Answer
\[
\mathcal{P}(A) \not\subseteq A \quad \text{(the power set of any set A}\\\text{ is strictly larger than A)}
\]
This mirrors **Russell’s paradox** and is equivalent to **Cantor’s theorem**:
> There is **no surjection** from a set to its power set.
Work Step by Step
We are asked to prove that for **any set** \( A \), the **power set** \( \mathcal{P}(A) \not\subseteq A \), using a method like **Russell’s Paradox**.
---
### 🧠 Key Goal:
\[
\boxed{
\mathcal{P}(A) \not\subseteq A \quad \text{(the power set of } A \text{ is not contained in } A)
}
\]
That means there is **at least one subset of \( A \)** that is **not an element of \( A \)**.
---
### 🔁 Use the Diagonal Argument (Russell/Cantor Style):
Assume, for contradiction, that:
\[
\mathcal{P}(A) \subseteq A \quad \text{(every subset of } A \text{ is actually an element of } A)
\]
Then define the set:
\[
R = \{ x \in A \mid x \notin x \}
\]
This is the classic **Russell set**: the set of all elements of \( A \) that do **not** contain themselves.
Since \( R \subseteq A \), then \( R \in \mathcal{P}(A) \).
But by assumption, \( \mathcal{P}(A) \subseteq A \), so **\( R \in A \)**.
Now ask:
**Is \( R \in R \)?**
- If \( R \in R \), then by definition of \( R \), \( R \notin R \) — contradiction.
- If \( R \notin R \), then by definition of \( R \), \( R \in R \) — contradiction.
Thus, assuming \( \mathcal{P}(A) \subseteq A \) leads to a contradiction.
