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Topic: About the boundary of the bounded region (Read 3530 times) 

immanuel78
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About the boundary of the bounded region
« on: Jun 14^{th}, 2007, 12:08am » 
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Let G be a bounded region in C. Then the boundary of G is the union of closed paths. At a glance, it is trivial but I can't prove it.


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Obob
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Re: About the boundary of the bounded region
« Reply #1 on: Jun 14^{th}, 2007, 10:33am » 
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For one thing, it is necessary that you consider a single point to be a closed path, in order for a punctured disc to have boundary that is a union of paths. But then the boundary is the union of its points. Of course, this is not what is intended. So the first thing to do is to more precisely define what is meant by the boundary being the union of closed paths. Perhaps each component of the boundary can be parameterized by a closed path, or something like that.


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Michael Dagg
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Re: About the boundary of the bounded region annulus.PNG
« Reply #2 on: Jun 14^{th}, 2007, 11:07am » 
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Here is an idea...


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Obob
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Re: About the boundary of the bounded region
« Reply #3 on: Jun 14^{th}, 2007, 12:01pm » 
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I think in the case of the annulus the way to do it is that the boundary is the union of the two boundary circles. This shouldn't be an issue of making the region the interior of the regions, and we shouldn't need to orient the paths and have cancellation either.


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immanuel78
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Re: About the boundary of the bounded region
« Reply #4 on: Jun 14^{th}, 2007, 10:31pm » 
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There is a modification in the problem. Let G be a bounded region in C. Then the boundary of G is the union of a finite number of closed paths. (Here we assume that a single point is not a closed path and a path is a continuous function on [a,b])

« Last Edit: Jun 14^{th}, 2007, 10:43pm by immanuel78 » 
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Obob
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Re: About the boundary of the bounded region
« Reply #5 on: Jun 15^{th}, 2007, 12:09pm » 
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Under this modification, the only way this could even possibly be true is if Michael_Dagg's interpretation of the problem is intended. But I'm not sure that will fix things either. For instance let A be the Cantor set obtained from the unit interval by removing middle thirds successively. Then (0,1) x (0,1)  A x A is a region in the complex plane, after identifying C with R^2, since the Cantor set is closed. But I think you'll have one heck of a time writing the boundary as a finite number of closed paths, in any interpretation. Maybe it can be done, but I doubt it.

« Last Edit: Jun 15^{th}, 2007, 12:10pm by Obob » 
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Michael Dagg
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Re: About the boundary of the bounded region
« Reply #6 on: Jun 15^{th}, 2007, 3:58pm » 
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My previous remark was intended to promote discussion. Obob, your remark is absolutely correct: if onepoint sets are to count as closed paths, then every set whatsoever is the union of closed paths.


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Felix.R
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Re: About the boundary of the bounded region
« Reply #7 on: Jun 15^{th}, 2007, 11:15pm » 
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I don't think I have ever seen this problem demonstrated as it is stated in the first version explicitly. But not knowing very much about a Cantor set as you describe, but if you make one or more holes on the real line as in removing the middle thirds (I don't know what you mean by "successively" because nothing will change regardless of the order in which they are removed) that individual point sets must surely be closed. For example, if we take each individual point of an arc in C as a closed region then this arc is a closed region. Is it simply connected? Looks like it is.


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Felix.R
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Re: About the boundary of the bounded region
« Reply #8 on: Jun 16^{th}, 2007, 5:22pm » 
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I know now what you mean by removing middle thirds successively. Isn't the region (0,1) x (0,1) really different than the product of the Cantor set A? I mean A doesn't contain certain points while (0,1) x (0,1) contains all points bounded by (0,1) x (0,1). This is an interest topic. The general result seems to only make sense when the situation is exactly like what the other member posted about one point sets being closed paths. This leads me to think that there is an equivalence relation between one point sets and disjoint one point sets in this same context. After all, how much different is a disk with multiple (nonoverlapping) punctures on the real line that meets the requirements of your A x A region compared to a disk with a single puncture?

« Last Edit: Jun 16^{th}, 2007, 5:23pm by Felix.R » 
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Felix.R
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Re: About the boundary of the bounded region
« Reply #9 on: Jun 16^{th}, 2007, 5:53pm » 
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I realized I actually mean overlapping punctures for then the region would not be equivalent to your A x A region.


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Obob
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Re: About the boundary of the bounded region
« Reply #10 on: Jun 16^{th}, 2007, 7:33pm » 
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I think that the correct statement is that every component of the boundary can be parameterized by a closed path. In general, uncountably infinitely many such paths will be needed, as in the example of (0,1) x (0,1)  A x A, where A is the Cantor set. Moreover, closed points are closed paths in this interpretation.


