To prove that $A \times (B \cup C) = (A \times B) \cup (A \times C)$, we will use set theory and the properties of Cartesian products.

Definitions:

1. Cartesian product $A \times B$ means the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. $A \times B = \{ (a, b) \mid a \in A, b \in B \}$
2. The union $B \cup C$ is the set of all elements that are in $B$ or in $C$ (or in both): $B \cup C = \{ b \mid b \in B \text{ or } b \in C \}$

Now, let’s prove the equality $A \times (B \cup C) = (A \times B) \cup (A \times C)$.

Step 1: Prove $A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$

Take any element $(a, x) \in A \times (B \cup C)$. By definition of the Cartesian product:

• $a \in A$
• $x \in B \cup C$, meaning that $x$ is either in $B$ or in $C$.

So, there are two cases to consider:

• Case 1: If $x \in B$, then $(a, x) \in A \times B$, so $(a, x) \in (A \times B) \cup (A \times C)$.
• Case 2: If $x \in C$, then $(a, x) \in A \times C$, so $(a, x) \in (A \times B) \cup (A \times C)$.

Thus, for every element $(a, x) \in A \times (B \cup C)$, we have $(a, x) \in (A \times B) \cup (A \times C)$. Therefore, $A \times (B \cup C) \subseteq (A \times B) \cup (A \times C).$

Step 2: Prove $(A \times B) \cup (A \times C) \subseteq A \times (B \cup C)$

Take any element $(a, x) \in (A \times B) \cup (A \times C)$. By definition of the union, this means that $(a, x)$ must be in either $A \times B$ or $A \times C$:

• Case 1: If $(a, x) \in A \times B$, then $a \in A$ and $x \in B$. Since $B \subseteq B \cup C$, we have $x \in B \cup C$. Therefore, $(a, x) \in A \times (B \cup C)$.
• Case 2: If $(a, x) \in A \times C$, then $a \in A$ and $x \in C$. Since $C \subseteq B \cup C$, we have $x \in B \cup C$. Therefore, $(a, x) \in A \times (B \cup C)$.

Thus, for every element $(a, x) \in (A \times B) \cup (A \times C)$, we have $(a, x) \in A \times (B \cup C)$. Therefore, $(A \times B) \cup (A \times C) \subseteq A \times (B \cup C).$

Conclusion:

Since we have shown both inclusions: $A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$ and $(A \times B) \cup (A \times C) \subseteq A \times (B \cup C),$ we conclude that $A \times (B \cup C) = (A \times B) \cup (A \times C).$

Questions for further exploration:

1. How does distributivity of Cartesian products over unions relate to other set operations like intersections or differences?
2. How can this result be generalized to more than two sets, such as $A \times (B \cup C \cup D)$?
3. How does this property compare with the distributive property for regular set operations like intersection or union?
4. Can you provide a real-world example where this distributive property is used?
5. How would the proof change if the sets were finite, and what difference would that make in terms of computational efficiency?

Tip: This distributive property for Cartesian products is very useful in computer science, especially in databases and when working with product spaces in mathematics!