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 <= t <= 1) must lie in A union B.
However this is clearly not the case since A intersect B is the null set. Therefore A union B is not convex in general.
Example:
A = [0,1], B = [2,3]. Choosing any point in A and any point in B, we find that there will be points on the line tx + (1-t)y that are not in A or B.
yes
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 set A union B can be decomposed into three disjoint sub sets A\ (A int B), B\(A int B), and (A int B). So in this case (A union B) and (A int B) are fixed but "moving" elelments from A\ (A int B) into B\(A int B) will not affect (A union B) and (A int B). You should be able to fill in the details now.
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.
yes
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 set A union B can be decomposed into three disjoint sub sets A\ (A int B), B\(A int B), and (A int B). So in this case (A union B) and (A int B) are fixed but "moving" elelments from A\ (A int B) into B\(A int B) will not affect (A union B) and (A int B). You should be able to fill in the details now.
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.
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.
To create a Venn diagram using the union of sets A and B, you would first draw two overlapping circles to represent sets A and B. The union of sets A and B, denoted as A ∪ B, includes all elements that are in either set A, set B, or both. Therefore, in the Venn diagram, you would shade the region where the circles overlap to represent the elements that are in both sets A and B, as well as the individual regions of each circle to represent elements unique to each set.