Answer
See explanation
Work Step by Step
We are asked to determine whether the following statement is **true or false**:
> For all sets \(A\) and \(B\),
> \[
\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B)
\]
---
### ✅ Understanding the statement:
- \(\mathcal{P}(A)\): the power set of \(A\), i.e., all subsets of \(A\)
- \(\mathcal{P}(B)\): all subsets of \(B\)
- \(\mathcal{P}(A \cup B)\): all subsets of the union of \(A\) and \(B\)
So the union \(\mathcal{P}(A) \cup \mathcal{P}(B)\) consists of:
- All subsets of \(A\)
- All subsets of \(B\)
And the right-hand side, \(\mathcal{P}(A \cup B)\), consists of:
- **All subsets** of the combined elements of \(A\) and \(B\)
---
### 🔍 Is the inclusion always true?
Let’s pick any \(X \in \mathcal{P}(A) \cup \mathcal{P}(B)\).
Then:
- Either \(X \subseteq A\), or \(X \subseteq B\)
In **either** case, we have:
- \(X \subseteq A \cup B\)
So \(X \in \mathcal{P}(A \cup B)\)
✅ Therefore, every element in the left-hand side is in the right-hand side.
---
### 🧠 Optional Follow-up (Is the reverse inclusion true?)
No, in general:
\[
\mathcal{P}(A \cup B) \nsubseteq \mathcal{P}(A) \cup \mathcal{P}(B)
\]
Counterexample:
Let \(A = \{1\},\; B = \{2\}\)
Then:
- \(\{1, 2\} \in \mathcal{P}(A \cup B)\),
- but \(\{1, 2\} \notin \mathcal{P}(A)\) and \(\{1, 2\} \notin \mathcal{P}(B)\)
So the reverse is false.
But the original direction **is correct**. ✅
