Answer
See explanation
Work Step by Step
We are given the identity to prove:
\[
\bigcup_{i=1}^{n}(A_i - B) = \left( \bigcup_{i=1}^{n} A_i \right) - B
\]
for any sets \(A_1, A_2, \ldots, A_n\) and \(B\), and integer \(n \geq 1\).
---
## ✅ Goal:
Use an **element argument** to prove both sides are equal by showing:
1. \(\displaystyle \bigcup_{i=1}^{n}(A_i - B) \subseteq \left( \bigcup_{i=1}^{n} A_i \right) - B\)
2. \(\displaystyle \left( \bigcup_{i=1}^{n} A_i \right) - B \subseteq \bigcup_{i=1}^{n}(A_i - B)\)
---
### 🔹 Part 1: Left ⊆ Right
Let \(x \in \displaystyle \bigcup_{i=1}^{n}(A_i - B)\).
Then:
- There exists some \(j\) such that \(x \in A_j - B\),
- So \(x \in A_j\) and \(x \notin B\),
- Then \(x \in \bigcup_{i=1}^{n} A_i\) and \(x \notin B\),
- So \(x \in \left( \bigcup_{i=1}^{n} A_i \right) - B\)
✅ Therefore,
\[
\bigcup_{i=1}^{n}(A_i - B) \subseteq \left( \bigcup_{i=1}^{n} A_i \right) - B
\]
---
### 🔹 Part 2: Right ⊆ Left
Let \(x \in \left( \bigcup_{i=1}^{n} A_i \right) - B\)
Then:
\(x \in \bigcup_{i=1}^{n} A_i\) and \(x \notin B\),
- So there exists some \(j\) such that \(x \in A_j\),
- Since \(x \notin B\), we get \(x \in A_j - B\),
- So \(x \in \bigcup_{i=1}^{n}(A_i - B)\)
✅ Therefore,
\[
\left( \bigcup_{i=1}^{n} A_i \right) - B \subseteq \bigcup_{i=1}^{n}(A_i - B)
\]
---
### ✅ Final Conclusion:
Since both inclusions hold, we conclude:
\[
\boxed{
\bigcup_{i=1}^{n}(A_i - B) = \left( \bigcup_{i=1}^{n} A_i \right) - B
}
\]
✔️ Proven by element argument.
