Understanding Set Equality Conditions

We are given three non-empty sets A, B, and C, and two conditions relating their intersections and unions:

• Condition 1: $\text{A} \cap \text{B} = \text{A} \cap \text{C}$
• Condition 2: $\text{A} \cup \text{B} = \text{A} \cup \text{C}$

We need to determine the relationship between sets B and C based on these conditions.

Proving B = C from the Given Conditions

To prove that B = C, we need to show that every element in B is also in C (i.e., $\text{B} \subseteq \text{C}$) and every element in C is also in B (i.e., $\text{C} \subseteq \text{B}$).

Showing B is a Subset of C ($\text{B} \subseteq \text{C}$)

Let's take an arbitrary element $x \in \text{B}$. We need to show that $x \in \text{C}$.

Consider two cases for the element $x$:

• Case 1: $x \in \text{A}$
• If $x \in \text{B}$ and $x \in \text{A}$, then $x \in \text{A} \cap \text{B}$.
• From Condition 1, we know that $\text{A} \cap \text{B} = \text{A} \cap \text{C}$. Therefore, $x \in \text{A} \cap \text{C}$.
• If $x \in \text{A} \cap \text{C}$, it means $x \in \text{A}$ and $x \in \text{C}$.
• So, in this case, if $x \in \text{B}$, then $x \in \text{C}$.
• Case 2: $x \notin \text{A}$
• If $x \in \text{B}$, then $x$ must be in the union of A and B, i.e., $x \in \text{A} \cup \text{B}$.
• From Condition 2, we know that $\text{A} \cup \text{B} = \text{A} \cup \text{C}$. Therefore, $x \in \text{A} \cup \text{C}$.
• If $x \in \text{A} \cup \text{C}$, it means $x \in \text{A}$ or $x \in \text{C}$.
• We assumed in this case that $x \notin \text{A}$. Since $x \in \text{A} \cup \text{C}$ and $x \notin \text{A}$, it must be true that $x \in \text{C}$.
• So, in this case, if $x \in \text{B}$, then $x \in \text{C}.

In both cases, if $x \in \text{B}$, then $x \in \text{C}$. This shows that every element of B is also an element of C, which means $\text{B} \subseteq \text{C}$.

Showing C is a Subset of B ($\text{C} \subseteq \text{B}$)

Now, let's take an arbitrary element $y \in \text{C}$. We need to show that $y \in \text{B}$.

Consider two cases for the element $y$:

• Case 1: $y \in \text{A}$
• If $y \in \text{C}$ and $y \in \text{A}$, then $y \in \text{A} \cap \text{C}$.
• From Condition 1, we know that $\text{A} \cap \text{C} = \text{A} \cap \text{B}$. Therefore, $y \in \text{A} \cap \text{B}$.
• If $y \in \text{A} \cap \text{B}$, it means $y \in \text{A}$ and $y \in \text{B}$.
• So, in this case, if $y \in \text{C}$, then $y \in \text{B}$.
• Case 2: $y \notin \text{A}$
• If $y \in \text{C}$, then $y$ must be in the union of A and C, i.e., $y \in \text{A} \cup \text{C}$.
• From Condition 2, we know that $\text{A} \cup \text{C} = \text{A} \cup \text{B}$. Therefore, $y \in \text{A} \cup \text{B}$.
• If $y \in \text{A} \cup \text{B}$, it means $y \in \text{A}$ or $y \in \text{B}$.
• We assumed in this case that $y \notin \text{A}$. Since $y \in \text{A} \cup \text{B}$ and $y \notin \text{A}$, it must be true that $y \in \text{B}$.
• So, in this case, if $y \in \text{C}$, then $y \in \text{B}$.

In both cases, if $y \in \text{C}$, then $y \in \text{B}$. This shows that every element of C is also an element of B, which means $\text{C} \subseteq \text{B}$.

Conclusion: B = C

Since we have shown that $\text{B} \subseteq \text{C}$ and $\text{C} \subseteq \text{B}$, we can conclude that the sets B and C are equal. $\text{B} = \text{C}$.

This is a fundamental property in set theory: if the intersection and union of a set A with two other sets B and C are equal, then B and C must be equal themselves.