Harald for the example. – Jaikrishnan Nov 9 2009 at 17:57

Greg Kuperberg · Answer 1 · 2009-11-09 19:41:15Z UTC

I can think of an important necessary condition: The boundary has to be locally contractible; in particular, locally connected. The topologist's sine curve is not locally connected, while the Hawaiian earring is not locally simply connected, and both occur in boundaries of open sets.

If you throw in the condition that the open set should be the interior of its closure (although that does not always happen even if the boundary is a union of Jordan curves), then at the moment I can't think of a counterexample.

Harald Hanche-Olsen · Answer 2 · 2009-11-09 18:00:34Z UTC

I am going to throw caution to the wind and suggest an Answer, based on the previous comments: No, there is no useful characterization of the regions you seek, other than the requirement stated. On the one hand, there is the region $\{(x,y):0\lt x\lt 1, -2\lt y\lt \sin x^{-1}\}$, and on the other, you can make part of the boundary an Osgood curve, by which I mean a Jordan arc of positive area. (See, e.g., W. F. Osgood, A Jordan curve of positive area. Trans. Amer. Math. Soc. 4 (1903) 107–112). Between these two examples, I think you'll be hard pressed to find an easily checkable local condition on the boundary that will guarantee the “Jordan-ness” of the boundary curve.

some guy on the street · Answer 3 · 2009-11-09 16:30:53Z UTC

Videtur I can't post comments of my own? This is not a complete answer.

@buzzard, I'd say yours probably isn't a facetious comment, in that I can imagine a union of two Jordan curves --- that is, an intersection of two connected open planar sets --- looking particularly wild. For example, take one a JC with positive measure, and for the second take a small isotopy of the first.

With Harald, I prefer to assume that "region" means "open subset"; this simplifies distinctions between "(connected) component" and "maximal connected subset".

Obvious note: neither the region nor its complement can have an infinite sequence of separated components; in other words, the (open) region's closure and its (closed) complement both have a finite number of components. But I don't believe this is sufficient; again, I'm thinking of rather fractalous regions, but they're trickier to describe.

Clinton Curry · Answer 4 · 2009-11-12 22:37:39Z UTC

Local connectivity of the boundary provides much of the topological structure that such a domain would have. If you specify the following two conditions, you have that the boundary of a domain $U$ is a finite union of simple closed curves, I think.

$\partial U$ is locally connected.
$\overline U$ has finitely many complementary components.

The thrust is this: if $V$ is a complementary component of $\overline U$, then $\partial V$ is the common boundary of two simply connected domains ($V$ and the component of the complement of $\overline V$ containing $U$). A locally connected plane continuum which is the common boundary of two connected open sets is always a circle.

Boundary of planar region

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