22 I. Convex Sets at Large
is non-empty. Since the sets Ar and Ab are open, their intersection is an open set;
therefore an intersection of sets Ar and Ab is non-empty if and only if it contains
a point (c, α) with c = 0. Hence for any subset S R B, the intersection
r∈S∩R
Ar
b∈S∩B
Ab
is not empty if and only if there is a point (c, α) Rd+1 such that the hyperplane
H = x
Rd
: c, x = α
strictly separates the sets B S and R S. This completes the proof.
For example, if two sets of points in the plain cannot be separated by a straight
line, one of the three configurations of Figure 6 must occur.
1 ) 2 ) 3 )
Figure 6. The three reasons points cannot be separated in the plane
PROBLEMS.
1. Prove that if a convex set is contained in the union of a finite family of
halfspaces in Rd (sometimes we say covered by a finite family of halfspaces; see
Section 5.2), then it is contained in the union of some d + 1 (or fewer) halfspaces
from the family (covered by some d + 1 subspaces).
2. Let I1,... , Im be parallel line segments in R2, such that for every three
Ii1 , Ii2 , Ii3 there is a straight line that intersects all three. Prove that there is a
straight line that intersects all the segments I1,... , Im.
3. Let Ai : i = 1, . . . , m be convex sets in R2 such that for every two sets Ai
and Aj there is a line parallel to the x-axis which intersects them both. Prove that
there is a line parallel to the x-axis which intersects all the sets Ai.
(5.2) The center point. Let us fix a Borel probability measure μ on
Rd.
This
means, roughly speaking, that for any “reasonable” subset A
Rd
a non-negative
number μ(A) is assigned which satisfies some additivity and continuity properties
and such that
μ(Rd)
= 1. We are not interested in rigorous definitions here; the
following two examples are already of interest:
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