dwsmith said:$(A - B)\cup B = A$ iff $B\subseteq A$.Suppose $B\subseteq A$.
$$
\begin{array}{lcl}
(A - B)\cup B & = & (A\cap B^c)\cup B\\
& = & (A\cup B)\cap (B^c\cup B)\\
& = & A\cap B\\
& = & B
\end{array}
$$
I keep getting = B not A.
The notation $(A - B)\cup B$ represents the set of all elements that are in either set A or set B, but not both. This is also known as the symmetric difference of A and B.
In order to prove this statement, you must show that both sides are subsets of each other. First, assume that $B\subseteq A$. This means that all elements in B are also in A. Therefore, the symmetric difference of A and B will only contain elements that are in A, making $(A - B)\cup B$ equal to A. On the other hand, if $(A - B)\cup B = A$, then all elements in the symmetric difference are also in A. This means that all elements in B are in A, making $B\subseteq A$.
Sure, let's say we have two sets: A = {1, 2, 3, 4, 5} and B = {3, 4, 6, 7}. The symmetric difference of A and B would be (A - B)∪B = {1, 2, 3, 4, 5, 6, 7}. However, if we take B as a subset of A (B⊆A), then the symmetric difference would only contain elements that are in A, making (A - B)∪B = A = {1, 2, 3, 4, 5}.
No, this statement is only true if both sets are finite. If either set is infinite, then the symmetric difference may not equal the original set. For example, if A = {1, 2, 3, ...} and B = {4, 5, 6, ...}, then the symmetric difference would be (A - B)∪B = {1, 2, 3, 4, 5, 6, ...} which is not equal to A.
Yes, this proof has applications in database management and computer science, particularly in operations on sets. It can also be used in situations where we need to compare two sets and determine if one is a subset of the other.