Key Concepts and Formulas

• Set Equality: Two sets $X$ and $Y$ are equal ($X=Y$) if and only if $X \subseteq Y$ and $Y \subseteq X$. This means every element in $X$ is in $Y$, and every element in $Y$ is in $X$.
• Set Decomposition: Any set $X$ can be decomposed with respect to another set $A$ using the identity $X = (X \cap A) \cup (X \cap A^c)$, where $A^c$ is the complement of $A$. This partitions $X$ into two disjoint parts: elements common to $X$ and $A$, and elements in $X$ but not in $A$.
• Properties of Set Operations: We will use the commutative property of intersection ($X \cap Y = Y \cap X$) and the definition of union and set difference.

Step-by-Step Solution

We are given two conditions for sets $A, B,$ and $C$:

1. $A \cap B = A \cap C$
2. $A \cup B = A \cup C$

We need to determine the relationship between $B$ and $C$.

Step 1: Decompose Sets $B$ and $C$ using Set $A$

To prove that $B=C$, we will show that their corresponding components, when decomposed with respect to $A$, are equal. Using the set decomposition identity, we can write: $B = (B \cap A) \cup (B \cap A^c)$ (Equation 1) $C = (C \cap A) \cup (C \cap A^c)$ (Equation 2)

Our goal is to prove that $B \cap A = C \cap A$ and $B \cap A^c = C \cap A^c$.

Step 2: Utilize the First Given Condition ($A \cap B = A \cap C$)

The first given condition directly provides the equality for the first part of our decomposition: $A \cap B = A \cap C$ Since set intersection is commutative, we can rewrite this as: $B \cap A = C \cap A$ (Result 1) This establishes the equality of the first components.

Step 3: Utilize the Second Given Condition ($A \cup B = A \cup C$) to Prove Equality of the Second Components

Now we need to show that $B \cap A^c = C \cap A^c$. This is equivalent to showing $B \setminus A = C \setminus A$. We will prove this by showing mutual inclusion: $B \setminus A \subseteq C \setminus A$ and $C \setminus A \subseteq B \setminus A$.

• Part 3a: Prove $B \setminus A \subseteq C \setminus A$ Let $x$ be an arbitrary element such that $x \in B \setminus A$. By the definition of set difference, this means $x \in B$ and $x \notin A$. Since $x \in B$, it must also be an element of $A \cup B$ (because $B \subseteq A \cup B$). We are given that $A \cup B = A \cup C$. Therefore, $x \in A \cup C$. The condition $x \in A \cup C$ implies that $x \in A$ or $x \in C$. However, we know from our initial assumption that $x \notin A$. Therefore, if $x \in A \cup C$ and $x \notin A$, it must be that $x \in C$. So, we have $x \in C$ and $x \notin A$. By the definition of set difference, this means $x \in C \setminus A$. Thus, we have shown that if $x \in B \setminus A$, then $x \in C \setminus A$. This implies $B \setminus A \subseteq C \setminus A$.

• Part 3b: Prove $C \setminus A \subseteq B \setminus A$ Let $y$ be an arbitrary element such that $y \in C \setminus A$. By the definition of set difference, this means $y \in C$ and $y \notin A$. Since $y \in C$, it must also be an element of $A \cup C$. We are given that $A \cup C = A \cup B$. Therefore, $y \in A \cup B$. The condition $y \in A \cup B$ implies that $y \in A$ or $y \in B$. However, we know from our initial assumption that $y \notin A$. Therefore, if $y \in A \cup B$ and $y \notin A$, it must be that $y \in B$. So, we have $y \in B$ and $y \notin A$. By the definition of set difference, this means $y \in B \setminus A$. Thus, we have shown that if $y \in C \setminus A$, then $y \in B \setminus A$. This implies $C \setminus A \subseteq B \setminus A$.

Since $B \setminus A \subseteq C \setminus A$ and $C \setminus A \subseteq B \setminus A$, we can conclude: $B \setminus A = C \setminus A$ Using the identity $X \setminus A = X \cap A^c$, we get: $B \cap A^c = C \cap A^c$ (Result 2) This establishes the equality of the second components.

Step 4: Combine Results to Conclude Set Equality

We have established the equality of both corresponding components:

• From Result 1: $B \cap A = C \cap A$
• From Result 2: $B \cap A^c = C \cap A^c$

Substitute these equalities back into the decomposition equations (Equation 1 and Equation 2): $B = (B \cap A) \cup (B \cap A^c) = (C \cap A) \cup (C \cap A^c)$ $C = (C \cap A) \cup (C \cap A^c)$

Since both $B$ and $C$ are equal to the same expression $(C \cap A) \cup (C \cap A^c)$, we can conclude that $B=C$.

Common Mistakes & Tips

• Cancellation Law for Sets: Do not assume that $A \cap B = A \cap C$ implies $B=C$ on its own. Both intersection and union conditions are necessary.
• Element-wise Proof: For proving set equality or subset relations, starting with an arbitrary element and showing it belongs to the other set is a rigorous and universally applicable method.
• Decomposition Strategy: Breaking down sets based on another set ($A$ in this case) into disjoint parts is a powerful technique when dealing with conditions involving intersections and unions with that set.

Summary

By decomposing sets $B$ and $C$ into components relative to set $A$ (namely, elements within $A$ and elements outside $A$), we used the given conditions to prove that the corresponding components of $B$ and $C$ are equal. Specifically, $A \cap B = A \cap C$ directly gave us $B \cap A = C \cap A$, and the condition $A \cup B = A \cup C$, combined with an element-wise proof, allowed us to show that $B \setminus A = C \setminus A$, which implies $B \cap A^c = C \cap A^c$. Since both disjoint components of $B$ are equal to the corresponding components of $C$, we conclude that $B=C$.

The final answer is $\boxed{B}$.