Answer
See explanation
Work Step by Step
We are given the identity to prove:
\[
\bigcap_{i=1}^{n}(A_i - B) = \left( \bigcap_{i=1}^{n} A_i \right) - B
\]
for all integers \(n \geq 1\), and any sets \(A_1, A_2, \dots, A_n\), and \(B\).
---
## ✅ Goal:
Use an **element argument** to prove the two sets are equal by showing:
1. \(\displaystyle \bigcap_{i=1}^{n}(A_i - B) \subseteq \left( \bigcap_{i=1}^{n} A_i \right) - B\)
2. \(\displaystyle \left( \bigcap_{i=1}^{n} A_i \right) - B \subseteq \bigcap_{i=1}^{n}(A_i - B)\)
---
### 🔹 Part 1: Left ⊆ Right
Let \(x \in \displaystyle \bigcap_{i=1}^{n}(A_i - B)\).
Then:
- For all \(i = 1, 2, \dots, n\), we have \(x \in A_i - B\),
- So for each \(i\), \(x \in A_i\) and \(x \notin B\)
Thus:
- \(x \in \bigcap_{i=1}^{n} A_i\),
- And \(x \notin B\)
So:
\[
x \in \left( \bigcap_{i=1}^{n} A_i \right) - B
\]
✅ Therefore:
\[
\bigcap_{i=1}^{n}(A_i - B) \subseteq \left( \bigcap_{i=1}^{n} A_i \right) - B
\]
---
### 🔹 Part 2: Right ⊆ Left
Let \(x \in \left( \bigcap_{i=1}^{n} A_i \right) - B\)
Then:
- \(x \in \bigcap_{i=1}^{n} A_i\) ⇒ for all \(i\), \(x \in A_i\),
- And \(x \notin B\)
So for each \(i\):
- \(x \in A_i\) and \(x \notin B\) ⇒ \(x \in A_i - B\)
Thus:
\[
x \in \bigcap_{i=1}^{n}(A_i - B)
\]
✅ Therefore:
\[
\left( \bigcap_{i=1}^{n} A_i \right) - B \subseteq \bigcap_{i=1}^{n}(A_i - B)
\]
---
## ✅ Final Conclusion:
Since both inclusions hold:
\[
\boxed{
\bigcap_{i=1}^{n}(A_i - B) = \left( \bigcap_{i=1}^{n} A_i \right) - B
}
\]
✔️ Proven by element argument.
