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(0)
Let P and Q be points in A + B. Then P = x1 + y1 and Q= x2 + y2 where x1 and x2 are in A and y1 and y2 are in B. Thus, because A and B are convex sets, then the line segments joining their respective pairs of points are contained in A and B, respectively. Thus, these line segments can be parametrized by the following equations:
.
where for every value of t, the values of the above equations represent points in A and B, respectively. Thus, for any value of t, the sum of these equations yields a point in the set A+B. Thus, one can add these two equations together and rearrange as follows:
and so for every value of t, the value of the above equation represents points in A+B; therefore, A+B is convex.
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(2)
Suppose that 
are convex sets in the xy-plane, and suppose 
is the intersection of these convex sets. Let 
and let 
. Thus, 
. Thus, since 
then AB is contained in 
, because is convex. Therefore, since is arbitrary, is convex since AB is contained in S. This same argument works for the intersection of any indexed set of convex planar sets.
(3)
Suppose that are convex sets in the xy-plane, that 
(that is,are nested), and that 
. Observe that 
as a result of all the 
being nested inside 
. Since is a convex set, then H is a convex set as well. This same argument works for a countably infinite collection of nested convex sets.
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Let p be the support function of K+L. By definition,
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Definition: The unit sphere, denoted Sn, is the set 
. The unit ball, denoted Bn, is the set 
.
Let On=surface area of Sn and let Kn= volume of Bn.
