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Heine Borel

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This video illustrates a proof of the Heine Borel Theorem: "Every closed bounded set in R^n is compact." The proof goes like this:Assume the contary. Let X be a bounded closed set (the disk in the video) that has an infinite open covering (symbolized by the colored small disks in the video) with no finite subcovering. Now devide X and the covering in two halfs. Now at least one sides does not admit a finite subcovering (otherwise we could construct a finite subcovering of X). Now subdivide the side without finite subcovering, and repeat the construction. We obain a sequence of subdivisions that do not have a finite subcovering of the given infinite covering.If we now choose a point inside each of the subdivisions we obtain a cauchy sequence, which must have a limit point P since R^n is complete. P lies in every subdivision. Now choose an open set U (yellow in the movie) of the covering which contains P. Since U has a finite size it contains all subdivisions which are small enough. But then U is a finite subcovering for these subdivisions which contradicts our construction. Therefore our assumption that allowed the construction must be wrong and the Heine-Borel-Theorem correct. q.e.d This Video was produces for a topology seminar at the Leibniz Universitaet Hannover. http://www-ifm.math.uni-hannov... film is #7 of our geometric animations calendar http://www.calendar.algebraics...

Channel: Howto & Style
Uploaded: November 30, 1999 at 12:00 am
Author: bothmer

Length: 01:33
Rating: 4.46
Views: 9239

Tags: advent  Borel  bothmer  calendar  cg  compactness  covering  fugru  geometry  Heine  limit  mathematics  open  topology  

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Video Comments

mediteight (November 30, 1999 at 12:00 am)
..and this is the contradiction the video illustrates. Hope this explanation is useful - very interested to hear if any of it isn't clear, or is wrong in some way.
mediteight (November 30, 1999 at 12:00 am)
This results in an infinite sequence of subsets of X (little squares), each of which cannot be covered by a finite number of discs. Choosing a point from each of these little squares, gives a Cauchy sequence which converges to a point inside all of them. The video then shows that any of the coloured discs that contain this point, will cover at least one (in fact infintely many) of the little squares - all of which should not be coverable with an infinite number of discs (yet alone one!).
mediteight (November 30, 1999 at 12:00 am)
doh! - sincere apologies, please ignore my above ramblings. Consider the covering formed by the different coloured open discs. X is compact if and only if there exists a finite subset of these discs that also covers X. The video supposes that no such finite subset of discs exists, and proceeds to demonstrate a contradiction. Under the assumption, each time a subset of X is split in two, we can guarantee at least one of the resulting subsets cannot be covered by a finite number of the discs.
TQCKyle (November 30, 1999 at 12:00 am)
I think you're under the impression that I'm not familiar with topology or, in this case, analysis. Quite the contrary. I'm only curious as to how compactness is being shown here what with the infinite division of this set with squares. I see that the set of (different-colored) open discs is forming a covering of the larger set.
mediteight (November 30, 1999 at 12:00 am)
Just curious (to check my understanding) about this discussion - would you agree that:1) U must be open and non-empty2) U contains infinitely many points3) U does not have to be bounded.
mediteight (November 30, 1999 at 12:00 am)
Most mathematical proofs involve certain properties of sets (that mathematicians, for some reason or other, have decided are useful or interesting in some way) - and one of these important properties is "compactness".The definition of compactness (from topology) is slightly abstract and fiddly - and the theorem shown in this video gives mathematicians an alternative (but equivalent) way of handling compactness (in certain cases) that might help them reason about it in their research.
mediteight (November 30, 1999 at 12:00 am)
Thanks for the wonderful animation and description of the proof - very beautiful.It would be really neat to have a collection of such animations to illustrate each and every proof in analysis, geometry and topology... when you find the time :)
black0jackass (November 30, 1999 at 12:00 am)
i don't think it shows anything at all just some punk being padentic and misleading.
black0jackass (November 30, 1999 at 12:00 am)
yes, 3 equivalent statements
nxn8690 (November 30, 1999 at 12:00 am)
I'm sorry. It's my mistake. I thought that looking at Heine-Borel is the best way to see Heine-Borel. But your comment seems to point out that instead helping, I merely insisted my own way. I apologize again. Best Regard.

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