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: $A \cup (B - A) = A \cup B$.