Re: Another Difficult Combinatorial Geometry Problem
- From: "David C. Ullrich" <dullrich@xxxxxxxxxxx>
- Date: Fri, 29 Aug 2008 14:30:01 +0000 (UTC)
On Thu, 28 Aug 2008 20:00:02 +0000 (UTC), ical12345@xxxxxxxxxxxxxx
wrote:
A set X in Rd is said to be m-convex , m.$(A!](B 2 , iff for every m
distinct points in X at least one of the line segments determined by
those points lies in X. Ramsey$(A!/(Bs theorem ,in graph theory, implies
[...]
THEOREM Let X1 and X2 both be simply connected planar, 3-convex sets
then X1 $(A!I(B X2 is 5-convex.
Some of your post is unintelligible to many readers - you might want
to repost using nothing but plain ASCII.
[Note from the moderator:
All posters, please make sure that your
postings are in ASCII. I am usually able to catch strange encodings,
but missed this one. Sorry.
Maarten Bergvelt]
[...]
points of local non-convexity with X1 $(A!I(B X2 not 4-convex see for
example [2] page 48.
LEMMA 2 is also an immediate consequence of LEMMA 1.
LEMMA 2. If x1$(A!-(Bx5 are five points of X1$(A!I(BX2 with none of the
associated line segments lying in X1$(A!I(BX2 , in other words the x i are
5 visually independent points of the intersection, then no three of
the xi are collinear.
PROOF OF THEOREM. The proof now follows easily but tediously from
considering the convex hull of 5 supposedly visually independent
points of the simply connected X1 and X2 . The cases where the
convex hull is a triangle or pentagon are relatively straightforward
whereas the case of a quadrilateral convex hull requires the
consideration of the two cases : where two non-adjacent sides belong
to each Xi or three belong to one Xi .
REMARK 2. It should be noted that my original detailed proof in [2]
used a result of Erdos and Szekeres [5] that $(A!0(Bfrom five points in the
plane of which no three lie on the same straight line it is always
possible to select four points determining a convex quadrilateral$(A!1(B.
However that proof required consideration of the location of the fifth
point outside the convex hull of the other four.
REMARK 3. It is also noteworthy that X1 $(A!I(B X2 may fail to be 5-convex
if just one of the Xi fails to be simply connected. One somewhat
inelegant example of this is included in [2] pages 41and 42, where
both Xi are also unions of three convex sets.
REMARK 4. Let xi 1$(A!\(B i $(A!\(B 5 be five points on the moment curve with
xi = p($(A&A(Bi) where 0 < $(A&A(Bi $(A!\(B 1 and p($(A&A(B) = ($(A&A(B,$(A&A(B2,$(A&A(B3,$(A&A(B4) . As in
[3] it is possible to construct compact X1 and X2 ,3-convex sets
containing the xi with X1$(A!I(BX2 not 5-convex.
REFERENCES
1. M.Breen, Decomposition theorems for 3-convex subsets of the plane,
Pacific. J. Math. 53, (1974), 43-57.
2. D.I.Calvert, Generalisations of convexity, Doctoral Dissertation,
University of London, (1978).
3. H.G.Eggleston, Valentine convexity in n dimensions, Math Proc.
Camb. Phil. Soc., 80 (1976), 223-228.
4. H.G.Eggleston, Vector sums of Valentine convex sets, Math. Proc.
Camb. Phil. Soc., 92 (1982), 17-19.
5. P.Erdos and G.Szekeres, A combinatorial problem in geometry,
Compos. Math. 2 (1935), 463-470.
6. F.A.Valentine, A three point convexity property, Pacific J. Math.
7 (1957), 1227-1235.
7. F.A.Valentine, The intersection of two convex surfaces and
property P3, Proc. Amer. Math. Soc. 9 (1958), 47-54.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
.
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