Answer
The mistake is assuming $x \in A \Rightarrow x \in A - B$, which is false unless $x \notin B$. The proof reverses the logic and misuses the definition of set difference.
Work Step by Step
We are given a “proof” that:
\[
(A - B) \cup (A \cap B) \subseteq A
\] and asked to **find the mistake**.
---
## ✅ First: Is the Statement Itself True?
Let’s check:
\(A - B = \{ x \in A \mid x \notin B \}\)
\(A \cap B = \{ x \in A \mid x \in B \}\)
So:
\[
(A - B) \cup (A \cap B) = \{ x \in A \mid x \notin B \text{ or } x \in B \}
= \{ x \in A \}
= A
\]
✅ So in fact:
\[
\boxed{(A - B) \cup (A \cap B) = A}
\]
Hence, the **inclusion is true**, and the inclusion is actually an equality.
---
## ❌ What's Wrong with the “Proof”?
Let’s look closely at the “proof”:
> Suppose \(x \in (A - B) \cup (A \cap B)\).
> If \(x \in A\), then \(x \in A - B\).
> Then, by definition, \(x \in A\) and \(x \notin B\), so \(x \in A\)...
🔍 **Wait!**
The logic here is broken:
### The key issue:
> “If \(x \in A\), then \(x \in A - B\)” — ❌ This is **false** in general!
Why?
Because:
- \(x \in A\) **does not imply** \(x \in A - B\),
- unless you also know \(x \notin B\).
So the proof is **assuming the conclusion** (that \(x \in A\)) as a **premise**.
In fact, the proof **should** start with:
> Let \(x \in (A - B) \cup (A \cap B)\)
Then break into two cases:
1. If \(x \in A - B\):
Then \(x \in A\) and \(x \notin B\) ⇒ ✅ \(x \in A\)
2. If \(x \in A \cap B\):
Then \(x \in A\) and \(x \in B\) ⇒ ✅ \(x \in A\)
So in either case, \(x \in A\), and thus:
\[
(A - B) \cup (A \cap B) \subseteq A
\]
✅ **This** is the correct argument.
