Let $A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$ and $D = \{5, 6, 7, 8\}$. Verify that (i) $A imes (B \cap C) = (A imes B) \cap (A imes C)$. (ii) $A imes C$ is a subset of $B imes D$.

Question

Let $A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$ and $D = \{5, 6, 7, 8\}$. Verify that (i) $A 	imes (B \cap C) = (A 	imes B) \cap (A 	imes C)$. (ii) $A 	imes C$ is a subset of $B 	imes D$.

Accepted Answer

> [!TLDR]
> The Cartesian product distributes over intersection, so $A 	imes (B \cap C) = (A 	imes B) \cap (A 	imes C)$ holds. Also, since $A \subseteq B$ and $C \subseteq D$, every ordered pair in $A 	imes C$ is also in $B 	imes D$, making $A 	imes C$ a subset of $B 	imes D$.

## Why This Works — The Core Idea

The Cartesian product $X 	imes Y$ is the set of all ordered pairs $(x, y)$ where $x \in X$ and $y \in Y$. When we intersect two sets before taking the product, we are restricting the second coordinate to elements that belong to *both* sets. On the other hand, taking the product first and then intersecting keeps only those pairs whose second coordinate is common to both original sets. These two processes yield exactly the same collection of pairs — that’s the distributive law.

For the subset part, if every element of $A$ is also in $B$, and every element of $C$ is also in $D$, then any pair $(a, c)$ automatically has its first coordinate in $B$ and its second in $D$, so it belongs to $B 	imes D$.

Let’s verify both statements concretely.

---

## (i) $A 	imes (B \cap C) = (A 	imes B) \cap (A 	imes C)$

**Step 1: Compute $B \cap C$**  
$B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$. These sets have no common element, so  
$$B \cap C = \varnothing$$

**Step 2: Compute $A 	imes (B \cap C)$**  
The Cartesian product with an empty set is empty:  
$$A 	imes \varnothing = \varnothing$$

**Step 3: Compute $A 	imes B$**  
$A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$  
$$A 	imes B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)\}$$

**Step 4: Compute $A 	imes C$**  
$A = \{1, 2\}$, $C = \{5, 6\}$  
$$A 	imes C = \{(1,5), (1,6), (2,5), (2,6)\}$$

**Step 5: Intersect $A 	imes B$ and $A 	imes C$**  
Look for pairs that appear in both lists. The first set has second coordinates $1,2,3,4$; the second set has second coordinates $5,6$. No pair can match because the second coordinates are disjoint.  
$$(A 	imes B) \cap (A 	imes C) = \varnothing$$

**Step 6: Com

Q.Let $A = \{1, 2\}$ , $B = \{1, 2, 3, 4\}$ , $C = \{5, 6\}$ and $D = \{5, 6, 7, 8\}$ . Verify that

Concept understanding — Cartesian Product

Cartesian Product: From Intuition to Definition

The Intuition

The Precise Definition

Key Properties to Remember

Examples to Cement the Idea

Why This Matters

Why This Works — The Core Idea

(i) $A \times (B \cap C) = (A \times B) \cap (A \times C)$

(ii) $A \times C$ is a subset of $B \times D$

Q.Let A={1,2}A = \{1, 2\}A={1,2}, B={1,2,3,4}B = \{1, 2, 3, 4\}B={1,2,3,4}, C={5,6}C = \{5, 6\}C={5,6} and D={5,6,7,8}D = \{5, 6, 7, 8\}D={5,6,7,8}. Verify that

Why This Works — The Core Idea

(i) A×(B∩C)=(A×B)∩(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C)A×(B∩C)=(A×B)∩(A×C)

(ii) A×CA \times CA×C is a subset of B×DB \times DB×D

🎓Unlock everything free for 14 days

Q.Let $A = \{1, 2\}$ , $B = \{1, 2, 3, 4\}$ , $C = \{5, 6\}$ and $D = \{5, 6, 7, 8\}$ . Verify that

(i) $A \times (B \cap C) = (A \times B) \cap (A \times C)$

(ii) $A \times C$ is a subset of $B \times D$

Unlock everything free for 14 days