Understanding the Set Operation A - (A - B)

The question asks us to simplify the expression \(A - (A - B)\) for any two sets A and B. This involves understanding basic set operations, specifically set difference and set intersection. The expression \(A - (A - B)\) represents taking set A and removing all elements that are present in the set \((A - B)\).

What is Set Difference?

The set difference \(A - B\) is the set of elements that are in set A but are not in set B. Mathematically, it is defined as:

\(A - B = \{x \mid x \in A \text{ and } x \notin B\}\)

Think of it as removing the part of A that overlaps with B.

Breaking Down A - (A - B)

Now let's consider the expression \(A - (A - B)\). This means we are taking set A and subtracting the set \((A - B)\). According to the definition of set difference, \(A - (A - B)\) is the set of elements that are in A but are not in \((A - B)\).

\(A - (A - B) = \{x \mid x \in A \text{ and } x \notin (A - B)\}\)

Let's analyze the condition \(x \notin (A - B)\). An element \(x\) is NOT in the set \((A - B)\) if it is not the case that \(x \in A\) and \(x \notin B\). This means either \(x \notin A\) or \(x \in B\) (or both).

So, for an element \(x\) to be in \(A - (A - B)\), it must satisfy two conditions:

1. \(x \in A\)
2. \(x \notin (A - B)\), which means (\(x \notin A\) or \(x \in B\))

Combining these two conditions:

\(x \in A\) AND (\(x \notin A\) or \(x \in B\))

Since \(x\) must be in A (condition 1), the possibility \(x \notin A\) in the second part of the condition cannot be true. Therefore, the second part simplifies to just \(x \in B\).

So, for an element \(x\) to be in \(A - (A - B)\), it must satisfy:

1. \(x \in A\)
2. \(x \in B\)

This is exactly the definition of set intersection.

Relating to Set Intersection

The set intersection \(A \cap B\) is the set of all elements that are in both set A and set B. Mathematically, it is defined as:

\(A \cap B = \{x \mid x \in A \text{ and } x \in B\}\)

Comparing our derivation for \(A - (A - B)\) with the definition of \(A \cap B\), we see that they are the same.

Example to Illustrate A - (A - B)

Let's use a simple example to understand the operation \(A - (A - B)\).

Suppose Set A = \(\{1, 2, 3, 4\}\) and Set B = \(\{3, 4, 5, 6\}\).

First, find \(A - B\):

\(A - B\) contains elements in A but not in B. So, \(A - B = \{1, 2\}\).

Now, find \(A - (A - B)\):

\(A - (A - B)\) contains elements in A but not in \((A - B)\). The set \((A - B)\) is \(\{1, 2\}\).

So, we look at A (\(\{1, 2, 3, 4\}\)) and remove elements that are in \(\{1, 2\}\). The elements to remove are 1 and 2.

\(A - (A - B) = \{1, 2, 3, 4\} - \{1, 2\} = \{3, 4\}\)

Now, let's find \(A \cap B\) for our example sets:

\(A \cap B\) contains elements that are in both A and B. The elements common to \(\{1, 2, 3, 4\}\) and \(\{3, 4, 5, 6\}\) are 3 and 4.

\(A \cap B = \{3, 4\}\)

As you can see from this example, \(A - (A - B) = \{3, 4\}\) and \(A \cap B = \{3, 4\}\). This example supports our derivation that \(A - (A - B)\) equals \(A \cap B\).

Summary of A - (A - B)

In summary, the set difference \(A - B\) removes the elements of B from A. When we then take \(A - (A - B)\), we are taking the elements of A and removing those elements that were unique to A (the ones not in B). The elements of A that were NOT unique to A are precisely the elements that A has in common with B. These common elements form the set intersection, \(A \cap B\).

Therefore, for any two sets A and B, \(A - (A - B)\) equals \(A \cap B\).