Analyzing Set Equality Statements

The question asks us to evaluate the correctness of two statements related to set operations and equality. We need to determine if having the same intersection or union with a set A implies that sets B and C are equal.

Let's examine each statement carefully.

Statement 1: Intersection Implication

The first statement says:

\( A \cap B = A \cap C \implies B = C \)

This statement claims that if the intersection of set A with set B is equal to the intersection of set A with set C, then it must be true that set B is equal to set C.

To check if this statement is correct, we can try to find a counterexample. A counterexample is a specific case where the condition (\( A \cap B = A \cap C \)) is true, but the conclusion (\( B = C \)) is false.

Consider the following sets:

• Set A = \{1\}
• Set B = \{1, 2\}
• Set C = \{1, 3\}

Let's find the intersections:

• \( A \cap B = \{1\} \cap \{1, 2\} = \{1\} \)
• \( A \cap C = \{1\} \cap \{1, 3\} = \{1\} \)

In this case, \( A \cap B = A \cap C \) is true, because both intersections equal \{1\}.

Now let's check if \( B = C \):

• \( B = \{1, 2\} \)
• \( C = \{1, 3\} \)

Clearly, \( B \neq C \).

Since we found a case where \( A \cap B = A \cap C \) is true but \( B = C \) is false, the implication \( A \cap B = A \cap C \implies B = C \) is false in general.

Therefore, Statement 1 is incorrect.

Statement 2: Union Implication

The second statement says:

\( A \cup B = A \cup C \implies B = C \)

This statement claims that if the union of set A with set B is equal to the union of set A with set C, then it must be true that set B is equal to set C.

Let's try to find a counterexample for this statement as well.

Consider the following sets:

• Set A = \{1, 2\}
• Set B = \{1, 3\}
• Set C = \{2, 3\}

Let's find the unions:

• \( A \cup B = \{1, 2\} \cup \{1, 3\} = \{1, 2, 3\} \)
• \( A \cup C = \{1, 2\} \cup \{2, 3\} = \{1, 2, 3\} \)

In this case, \( A \cup B = A \cup C \) is true, because both unions equal \{1, 2, 3\}.

Now let's check if \( B = C \):

• \( B = \{1, 3\} \)
• \( C = \{2, 3\} \)

Clearly, \( B \neq C \).

Since we found a case where \( A \cup B = A \cup C \) is true but \( B = C \) is false, the implication \( A \cup B = A \cup C \implies B = C \) is false in general.

Therefore, Statement 2 is incorrect.

Conclusion

Based on our analysis and counterexamples, both Statement 1 (\( A \cap B = A \cap C \implies B = C \)) and Statement 2 (\( A \cup B = A \cup C \implies B = C \)) are incorrect.

Thus, neither of the given statements is correct.

Revision Table: Set Theory Basics

Additional Information: When the Implications Hold

While the statements are not universally true, there are specific conditions under which they might hold or related properties that are true.

• The statement \( A \cap B = A \cap C \implies B = C \) is true if we also know that \( B \subseteq A \) and \( C \subseteq A \). In this case, \( A \cap B = B \) and \( A \cap C = C \), so \( B = C \) follows directly.
• The statement \( A \cup B = A \cup C \implies B = C \) is true if we also know that \( A \subseteq B \) and \( A \subseteq C \). In this case, \( A \cup B = B \) and \( A \cup C = C \), so \( B = C \) follows directly.
• A related property that is always true is the cancellation law for symmetric difference: \( A \Delta B = A \Delta C \implies B = C \), where \( A \Delta B = (A \cup B) \setminus (A \cap B) \).
• Another related property is that if \( A \cap B = A \cap C \) AND \( A \cup B = A \cup C \), THEN \( B = C \). This can be shown using distributive laws or Venn diagrams.

Understanding these nuances is important for mastering set theory.