Answer
See explanation
Work Step by Step
We are asked to prove the identity:
\[
\bigcap_{i=1}^{n}(A \times B_i) = A \times \left( \bigcap_{i=1}^{n} B_i \right)
\]
for any sets \(A, B_1, B_2, \dots, B_n\), and any integer \(n \geq 1\).
---
## ✅ Goal:
Prove that both sides contain exactly the same ordered pairs by showing:
1. \(\displaystyle \bigcap_{i=1}^{n}(A \times B_i) \subseteq A \times \left( \bigcap_{i=1}^{n} B_i \right)\)
2. \(\displaystyle A \times \left( \bigcap_{i=1}^{n} B_i \right) \subseteq \bigcap_{i=1}^{n}(A \times B_i)\)
---
### 🔹 Part 1: Left ⊆ Right
Let \((a, b) \in \displaystyle \bigcap_{i=1}^{n}(A \times B_i)\)
Then:
- For **all** \(i\), \((a, b) \in A \times B_i\)
- So for all \(i\): \(a \in A\), \(b \in B_i\)
Since this holds for all \(i\), we conclude:
- \(a \in A\)
- \(b \in \bigcap_{i=1}^{n} B_i\)
Thus:
\[
(a, b) \in A \times \left( \bigcap_{i=1}^{n} B_i \right)
\]
✅ Therefore:
\[
\bigcap_{i=1}^{n}(A \times B_i) \subseteq A \times \left( \bigcap_{i=1}^{n} B_i \right)
\]
---
### 🔹 Part 2: Right ⊆ Left
Let \((a, b) \in A \times \left( \bigcap_{i=1}^{n} B_i \right)\)
Then:
- \(a \in A\)
- \(b \in \bigcap_{i=1}^{n} B_i\) ⇒ \(b \in B_i\) for **all** \(i\)
So for each \(i\), we have:
- \(a \in A\), \(b \in B_i\) ⇒ \((a, b) \in A \times B_i\)
Since this is true for all \(i\), it follows:
\[
(a, b) \in \bigcap_{i=1}^{n}(A \times B_i)
\]
✅ Therefore:
\[
A \times \left( \bigcap_{i=1}^{n} B_i \right) \subseteq \bigcap_{i=1}^{n}(A \times B_i)
\]
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## ✅ Final Conclusion:
Since both inclusions hold:
\[
\boxed{
\bigcap_{i=1}^{n}(A \times B_i) = A \times \left( \bigcap_{i=1}^{n} B_i \right)
}
\]
✔️ Proven using an element argument.
