Is there a necessary and sufficient condition for the boundary of a planar region to be a finite union of Jordan curves?
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5$\begingroup$ 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. $\endgroup$– Kevin BuzzardNov 9, 2009 at 11:28
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1$\begingroup$ 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. $\endgroup$– JaikrishnanNov 9, 2009 at 12:34
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$\begingroup$ 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 ${(x,y): 0 < x < 1, 0 < y < \sin (1/x)}$ $\endgroup$– Martin M. W.Nov 9, 2009 at 14:21
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1$\begingroup$ @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 $-2<y$. $\endgroup$– Harald Hanche-OlsenNov 9, 2009 at 15:39
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$\begingroup$ @Harald: By region I mean a open and connected set. Thank you Martin and Harald for the example. $\endgroup$– JaikrishnanNov 9, 2009 at 17:57
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4 Answers
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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.
answered Nov 9, 2009 at 19:41
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$\begingroup$ I can't think of a counterexample either. Of course, I have a feeling that if you can prove that the boundary is locally contractible, you already know enough to decide the question anyhow. But it's a nice conjecture. $\endgroup$ Nov 9, 2009 at 21:48
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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.
answered Nov 9, 2009 at 18:00
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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.
answered Nov 9, 2009 at 16:02
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$\begingroup$ eeep! how do I comment instead of "answering"? $\endgroup$ Nov 9, 2009 at 16:03
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2$\begingroup$ You need 50 reputation to comment (see mathoverflow.net/faq#reputation), which is kind of a weird feature of the reputation system. Since you can't comment on other peoples' posts, nobody will mind that you left an answer instead of a comment. But if you have more to say (and you still don't have 50 rep), edit this answer rather than posting another one. $\endgroup$ Nov 9, 2009 at 16:45
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$\begingroup$ Martin's example as amended by me in the comments is an example of a connected, bounded region with a connected complement, whose boundary is not a finite union of Jordan curves. Moreover, the region is the interior of its closure. $\endgroup$ Nov 9, 2009 at 17:39
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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.
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$\begingroup$ Oh, yes -- It must also be that $\partial U$ is the boundary of the complement of $\overline U$. Otherwise, you could have something silly like $U$ being the complement of an arc. Incidentally, I think that one could then omit the second condition if one replaces local connectivity by local contractibility, as Greg suggested. $\endgroup$ Nov 12, 2009 at 22:41