Let $P$ denote the perimeter function. It's not hard to prove that for any rectangle $R$ in $\mathbb{R}^2$, $R$ can be partitioned into a countable collection of squares $\{Q_k\}_{k=1}^{\infty}$ such that $R = \cup_kQ_k$, $\mathring{Q_k} \cap \mathring{Q_l}=\emptyset$, and $$\Sigma_{k=1}^{\infty}P(Q_k)\le 6P(R)$$

Now instead consider a convex polygon $K$ in $\mathbb{R}^2$, is it possible to find a universal constant $C$ and a partition $\{Q_k\}_{k=1}^{\infty}$ such that $R = \cup_kQ_k$, $\mathring{Q_k} \cap \mathring{Q_l}=\emptyset$, and $$\Sigma_{k=1}^{\infty}P(Q_k)\le CP(K)?$$

Also, what is the optimal value for $C$?

How about cases in $\mathbb{R}^n$? I could not even give a proof for $n$-dimensional rectangles.

Motivation of these questions:

It's a long story. I'm working on a variational problem related to sets of finite perimeter. A set of finite perimeter can be approximated by open sets with polyhedron boundary. A polyhedron can be divided into countable disjoint union of components. Note the perimeter of each component would not increase by considering its convex hull instead, thus it suffices to consider convex polygon. It is nice to use cubes to approximate sets of finite perimeter without increasing the perimeter too much.

I proved some covering lemmas to successfully give some characterizations of some function spaces with respect to sets of finite perimeter, and constructed a counterexample to show some inclusion relationships between those function spaces. I find combinatorics and some basic geometry knowledge is very helpful in my study.

These questions I asked here are not quite related to the problem I'm working on, since the covering lemma instead of "partition lemma" is enough. However, I'm just curious about the sharp case: why not give a partition lemma? And what is the optimal constant?

I'm not sure whether these results are known. Maybe they are very easy questions, since I googled them but didn't find anything.

Can anyone give me some references? Thanks in advance!

  • $\begingroup$ I am afraid that triangle can not be partitioned into squares. Or do you consider open polygon? $\endgroup$ Jun 22, 2015 at 18:57
  • $\begingroup$ @Fedor Petrov, yes, I consider the open polygon. I'm frustrated because I could not either prove for triangles in $\mathbb{R}^2$ or for rectangles in $\mathbb{R}^3$. The thing is, I've no idea how to disprove it. $\endgroup$
    – student
    Jun 22, 2015 at 19:09
  • 1
    $\begingroup$ Can you say some more about where it comes up? It does seem intriguing. $\endgroup$ Jun 22, 2015 at 19:53
  • $\begingroup$ @Fedor Petrov. Now I proved the case for the triangle, by first dividing it into countable union of rectangles, and the convergence of geometric progression. $\endgroup$
    – student
    Jun 22, 2015 at 21:56
  • $\begingroup$ @student how do you do it for, say, right isosceles triangle? I try to remove a square, then I get two similar triangles with the same total perimeter. Such process leads to divergent sum of perimeters. $\endgroup$ Jun 24, 2015 at 4:42

1 Answer 1


Well, let me prove that the answer is negative even for triangle. Rotating coordinate system we may suppose that vertical lines are not parallel to sides of triangle and to sides of all squares (as there are only countably many of them), also, if the triangle has two perpendicular sides, vertical lines can not meet both (because third side is less or more horizontal).

Now project all squares onto OX. I claim that all but countably many many points in the segment $[a,b]$, which is projection of our triangle are covered infinitely many times. This means that total sum of projections is infinite, hence total sum of perimeters of squares is infinite. indeed, assume that point $c\in (a,b)$ is not a projection of any angle of any square. Then the vertical section $\{x=c\}$ of our triangle is covered by infinitely many squares, right?

  • $\begingroup$ I checked my proof,and found a mistake. I used a result here link.springer.com/article/10.1007/BF01263495, but I messed up with the direction of inequalities. Just one thing I'm not quite sure about, is the sentence "This means that total sum of projections is infinite". $\endgroup$
    – student
    Jun 24, 2015 at 21:12
  • $\begingroup$ Could you explain why the fact that the projection of triangle is covered infinitely many times implies the total sum of projections is infinite? $\endgroup$
    – student
    Jun 24, 2015 at 21:13
  • 1
    $\begingroup$ Sum of projections is integral of multiplicity function, it is a case of Fubini theorem: $\sum \int f_i=\int \sum f_i$, where $f_i$'s are characteristic functions of projections. $\endgroup$ Jun 24, 2015 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.