Answer
The mistake is assuming that $x \notin A \text{ or } x \notin B \Rightarrow x \notin A \cup B$. This is false. The inclusion $A^c \cup B^c \subseteq (A \cup B)^c$ is not always true.
Instead, De Morgan’s law tells us:
\[
\boxed{(A \cup B)^c = A^c \cap B^c}
\] So the direction should be:
\[
(A \cup B)^c \subseteq A^c \cup B^c \quad \text{(False)} \\
(A \cup B)^c \subseteq A^c \cap B^c \quad \text{(True)}
\]
Work Step by Step
The theorem being “proven” is:
For all sets \(A\) and \(B\),
\[
A^c \cup B^c \subseteq (A \cup B)^c
\] We are asked to find the **mistake** in the given "proof."
---
## ✅ Step 1: What’s Actually True?
This identity is **not true in general**. In fact, the **opposite** direction holds:
\[
\boxed{(A \cup B)^c = A^c \cap B^c}
\]
Thus, by De Morgan’s laws:
\[
A^c \cup B^c \supseteq (A \cup B)^c
\]
**So the claim \(A^c \cup B^c \subseteq (A \cup B)^c\)** is **false**, meaning the whole proof is trying to prove a false statement.
---
## ❌ Step 2: What's the mistake in the proof?
Let’s summarize the flawed logic:
1. Let \(x \in A^c \cup B^c\)
2. Then \(x \in A^c\) or \(x \in B^c\)
3. So \(x \notin A\) or \(x \notin B\)
4. Therefore \(x \notin A \cup B\)
5. So \(x \in (A \cup B)^c\)
The mistake is in step 4:
> From “\(x \notin A\) or \(x \notin B\)” ⇒ conclude “\(x \notin A \cup B\)”
This is **not valid** logic!
---
### ❌ Why it's wrong:
Let’s take a counterexample.
Let:
- \(A = \{1\}\)
- \(B = \{2\}\)
- Universal set \(U = \{1, 2, 3\}\)
Then:
- \(A^c = \{2, 3\}\)
- \(B^c = \{1, 3\}\)
- \(A^c \cup B^c = \{1, 2, 3\} = U\)
- \((A \cup B)^c = \{3\}\)
So:
\[
A^c \cup B^c = U \nsubseteq \{3\} = (A \cup B)^c
\]
**Counterexample disproves the theorem**, and shows that the proof uses **invalid logic**.
---
### 🔎 What’s the specific mistake?
From:
\[
x \notin A \text{ or } x \notin B
\]
You **cannot** conclude:
\[
x \notin A \cup B
\]
A simple example:
- Let \(x = 1\), \(x \in A\), \(x \notin B\)
- So \(x \notin B\) is true ⇒ \(x \in B^c\)
- But \(x \in A\) ⇒ \(x \in A \cup B\)
- So \(x \notin A \cup B\) is **false**
This proves that the step:
> "If \(x \in A^c\) or \(x \in B^c\), then \(x \notin A \cup B\)"
is **logically invalid**.
