Answer
See explanation
Work Step by Step
We are asked to prove the identity:
\[
\bigcup_{i=1}^{n}(A \times B_i) = A \times \left( \bigcup_{i=1}^{n} B_i \right)
\]
for all integers \(n \geq 1\), where \(A, B_1, B_2, \dots, B_n\) are any sets.
---
## ✅ Goal:
Use an **element argument** to show that both sides contain exactly the same ordered pairs.
That is, show:
1. \(\displaystyle \bigcup_{i=1}^{n}(A \times B_i) \subseteq A \times \left( \bigcup_{i=1}^{n} B_i \right)\)
2. \(\displaystyle A \times \left( \bigcup_{i=1}^{n} B_i \right) \subseteq \bigcup_{i=1}^{n}(A \times B_i)\)
---
### 🔹 Part 1: Left ⊆ Right
Let \((a, b) \in \displaystyle \bigcup_{i=1}^{n}(A \times B_i)\)
Then:
- ∃ some \(j\) such that \((a, b) \in A \times B_j\)
- So \(a \in A\), and \(b \in B_j\)
- Then \(b \in \bigcup_{i=1}^{n} B_i\)
So:
\[
(a, b) \in A \times \left( \bigcup_{i=1}^{n} B_i \right)
\]
✅ Thus:
\[
\bigcup_{i=1}^{n}(A \times B_i) \subseteq A \times \left( \bigcup_{i=1}^{n} B_i \right)
\]
---
### 🔹 Part 2: Right ⊆ Left
Let \((a, b) \in A \times \left( \bigcup_{i=1}^{n} B_i \right)\)
Then:
- \(a \in A\),
- \(b \in \bigcup_{i=1}^{n} B_i\)
⇒ ∃ some \(j\) such that \(b \in B_j\)
So:
- \((a, b) \in A \times B_j\)
- ⇒ \((a, b) \in \bigcup_{i=1}^{n}(A \times B_i)\)
✅ Therefore:
\[
A \times \left( \bigcup_{i=1}^{n} B_i \right) \subseteq \bigcup_{i=1}^{n}(A \times B_i)
\]
---
## ✅ Final Conclusion:
Since both inclusions hold:
\[
\boxed{
\bigcup_{i=1}^{n}(A \times B_i) = A \times \left( \bigcup_{i=1}^{n} B_i \right)
}
\]
✔️ Proven using an element argument.
