Understanding Set Theory Statements

This question asks us to evaluate the correctness of three different statements involving basic set operations: union (\(\cup\)) and difference (\(-\)). We need to determine which of these statements are universally true for any sets A and B.

Analyzing Statement 1: \(A = (A \cup B) \cup (A - B)\)

Let's break down this statement. The right side of the equation is the union of two sets:

• \(A \cup B\): This set contains all elements that are in A, or in B, or in both.
• \(A - B\): This set contains all elements that are in A but not in B.

Now, consider the union of these two sets: \((A \cup B) \cup (A - B)\). If an element is in \(A - B\), it is by definition in A. If an element is in A, it is also in \(A \cup B\). Therefore, the set \(A - B\) is always a subset of \(A \cup B\). The union of a set with its subset is simply the larger set.

So, \((A \cup B) \cup (A - B) = A \cup B\).

The statement becomes \(A = A \cup B\). This is only true if set B is a subset of set A (\(B \subseteq A\)). However, the statement claims it is true for any sets A and B. Consider an example: Let A = {1, 2} and B = {2, 3}. Then \(A \cup B = \{1, 2, 3\}\). Here, \(A \neq A \cup B\).

Thus, Statement 1 is incorrect.

Analyzing Statement 2: \(A \cup (B - A) = (A \cup B)\)

Let's analyze the left side: \(A \cup (B - A)\).

• \(A\): The set A.
• \(B - A\): This set contains all elements that are in B but not in A.

The union \(A \cup (B - A)\) contains all elements that are either in A, or in B but not in A. This covers all elements that are in A, and all elements that are in B that are outside of A. Together, these comprise all elements that are in A or in B (or both, as the part of B that is in A is already covered by A). This is precisely the definition of \(A \cup B\).

So, \(A \cup (B - A) = A \cup B\). This identity is always true for any sets A and B.

Let's use an example: Let A = {1, 2} and B = {2, 3}. Then \(B - A = \{3\}\). \(A \cup (B - A) = \{1, 2\} \cup \{3\} = \{1, 2, 3\}\). Also, \(A \cup B = \{1, 2\} \cup \{2, 3\} = \{1, 2, 3\}\). The statement holds.

Thus, Statement 2 is correct.

Analyzing Statement 3: \(B = (A \cup B) - (A - B)\)

Let's analyze the right side: \((A \cup B) - (A - B)\). This means taking the set \(A \cup B\) and removing all elements that are in the set \(A - B\).

• \(A \cup B\): Contains elements in A or B.
• \(A - B\): Contains elements in A but not in B.

We are removing the part of A that is outside of B from the combined set \(A \cup B\). What remains?

The elements in \(A \cup B\) are either exclusively in A (part of \(A - B\)), exclusively in B (part of \(B - A\)), or in both (part of \(A \cap B\)).

When we remove \(A - B\) from \(A \cup B\), we are removing the elements that are only in A. The elements that remain are those in \(A \cup B\) that are not exclusively in A. These are the elements that are in B (which includes elements in \(A \cap B\) and elements in \(B - A\)).

So, \((A \cup B) - (A - B)\) is the set of elements in \(A \cup B\) that are not in \(A - B\). These are exactly the elements in B.

Thus, \((A \cup B) - (A - B) = B\). This identity is always true for any sets A and B.

Let's use an example: Let A = {1, 2} and B = {2, 3}. Then \(A \cup B = \{1, 2, 3\}\). \(A - B = \{1\}\). \((A \cup B) - (A - B) = \{1, 2, 3\} - \{1\} = \{2, 3\}\). This is equal to B.

Thus, Statement 3 is correct.

Conclusion on Set Theory Statements

Based on our analysis:

• Statement 1: \(A = (A \cup B) \cup (A - B)\) is Incorrect.
• Statement 2: \(A \cup (B - A) = (A \cup B)\) is Correct.
• Statement 3: \(B = (A \cup B) - (A - B)\) is Correct.

Therefore, the correct statements are 2 and 3 only.

Revision Table: Key Set Theory Concepts

Reviewing basic set operations is crucial for solving such problems.

Additional Information: Proving Set Identities

Set identities can be proven using element-wise arguments or by using known set algebra laws. For example, to prove \(A \cup (B - A) = A \cup B\):

Element-wise proof:

1. Show \(A \cup (B - A) \subseteq A \cup B\): Let \(x \in A \cup (B - A)\). This means \(x \in A\) or \(x \in (B - A)\).
   • If \(x \in A\), then \(x \in A \cup B\).
   • If \(x \in (B - A)\), then \(x \in B\) and \(x \notin A\). If \(x \in B\), then \(x \in A \cup B\).
   In either case, if \(x \in A \cup (B - A)\), then \(x \in A \cup B\). Thus, \(A \cup (B - A) \subseteq A \cup B\).
2. Show \(A \cup B \subseteq A \cup (B - A)\): Let \(x \in A \cup B\). This means \(x \in A\) or \(x \in B\).
   • If \(x \in A\), then \(x \in A \cup (B - A)\) (since \(A\) is part of the union).
   • If \(x \in B\), we consider two cases: \(x \in A\) or \(x \notin A\).
     • If \(x \in A\), then \(x \in A \cup (B - A)\) (since \(A\) is part of the union).
     • If \(x \notin A\), then \(x \in B\) and \(x \notin A\), which means \(x \in (B - A)\). If \(x \in (B - A)\), then \(x \in A \cup (B - A)\) (since \(B-A\) is part of the union).
   In all cases, if \(x \in A \cup B\), then \(x \in A \cup (B - A)\). Thus, \(A \cup B \subseteq A \cup (B - A)\).

Since both set inclusions hold, the sets are equal: \(A \cup (B - A) = A \cup B\).