Answer
See explanation
Work Step by Step
We are asked to prove the following identity for all integers \(n \geq 1\):
\[
A \cap \left( \bigcup_{i=1}^{n} B_i \right) = \bigcup_{i=1}^{n} (A \cap B_i)
\]
This is a standard **distributive law** of set theory.
---
## ✅ Goal:
Prove both directions:
1. \(A \cap \left( \bigcup_{i=1}^{n} B_i \right) \subseteq \bigcup_{i=1}^{n} (A \cap B_i)\)
2. \(\bigcup_{i=1}^{n} (A \cap B_i) \subseteq A \cap \left( \bigcup_{i=1}^{n} B_i \right)\)
---
### 🔹 Part 1: Left ⊆ Right
Let \(x \in A \cap \left( \bigcup_{i=1}^{n} B_i \right)\).
Then:
\(x \in A\),
\(x \in \bigcup_{i=1}^{n} B_i\)
⇒ ∃ some \(j\) (with \(1 \leq j \leq n\)) such that \(x \in B_j\)
Then:
\(x \in A \cap B_j\),
So \(x \in \bigcup_{i=1}^{n} (A \cap B_i)\)
✅ So:
\[
x \in \bigcup_{i=1}^{n} (A \cap B_i)
\]
Thus:
\[
A \cap \left( \bigcup_{i=1}^{n} B_i \right) \subseteq \bigcup_{i=1}^{n} (A \cap B_i)
\]
---
### 🔹 Part 2: Right ⊆ Left
Let \(x \in \bigcup_{i=1}^{n} (A \cap B_i)\).
Then:
- ∃ some \(j\) such that \(x \in A \cap B_j\)
⇒ \(x \in A\) and \(x \in B_j\)
Then:
- \(x \in \bigcup_{i=1}^{n} B_i\)
- So \(x \in A \cap \left( \bigcup_{i=1}^{n} B_i \right)\)
✅ Thus:
\[
\bigcup_{i=1}^{n} (A \cap B_i) \subseteq A \cap \left( \bigcup_{i=1}^{n} B_i \right)
\]
---
### ✅ Final Conclusion:
Since both inclusions hold, we conclude:
\[
\boxed{
A \cap \left( \bigcup_{i=1}^{n} B_i \right) = \bigcup_{i=1}^{n} (A \cap B_i)
}
\]
This completes the proof.
