Talk:Hadwiger's theorem

Can someone please clarify the point concerning the Euler characteristic? By self-similarity, the measure of a cube in R³ must be zero, but one can easily construct a finite union of cubes which is homotopy equivalent to the 2-sphere, and the latter has non-zero Euler characteristic by all definitions I know.--80.133.118.97 12:30, 22 Mar 2005 (UTC)

I think what you're missing is that the condition that the measure is additive does not mean that it is additive over union, but rather over disjoint union (think about it:

S\cup S=S

, but

m(S)+m(S)=2m(S)\,

). So the idea of building objects in 3-space using cubes as bricks simply does not work, as the disjoint union of two cubes is disconnected (conversely, since Rⁿ is Hausdorff, any nontrivial finite decomposition of a compact connected set in Rⁿ yields at least one non-compact subset; a cube minus one of its faces, if you like). --86.130.247.29 22:59, 25 July 2005 (UTC)[reply]

I see, thank you.--80.136.146.247 18:12, 10 August 2005 (UTC)[reply]

Is it correct to replace "not-necessarily-nonnegative set functions" by the more straightforward phrasing "real-valued functions" or am I missing something (which is quite likely as I know nothing about this subject)?

Crust 16:32, 30 August 2005 (UTC)[reply]

Why isn't the number of connected components a second example (after Euler characteristic) of a homogenous measure of degree 0 (contrary to the asserted uniqueness)? (Maybe the theorem is only supposed to refer to disjoint unions not general unions of finitely many convex sets. If so, the Euler characteristic is just the number of connected components in this case.)

Crust 16:50, 31 August 2005 (UTC)[reply]

You'll notice the cited book by Klain and Rota. The latter author died in 1999, but perhaps the former could answer your question. Try googling his name and see if you find his email address. Michael Hardy 02:36, 31 August 2007 (UTC)[reply]

Because the number of connected-components doesn't satisfy the additivity condition. For example, let C be a square (not filled in). Let A be the union of two adjacent edges, and B the union of the other two adjacent edges. Then A and B each have 1 connected-component. Also, A ∩ B is a pair of points, and so has 2 connected-components. So, writing n(X) for the number of connected-components of a set X, we have n(A) + n(B) - n(A ∩ B) = 1 + 1 - 2 = 0. On the other hand, A ∪ B = C, and C has 1 connected-component, so n(A ∪ B) = 1. --Tom Leinster 21:13, 6 October 2007 (UTC)[reply]

I've split the first paragraph into two, so that the new first para contains a cleaner statement of the theorem and the new second para explains what "measure" means in this context. I added to the definition of "measure" the requirements that it be rotation-invariant and continuous, since these are both hypotheses in Hadwiger's Theorem. I removed the reference to scalar multiplication, since this is already included within the term "linear combination".

130.209.6.40 16:36, 30 January 2007 (UTC) (Tom Leinster)[reply]

Is there an error in the definition?

I don't know anything about this subject, but it seems to me like the homogeneous measure of degree n-1 would have to be able to take infinite values (any such measure would take the value +infinity or -infinity, not both.) This is because if you take a compact set K whose boundary is a fractal with infinite surface, then m(K) should be infinite as well by the continuity hypothesis.

So my comment is, should these measures not take value in the extended real line? (I.e., the reals union {-infinity,+infinity}?) If so, you have to take care when you define the additivity, so you don't subtract infinity from infinity.

Loisel 06:22, 24 August 2007 (UTC)[reply]

I'm not sure of the answer but I wonder if your proposed example would be "measurable" in the relevant sense. Michael Hardy 02:38, 31 August 2007 (UTC)[reply]

My mind must have skipped a beat. I totally missed the "convex" bit. Loisel 21:10, 31 August 2007 (UTC)[reply]

Removed my expert-verify tag. With convex, I can believe that what it says is true. Loisel 04:15, 1 September 2007 (UTC)[reply]

Mean width

The mean width is described as "mysterious" and a "misnomer". Could someone with a clue expand on what it means/what properties it has, what makes it mysterious, and why it's a misnomer? Dfeuer 02:07, 29 September 2007 (UTC)[reply]

I have exactly the same questions as Dfeuer, and I couldn't find any other sources for this "mean width". Can anyone elaborate, please? Jose Brox —Preceding unsigned comment added by 217.127.9.138 (talk) 17:52, 30 January 2009 (UTC)[reply]

Revision

Hello fellow editors,

I have revised the article, with the hope that now it is self-contained and more intelligible. Please comment (esp. if you find the previous epithets are not yet appropriate).

Sasha (talk) 21:21, 2 September 2011 (UTC)[reply]