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The proof of this theorem is by contradiction. Suppose for convex sets S and T there are elements a and b such that a and b both belong to S∩T, i.e., a belongs to S and T and b belongs to S and T and there is a point c on the straight line between a and b that does not belong to S∩T. This would mean that c does not belong to one of the sets S or T or both. For whichever set c does not belong to this is a contradiction of that set's convexity, contrary to assumption. Thus no such c and a and b can exist and hence S∩T is convex.
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Prove that intersection of two convex polgons is a convex as well?
I will prove a more general theorem from which your answer follows immediately. Theorem: The intersection of any number (including 2) of convex polygons is convex.ProofLet C be the intersection of Ci which is a set of iconvex polygons. By definition of intersection, if two points A and B belong to C then they belong to every one of the Ci . But the convexity of each of the Ci tells us that line segment AB is contained in Ci . Therefore, the line segment AB is in C and because ABwas arbitrary we conclude that C is convex
What is the union intersection set?
Given two or more sets there is a set which is their union and a set which is there intersection. But, there is no such thing as a "union intersection set", as required for the answer to the question.
What is the maximum possible finte number of intersection points of two non-convex quadrilaterals?
12
What are the two ways in naming a SET?
the other one is intersection
What is the meaning of intersection of two sets?
The intersection of two sets S and T is the set of all elements that belong to both S and T.