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immanuel78
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Re: About the boundary of the bounded region
« Reply #11 on: Jun 17^{th}, 2007, 10:03am » 
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Obob, I have a question. Is (0,1) x (0,1)  A x A connected?


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Obob
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Re: About the boundary of the bounded region
« Reply #12 on: Jun 17^{th}, 2007, 11:20am » 
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Yes it is, Immanuel. Suppose (a,b) and (c,d) are in S = (0,1) x (0,1)  A x A. Then either a or b is not in A. If a is not in A, then {a} x (0,1) is contained in S. In particular, (a,b) is connected by a path to some point (a,e) in S, where e is not in A. Then (0,1) x {e} is contained in S. Next note that as before either {c} x (0,1) or (0,1) x {d} is contained in S. But {c} x (0,1) intersects (0,1) x {e}, and (0,1) x {d} intersects {a} x (0,1). So in the case where a is not in A, we can join (a,b) to (c,d) by traveling along horizontal or vertical paths alone. The same result holds if b is not in A. Therefore S is path connected, and its boundary has uncountably infinitely many components.


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Obob
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Re: About the boundary of the bounded region
« Reply #13 on: Jun 17^{th}, 2007, 11:56am » 
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More generally, if A is any subset of a pathconnected topological space X, then X x X  A x A is pathconnected (possibly empty), by the same argument as above.


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Eigenray
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Re: About the boundary of the bounded region
« Reply #14 on: Jun 20^{th}, 2007, 8:39pm » 
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on Jun 15^{th}, 2007, 12:09pm, Obob wrote:For instance let A be the Cantor set obtained from the unit interval by removing middle thirds successively. Then (0,1) x (0,1)  A x A is a region in the complex plane, after identifying C with R^2, since the Cantor set is closed. But I think you'll have one heck of a time writing the boundary as a finite number of closed paths, in any interpretation. Maybe it can be done, but I doubt it. 
 A simpler example (but with only countably many components) is to let X = {0} u {1/n : n=1,2,...}, or more generally any compact set with infinitely many components. Then DX has boundary a circle plus infinitely many points. on Jun 16^{th}, 2007, 7:33pm, Obob wrote:I think that the correct statement is that every component of the boundary can be parameterized by a closed path. 
 This doesn't work either. If X is compact, connected, but not pathconnected (e.g., the topologist's sine curve), and D is an open disc containing X, then the boundary of DX is the disjoint union of a circle and X, which is not the continuous image of a circle. On the other hand, if the requirement is only that the boundary is a (nondisjoint) union of paths, then it's trivial.


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Obob
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Re: About the boundary of the bounded region
« Reply #15 on: Jun 21^{st}, 2007, 4:03pm » 
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When I said component in my earlier post, I meant pathcomponent. The topologist's sine curve example does not violate this statement. However, given that the paths would have to cross themselves in order to parameterize a boundary component looking like the letter "X", for instance, I am not sure that this is necessarily interesting. For instance, space filling curves show that the unit square can be parameterized by a closed curve. Is it true that more generally any path connected compact subset of the plane is parameterized by a closed curve?

« Last Edit: Jun 21^{st}, 2007, 4:21pm by Obob » 
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Eigenray
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Re: About the boundary of the bounded region
« Reply #16 on: Jun 21^{st}, 2007, 11:04pm » 
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Path connected is not enough. For example, let C [0,1] be the Cantor set, and let X ^{2} be the union of line segments from the point (0, 1) to (c, 0) for c C. That is, X is the image of C x [0,1] under (x,y) > (x(1y), y). Then X is pathconnected and compact, but not locally connected, as any continuous image of [0,1] is. So this X has only 1 pathcomponent, but requires uncountably many paths to cover it (since each point of C requires an open interval of path to get to). However, the HahnMazurkiewicz theorem states: A Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected secondcountable space. So you need only replace 'pathconnected' with 'connected & locally connected'.

« Last Edit: Jun 22^{nd}, 2007, 2:02pm by Eigenray » 
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Obob
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Re: About the boundary of the bounded region
« Reply #17 on: Jun 22^{nd}, 2007, 11:00am » 
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In your counterexample Eigenray, where do the line segments emanate from?


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Obob
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Re: About the boundary of the bounded region
« Reply #18 on: Jun 22^{nd}, 2007, 12:41pm » 
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Thanks for the clarification, Eigenray. For some reason I was thinking that (0,1) denoted an interval.


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Eigenray
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Re: About the boundary of the bounded region
« Reply #19 on: Jun 22^{nd}, 2007, 2:10pm » 
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It's pretty annoying to use both ordered pairs and intervals at the same time. I've modified the post to be more clear.


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