Analyzing Set Theory Statements

The question asks us to evaluate the correctness of two given statements involving set operations on two non-empty sets, A and B. We will analyze each statement individually using fundamental set identities and properties.

Evaluation of Statement 1: $(A \cap B) \cup (A \cap B̅) \cup (A̅ \cap B) = A \cup B$

Let's analyze the left-hand side (LHS) of the equation: $(A \cap B) \cup (A \cap B̅) \cup (A̅ \cap B)$.

We can group the first two terms and apply the distributive property:

$(A \cap B) \cup (A \cap B̅) = A \cap (B \cup B̅)$

Since $B \cup B̅$ represents the union of set B and its complement, it equals the universal set, denoted by U.

$A \cap (B \cup B̅) = A \cap U = A$

So, the LHS simplifies to $A \cup (A̅ \cap B)$.

Now, let's analyze $A \cup (A̅ \cap B)$. We can again use the distributive property:

$A \cup (A̅ \cap B) = (A \cup A̅) \cap (A \cup B)$

Since $A \cup A̅$ represents the union of set A and its complement, it equals the universal set U.

$(A \cup A̅) \cap (A \cup B) = U \cap (A \cup B)$

The intersection of the universal set U with any set is that set itself.

$U \cap (A \cup B) = A \cup B$

Thus, the LHS simplifies to $A \cup B$, which is equal to the right-hand side (RHS).

Alternatively, we can think of the sets in terms of regions in a Venn diagram:

• $A \cap B$: Elements in both A and B.
• $A \cap B̅$: Elements in A but not in B (A only).
• $A̅ \cap B$: Elements in B but not in A (B only).

The union of these three disjoint regions covers all elements that are in A, or in B, or in both. This precisely describes the set $A \cup B$.

Therefore, Statement 1 is correct.

Evaluation of Statement 2: $(A \cup (A̅ \cap B̅)) = A \cup B$

Let's analyze the left-hand side (LHS) of the equation: $A \cup (A̅ \cap B̅)$.

By De Morgan's Law, the intersection of the complements $A̅ \cap B̅$ is equal to the complement of the union $(A \cup B)̅$.

So, the LHS can be written as $A \cup (A \cup B)̅$.

This expression represents the union of set A with all elements that are *outside* of the union of A and B ($A \cup B$).

Let's consider the regions in a Venn diagram:

• A: Elements in set A.
• $(A \cup B)̅$: Elements outside both A and B.

The union $A \cup (A \cup B)̅$ includes all elements in A and all elements that are neither in A nor in B. It does *not* include elements that are in B but not in A ($A̅ \cap B$).

For the statement to be true, $A \cup (A \cup B)̅$ must equal $A \cup B$. This would only happen if the region $A̅ \cap B$ was empty, which is not generally true for non-empty sets A and B.

For example, let A = {1}, B = {2}. Assume Universal Set U = {1, 2, 3}.

• $A \cup B = \{1, 2\}$
• $A̅ = \{2, 3\}$, $B̅ = \{1, 3\}$
• $A̅ \cap B̅ = \{3\}$
• LHS: $A \cup (A̅ \cap B̅) = \{1\} \cup \{3\} = \{1, 3\}$
• RHS: $A \cup B = \{1, 2\}$

Since $\{1, 3\} \neq \{1, 2\}$, the statement is false.

Therefore, Statement 2 is incorrect.

Conclusion

Based on our analysis, Statement 1 is correct, and Statement 2 is incorrect.

The option that states only Statement 1 is correct is the correct answer.

Revision Table: Key Set Identities

Additional Information: Visualizing Sets with Venn Diagrams

Venn diagrams are useful tools for visualizing set operations and identities. Each set is represented by a circle within a rectangle representing the universal set U. The regions created by overlapping circles represent intersections, and the areas covered by the circles represent unions.

• The region $A \cap B$ is the overlap of the circles for A and B.
• The region $A \cap B̅$ is the part of circle A that is outside circle B.
• The region $A̅ \cap B$ is the part of circle B that is outside circle A.
• The region $A̅ \cap B̅ = (A \cup B)̅$ is the area outside both circles A and B within the universal set.

Statement 1 corresponds to the union of the 'overlap' region, the 'A only' region, and the 'B only' region. Combining these three disjoint regions covers the entire area of both circles combined, which is $A \cup B$.

Statement 2 corresponds to the union of the entire circle A and the region outside both circles ($A̅ \cap B̅$). This union includes all of A and all elements outside of $A \cup B$, but it specifically excludes the region $A̅ \cap B$ (B only). Therefore, $A \cup (A̅ \cap B̅)$ is generally smaller than $A \cup B$, making Statement 2 incorrect.