Heine Borel

[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!

[mediteight]

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]

yes, 3 equivalent statements

[phil10300]

What you see here is a closed and bounded subset of R^2.

The spots are open sets covering the disc (they have to stick out a bit to contain the boundary) and some of them are very small and there might be an infinity of them.

The goal here is to show that we can select a finite number of these spots that will still cover our disc.

The video says that if we could not do this, ther would be a contradiction.

That is what the shiny light towards the end is.

[Audioslave6556]

could u explain this in english please

[nxn8690]

Let S is a subset of R^n. Then following propositions are equivalent. (i) S is closed & bounded. (ii) S is compact.

[TQCKyle]

The Heine-Borel Theorem...yeah. So what is this vid supposed to be showing?

[nxn8690]

W.Rudin, Principles of Mathematical Analysis, McGraw-Hill, p39,40 ISBN 0-07-054235-x.

[kaitenuser]

Sending a person to a formal analysis book is obviously a waste of time. They are asking for an intuitive understanding, not some "for a given e..."

[nxn8690]

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.

[black0jackass]

i don't think it shows anything at all just some punk being padentic and misleading.

[mediteight]

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.

[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]

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.

[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]

..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.

[FlickerEdge]

Thanks very much for the explanation!

[Einstein0GETBaC]

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

[systi1912]

seids ZUA

[MrMoto]

Why must U have a finite size?

[bothmer]

An open covering consists of, possibly infinitely many, open sets of finite size. U is one of these open sets and therefore also of finite size.

[MrMoto]

In the definition I am familiar with (e.g. on Mathworld or Wikipedia), the sets aren't required to be of finite size.

[bothmer]

The sets are required to be non empty and open. Every non empty _open_ set is of finite size. The only non empty sets of diameter 0 are _closed_ (i.e. points).

[fluffythekitten]

no, it isn't... (x,y):y real, 1<x<2

is open in both directions, but infinite

[bothmer]

I agree that "finite size" is confusing. What I mean is that the open set U is bigger that just the point P. More precisely: since U is open, there exist an e>0 such that U contains an open circle/ball of radius e around P. So as soon as the subdivisions have a diameter smaller than e they are completely contained in U.

[AnExtropian]

I believe the correct word is "bounded."

[bothmer]

Yes - Thanks!

[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.

[DjunioGA]

boa animação!!!

[gomubob]

Great animation. It would be better if there was a commentator guiding the "proof".