Re: math -- finite union of rectangular regions
- From: quasi <quasi@xxxxxxxx>
- Date: Tue, 01 Apr 2008 02:48:44 -0500
On Sun, 30 Mar 2008 23:24:36 -0700 (PDT), Mariano Suárez-Alvarez
<mariano.suarezalvarez@xxxxxxxxx> wrote:
Let A be a finite set of closed rectangles in the plane
whose sides are parallel to the coordinate axes whose
union is simply connected. Let us assume that
(*) for all R in A, the union of the rectangles
in A - { R } is not simply connected.
For each rectangle R in A, let
C(R) = R - union { R' in A : R' != R }
Let us call each of the connected components of C(R) such
that its closure is contained in the interior of R a *hole*
of A, and let us say that it belongs to R.
Yes, I follow.
A hole of A belongs to exactly one rectangle in R, which is
the only one which contains it. The holes are moreover disjoint.
Yes.
Conversely, our hypothesis (*) is equivalent to the
statement that to each rectangle in R belongs at least one
hole.
For each bounded subset X of the plane, let us call the sup
of the y coordinates of the points in X the *height* of X.
Fine.
Let as pick a hole H of minimal height among the
holes of A. The boundary B of H is a finite polygonal
closed arc, whose segments are parallel to the coordinate
axes. Pick one horizontal segment in B which has minimal
height among the horizontal segments of B. This segment
shares a subsegment of positive length with the boundary
of some rectangle S in A. Let H' be one of the holes in S.
It is clear that the height of H' is strictly less that the
weight of H. This is absurd.
Yes!
(although I assume you meant "height of H").
Kind of a "descent argument". If there's a hole, there has to be a
lower one.
Very nice.
quasi
.
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