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.
First, let's understand the notation A\B(A\B). The expression A\B represents the set of elements in A that are not in B. A\B(A\B) represents the set of elements in A that are not in B, but when again removing the elements in B that are not in A. In other words, A\B(A\B) eliminates the elements in B that don't belong to A.
Now, let's proceed with the proof:
1. Let x be an arbitrary element of A\B(A\B).
2. By the definition of A\B(A\B), x belongs to A and is not in B. This implies that x is in A but not in B.
3. Since x is in A but not in B, we know that x is an element of A\B.
4. Since x is an element of A\B, but not in B, we have x ∈ A\B(A\B).
5. Since x belongs to A\B(A\B), we need to show that x is also an element of B.
6. However, since x is not in B, it contradicts the assumption that x is in B.
7. Therefore, we conclude that A\B(A\B) ⊆ B.
In summary, we have shown that every element in A\B(A\B) is also in B.