Heine Borel

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-hannover.de/~fugru/?topologie_teil1

This film is #7 of our geometric animations calendar
http://www.calendar.algebraicsurface.net

Category:

Howto & Style

Tags:

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

[Einstein0GETBaC]

@Audioslave6556 Obayashi short elements as talent did not leave packages without package Imaichi Imaichi as Olympic.

[morningiggle]

HI :) VERY INTERESTING Video. What other shapes or methods are used to do the same job? I imagine their to be drawbacks in using a square. The number of times a line was added may be considerably higher than that of another shape/method. Thanks ..Thumbs up. '.)

[loottimo]

Thank you very much, now I've understood :)

[mikxer]

why there has to be a video error just when i need this video so much arg!

[FlickerEdge]

Thanks very much for the explanation!

[mediteight]

..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]

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]

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]

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]

Just curious (to check my understanding) about this discussion - would you agree that:

1) U must be open and non-empty

2) U contains infinitely many points

3) U does not have to be bounded.