Let A and B be sets. Show that A\B(A\B) ⊆ B.
To show that A\B(A\B) ⊆ B, we need to prove that every element in the set A\B(A\B) is also an element of B.
To begin, let's clarify what A\B(A\B) means. The expression A\B represents the set of elements that are in A but not in B. So, A\B(A\B) means taking the set A\B and then subtracting from it the elements that are in both A and B.
Now, to prove that A\B(A\B) ⊆ B, we need to show that if x is an element of A\B(A\B), then x is also an element of B.
Let's assume that x is an element of A\B(A\B). This means that x is in A but not in B. To prove x is also in B, we need to demonstrate a contradiction.
However, since x is not in B, it logically follows that x cannot be in both A and B. Therefore, the intersection of the sets A\B and (A\B) would be an empty set. Thus, A\B(A\B) would also be an empty set.
Given that the empty set is a subset of any set, we can conclude that A\B(A\B) ⊆ B.
In summary, we have shown that A\B(A\B) is an empty set, and the empty set is a subset of B. Therefore, A\B(A\B) ⊆ B.