Answer
See explanation
Work Step by Step
Let's prove both parts (a) and (b) using **element arguments** and basic set theory.
---
## **(a)**
Prove that:
\[
(A - B) \cup (B - A) \cup (A \cap B) = A \cup B
\]
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### ✅ Proof:
Let’s take any element \(x\). We will show both sides contain the same elements.
### 🔹 Left ⊆ Right
Suppose \(x \in (A - B) \cup (B - A) \cup (A \cap B)\). Then:
- If \(x \in A - B\):
Then \(x \in A\) and \(x \notin B\) ⇒ \(x \in A \cup B\)
- If \(x \in B - A\):
Then \(x \in B\) and \(x \notin A\) ⇒ \(x \in A \cup B\)
- If \(x \in A \cap B\):
Then \(x \in A\) and \(x \in B\) ⇒ \(x \in A \cup B\)
✅ So in all cases: \(x \in A \cup B\)
Thus:
\[
(A - B) \cup (B - A) \cup (A \cap B) \subseteq A \cup B
\]
---
### 🔹 Right ⊆ Left
Suppose \(x \in A \cup B\)
Now we consider cases:
- If \(x \in A\) and \(x \notin B\), then \(x \in A - B\)
- If \(x \in B\) and \(x \notin A\), then \(x \in B - A\)
- If \(x \in A\) and \(x \in B\), then \(x \in A \cap B\)
In all these cases:
\[
x \in (A - B) \cup (B - A) \cup (A \cap B)
\]
✅ So:
\[
A \cup B \subseteq (A - B) \cup (B - A) \cup (A \cap B)
\]
---
### ✅ Final Conclusion for (a):
Since both inclusions hold:
\[
\boxed{
(A - B) \cup (B - A) \cup (A \cap B) = A \cup B
}
\]
---
## **(b)**
Prove that the sets \(A - B\), \(B - A\), and \(A \cap B\) are **mutually disjoint**.
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### ✅ Proof:
Mutually disjoint means:
No element is in **more than one** of the sets.
Let’s check pairwise intersections:
1. **\((A - B) \cap (B - A)\)**
- If \(x \in A - B\), then \(x \in A\) and \(x \notin B\)
- If \(x \in B - A\), then \(x \in B\) and \(x \notin A\)
- So no element can satisfy both ⇒ intersection is empty
✅ \((A - B) \cap (B - A) = \varnothing\)
---
2. **\((A - B) \cap (A \cap B)\)**
- \(x \in A - B\) ⇒ \(x \in A\) and \(x \notin B\)
- \(x \in A \cap B\) ⇒ \(x \in A\) and \(x \in B\)
- Contradiction: can’t be both in and not in \(B\)
✅ \((A - B) \cap (A \cap B) = \varnothing\)
---
3. **\((B - A) \cap (A \cap B)\)**
- \(x \in B - A\) ⇒ \(x \in B\) and \(x \notin A\)
- \(x \in A \cap B\) ⇒ \(x \in A\) and \(x \in B\)
- Contradiction: can’t be both in and not in \(A\)
✅ \((B - A) \cap (A \cap B) = \varnothing\)
---
### ✅ Final Conclusion for (b):
\[
\text{The sets } A - B,\; B - A,\; \text{and } A \cap B \text{ are mutually disjoint.}
\]
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## ✅ Summary:
- (a) The three sets form a **partition** of \(A \cup B\).
- (b) They are **mutually disjoint**, and their union is exactly \(A \cup B\).
