In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0
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.
If B = {10111213} and C = {1213} then their intersection is the empty set, {}.The union of A with an empty set is set A.
The union of A and B.
if we have set A and B consider A={1,2,3,4}and B={3,4,5,6} the union of these sets is A and B={1,2,3,4,5,6}and the intersection is{3,4} the union and the intersection are same only if A=B
the union of two convex sets need not be a convex set.
In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0
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.
the union of two sets A and b is the set of elements which are in s in B,or in both A and B
If B = {10111213} and C = {1213} then their intersection is the empty set, {}.The union of A with an empty set is set A.
The union of A and B.
NO. The set of numbers in Set B and the set of numbers in Set C CAN be the same, but are not necessarily so.For example if Set A were "All Prime Numbers", Set B were "All Even Numbers", and Set C were "All numbers that end in '2'", A union B would equal A union C since 2 is the only even prime number and 2 is the only prime number that ends in 2. However, Sets B and C are not the same set since 4 exists in Set B but not Set C, for example.However, we note in this example and in any other possible example that where Set B and Set C are different, one will be a subset of the other. In the example, Set C is a subset of Set B since all numbers that end in 2 are even numbers.
if we have set A and B consider A={1,2,3,4}and B={3,4,5,6} the union of these sets is A and B={1,2,3,4,5,6}and the intersection is{3,4} the union and the intersection are same only if A=B
A set is a collection of well defined objects known as elements Opperatons of sets are 1)union - the union of sets A and B is the set that contains all elements in A and all elements in B. intersection - given two sets A and B, the intersection of A and B is a set that contains all elements in common between A and B. compliments - given set A, A compliment is the set of all elements in the universal set but not in A difference - A-B is a set containing all elements in A that are not in B. symmetric difference - it is the sum of A and B minus A intersection B.
The union of two sets A and B is a set that consists of all elements which are either in A, or in B or in both.
no
The concept of closure: If A and B are sets the intersection of sets is a set. Then if the intersection of two sets is a set and that set could be empty but still a set. The same for union, a set A union set Null is a set by closure,and is the set A.