A is a subset of a set B if every element of A is also an element of B.
wrong!
Let A be the set {1,2,3,4} B is {1,2} and B is a proper subset of A C is {1} and C is also a proper subset of A. B and C are proper subsets of the set A because they are strictly contained in A. necessarily excludes at least one member of A. The set A is NOT a proper subset of itself.
The definition of subset is ; Set A is a subset of set B if every member of A is a member of B. The null set is a subset of every set because every member of the null set is a member of every set. This is true because there are no members of the null set, so anything you say about them is vacuously true.
Perpendicular lines are a subset of intersecting lines. The only difference is that perpendicular lines intersect at an angle of 90o, whereas intersecting lines can converge at any angle.
A - B is null.=> there are no elements in A - B.=> there are no elements such that they are in A but not in B.=> any element in A is in B.=> A is a subset of B.
Not in general , This would be true if B intersect A is empty, in which case A set difference B is A itself.
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
A is a subset of a set B if every element of A is also an element of B.
first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers) P(A' intersect B') = P(B')P(A'|B') by definition = P(B')[1-P(A|B')] since 1 = P(A) + P(A') = P(B')[1 - P(A)] from the first proof * = P(B')P(A') since 1 = P(A) + P(A') conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition. ***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".
Since B is a subset of A, all elements of B are in A.If the elements of B are deleted, then B is an empty set, and also it is a subset of A, moreover B is a proper subset of A.
a subset is a group that is contained within the other group. So the set of letters {a, b} is a subset of {a, b, c} It is also worth noting that {a, b} is also a subset of itself {a, b}. In set arithmatic a subset I believe is defined like this: set1 is a subset of set2 if set1 + set2 = set2. {a, b} + {a, b, c} = {a, b, c}
Assume that set A is a subset of set B. If sets A and B are equal (they contain the same elements), then A is NOT a proper subset of B, otherwise, it is.
a is intersection b and b is a subset
Set "A" is said to be a subset of set "B" if it fulfills the following two conditions:A is a subset of B, andA is not equal to B
wrong!
A set "A" is said to be a subset of of set "B", if every element in set "A" is also an element of set "B". If "A" is a subset of "B" and the sets are not equal, "A" is said to be a proper subset of "B". For example: the set of natural numbers is a subset of itself. The set of square numbers is a subset (and also a proper subset) of the set of natural numbers.