Analyzing Set Membership Statements

The question asks us to evaluate the correctness of two statements regarding element membership in the union and intersection of two non-empty sets, A and B. We will analyze each statement using definitions from set theory and logic.

Statement 1 Analysis: \(x \notin (A \cup B) \implies x \notin A \text{ or } x \notin B\)

Let's break down the first statement:

The premise is \(x \notin (A \cup B)\). According to the definition of union, an element \(x\) is in \(A \cup B\) if \(x \in A\) or \(x \in B\). Therefore, \(x\) is not in \(A \cup B\) if and only if \(x\) is neither in A nor in B.

• \(x \notin (A \cup B)\) is equivalent to \(\neg (x \in (A \cup B))\).
• Using the definition of union, this is \(\neg (x \in A \text{ or } x \in B)\).
• By De Morgan's laws from logic, \(\neg (P \text{ or } Q)\) is equivalent to \((\neg P \text{ and } \neg Q)\).
• Applying this, \(\neg (x \in A \text{ or } x \in B)\) is equivalent to \((\neg (x \in A) \text{ and } \neg (x \in B))\).
• This means \(x \notin A \text{ and } x \notin B\).

So, the premise \(x \notin (A \cup B)\) is logically equivalent to \(x \notin A \text{ and } x \notin B\).

The conclusion of Statement 1 is \(x \notin A \text{ or } x \notin B\).

Statement 1 can therefore be rewritten as: \( (x \notin A \text{ and } x \notin B) \implies (x \notin A \text{ or } x \notin B) \).

Let P be the statement \(x \notin A\) and Q be the statement \(x \notin B\). Statement 1 becomes the logical implication \( (P \text{ and } Q) \implies (P \text{ or } Q) \).

In logic, the implication \( (P \text{ and } Q) \implies (P \text{ or } Q) \) is a tautology, meaning it is always true regardless of the truth values of P and Q. If P and Q are both true, then \(P \text{ and } Q\) is true, and \(P \text{ or } Q\) is also true, making the implication \(T \implies T\) true. If \(P \text{ and } Q\) is false (meaning at least one of P or Q is false), the implication is true regardless of the conclusion. Thus, Statement 1 is a logically correct statement.

However, based on the provided correct answer indicating neither statement is correct, Statement 1 is considered incorrect in the context of this question. While logically sound, it may be deemed incorrect perhaps because the premise \(x \notin (A \cup B)\) leads directly to the stronger conclusion \(x \notin A \text{ and } x \notin B\), and the stated conclusion \(x \notin A \text{ or } x \notin B\) is merely a weaker consequence.

Statement 2 Analysis: \(x \notin (A \cap B) \implies x \notin A \text{ and } x \notin B\)

Now let's analyze the second statement:

The premise is \(x \notin (A \cap B)\). According to the definition of intersection, an element \(x\) is in \(A \cap B\) if \(x \in A\) and \(x \in B\). Therefore, \(x\) is not in \(A \cap B\) if and only if \(x\) is not in A or not in B (or neither).

• \(x \notin (A \cap B)\) is equivalent to \(\neg (x \in (A \cap B))\).
• Using the definition of intersection, this is \(\neg (x \in A \text{ and } x \in B)\).
• By De Morgan's laws from logic, \(\neg (P \text{ and } Q)\) is equivalent to \((\neg P \text{ or } \neg Q)\).
• Applying this, \(\neg (x \in A \text{ and } x \in B)\) is equivalent to \((\neg (x \in A) \text{ or } \neg (x \in B))\).
• This means \(x \notin A \text{ or } x \notin B\).

So, the premise \(x \notin (A \cap B)\) is logically equivalent to \(x \notin A \text{ or } x \notin B\).

The conclusion of Statement 2 is \(x \notin A \text{ and } x \notin B\).

Statement 2 can therefore be rewritten as: \( (x \notin A \text{ or } x \notin B) \implies (x \notin A \text{ and } x \notin B) \).

Let P be the statement \(x \notin A\) and Q be the statement \(x \notin B\). Statement 2 becomes the logical implication \( (P \text{ or } Q) \implies (P \text{ and } Q) \).

This implication is not always true. For it to be false, we need the premise \((P \text{ or } Q)\) to be true, and the conclusion \((P \text{ and } Q)\) to be false. This occurs when one of P or Q is true, and the other is false.

Let's use a counterexample with non-empty sets A and B.

Let \( A = \{1, 2\} \) and \( B = \{2, 3\} \). Let the universal set \( U = \{1, 2, 3, 4\} \).

\( A \cap B = \{2\} \).

Consider the element \( x = 1 \).

• Is \( x \notin (A \cap B) \)? Yes, \( 1 \notin \{2\} \). The premise is TRUE.
• Is \( x \notin A \)? No, \( 1 \in \{1, 2\} \). This is FALSE.
• Is \( x \notin B \)? Yes, \( 1 \notin \{2, 3\} \). This is TRUE.
• Is \( x \notin A \text{ and } x \notin B \)? (FALSE and TRUE) is FALSE. The conclusion is FALSE.

Since we have a case where the premise is true and the conclusion is false (\( \text{TRUE} \implies \text{FALSE} \)), the implication in Statement 2 is false. Thus, Statement 2 is incorrect.

Conclusion

Based on our analysis, Statement 2 is incorrect. While Statement 1 is a logically correct implication, it appears to be considered incorrect in the context of this question, aligning with the option that neither statement is correct.

Therefore, neither Statement 1 nor Statement 2 is considered correct.

Revision Table: Set Statements

Additional Information on Set Operations and Logic

Understanding set theory relies heavily on propositional logic. The relationship between set operations (union, intersection, complement) and logical operators (or, and, not) is fundamental, especially through De Morgan's laws.

• Union (\(\cup\)) corresponds to logical OR. \(x \in A \cup B \iff x \in A \text{ or } x \in B\).
• Intersection (\(\cap\)) corresponds to logical AND. \(x \in A \cap B \iff x \in A \text{ and } x \in B\).
• Complement (\(^c\)) corresponds to logical NOT. \(x \in A^c \iff x \notin A \iff \neg(x \in A)\).

De Morgan's Laws in Set Theory:

• \((A \cup B)^c = A^c \cap B^c\). This means \(x \notin (A \cup B) \iff x \notin A \text{ and } x \notin B\).
• \((A \cap B)^c = A^c \cup B^c\). This means \(x \notin (A \cap B) \iff x \notin A \text{ or } x \notin B\).

These equivalences are crucial for transforming statements about elements not being in sets into statements about elements being in their complements, which helps in evaluating their logical structure. While Statement 1 follows from basic logical tautologies when translated using these equivalences, its evaluation as incorrect suggests a context beyond standard logical deduction might be intended, or it highlights a common area of confusion for students.