Re: math -- finite union of rectangular regions
- From: Chip Eastham <hardmath@xxxxxxxxx>
- Date: Mon, 31 Mar 2008 14:12:26 -0700 (PDT)
On Mar 31, 6:52 am, Chip Eastham <hardm...@xxxxxxxxx> wrote:
On Mar 31, 6:28 am, Chip Eastham <hardm...@xxxxxxxxx> wrote:
On Mar 31, 3:52 am, Robert Sheskey <rshes...@xxxxxxxxxx> wrote:
In article <b7a64593-403c-44e4-9cc9-da8fc669c...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, mariano.suarezalva...@xxxxxxxxx says...
On 30 mar, 00:31, quasi <qu...@xxxxxxxx> wrote:
Trying to avoid counterexamples, while still saying something
nontrivial ...
Conjecture:
If n closed, nondegenerate, rectangular regions in R^2, where n>1, and
where all rectangle sides are either horizontal or vertical, is such
that the union is contractible, then it is possible to omit one of the
regions from the union, so that the union of the remaining regions is
still contractible.
Remark:
If this conjecture manages to hold up, consider the generalization to
m-dimensional rectangular regions in R^m.
quasi
I was wondering, instead of rectangles, what happens with discs ;-)
In particular, I'd say a finite set of discs does have that property.
And I'd say, too, that a finite set of arbitrary ellipses does not,
in general.
Take 3 identical disks, arranged so that each is tangent to the
other 2. Add a copy of this 3-disk configuration, rotated 60 degrees.
Finally, add a small disk in the middle just large enough to cover
the hole. (Obviously you can't remove the middle disk. And if you
remove one of the 6 large disks you uncover a little pointy region.)
Now a finite set of disks all the same size would have the property
in question, I think.
RS
Even if we require the disks to be equal size,
we could have the usual six disks surrounding
one central disk, together with a rotated copy
of those six disks which cover the respective
"little pointy region[s]", and correspondingly
the first set of six surrounding disks cover
holes for the second set. The central disk
also covers a hole...
regards, chip
I realized my example requires elaboration, as
it will not suffice to have disks tangent to
one another, and hence something must be said
about how much overlap between disks suffices
so that one set of "six disks" covers the
holes left by the other set (and conversely).
Sorry for the confusion caused by my incomplete
description.
--c
Okay, if we start with a set of thirteen unit
disks, one surrounded by twelve equally spaced
tangentially placed disks, then expand all
radii to sec(pi/12), then the union of these
is contractible but removing any one of them
leaves a hole.
--c
.
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