# Minkowski sum of small connected sets

Suppose that the convex hull of the Minkowski sum of several compact connected sets in $\mathbb R^d$ contains the unit ball centered at the origin and the diameter of each set is less than $\delta$. If $\delta$ is very small (this smallness may depend on $d$ but on nothing else), does it follow that the sum itself contains the origin?

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WRONG TAGS --- it should be geometry or/and convex-geometry... –  Anton Petrunin Nov 23 '09 at 1:40
This was asked long before such tags existed and before I would be able to create them. I'm not really sure what are the best tags for this now (I chose these two because the question arose in a purely analytic setting and it seemed topological in nature) but I do not mind in the slightest if somebody more comfortable with the vast forest of the current tags will retag it. I am currently completely lost in the aforementioned forest, so I'll take no action myself. –  fedja Nov 23 '09 at 2:59

I finally figured it out. My solution is here. I would repost it on mathoverflow but until LaTeX is enabled, it is quite hard for me to communicate such things here...

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In dimension 1, why are the sets {-100}, {80},{120} not a counterexample?

(In particular, they are compact connected sets with 'diameter' 0; their Minkowski sum, going by the wikipedia definition, is {-20,20,200}, which does not contain 0, but whose convex hull [-20,200] contains a large ball around 0)

This example of course has nothing to do with dimension, and you can easily flush out the points into tiny balls, if you like. I suspect that this is based on a misreading of the question.

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The Minkowski sum of those sets is the set {100}. The Wikipedia article discusses the Minkowski sum of two sets. But this is a commutative and associative operation, so we can discuss the Minkowski sum of any finite number of sets. –  David Speyer Oct 27 '09 at 16:12
As David Speyer points out, the Minkowski sum is a singleton in this case. My only reason for making this comment is to say that I came up with exactly the same "counterexample" myself at one point, and even started writing an answer based on it. But then I realized my mistake. –  gowers Oct 29 '09 at 11:53

Argh! I tried to add this as a comment but when submitting it, I was told that there was a 600 character limit and all my typing just disappeared without trace.

Anyway, my point was that you can get "straight paths". The reason is that if $a=\sum_ i x_ i$ and $b=\sum_ i y_ i$ are in the sum, then the vectors $v_ i=y_ i-x_ i-(b-a)/n$ are small and add up to $0$. There is a cute result that then you can rearrange them in such order that all partial sums are small (just constant times larger than the vectors themselves). If you switch from $x_ i$ to $y_ i$ along the corresponding compact in this order, you get a curve that travels from $a$ to $b$ within a small neighborhood of the segment $[a,b]$.

Another observation that may be useful is that the sum is $d\delta$ dense in its convex hull. Indeed, if $a=\sum_ i v_ i$ and $v_ i$ are in the convex hull of $K_ i$, then we can start moving $v_ i$ until their representations as convex combinations of points in $K_ i$ get shorter and we can do it as long as there are at least $d+1$ vectors $v_ i$ that do not belong to $K_ i$ themselves (any $d+1$ vectors in $\mathbb R^d$ are linearly dependent). Thus, in the representation of every point in the convex hull as a sum, we need only $d$ vectors from convex hulls and the rest may be taken in $K_ i$.

I feel that these two observations put together should be enough and I just do not see how to add 2 and 2 here.

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