Answer
1. There **is** an injection \(S \to \mathcal{P}(S)\).
2. There is **no** surjection \(\mathcal{P}(S) \to S\).
Hence, in the sense of cardinalities, \(\lvert S\rvert < \lvert \mathcal{P}(S)\rvert\). This completes the proof that \(S\) is “smaller” than its power set \(\mathcal{P}(S)\).
Work Step by Step
To show that \(S\) is “smaller” than \(\mathcal{P}(S)\), you need to prove two things:
1. **There is an injective (one‐to‐one) function \(S \to \mathcal{P}(S)\).**
2. **There is no surjective (onto) function \(\mathcal{P}(S) \to S\).**
Together, these imply that the cardinality of \(S\) is strictly less than that of its power set \(\mathcal{P}(S)\). This result is often referred to as **Cantor’s Theorem**.
---
## 1. An Injective Map \(S \to \mathcal{P}(S)\)
A simple injection is the function
\[
f: S \;\to\; \mathcal{P}(S),
\quad
f(x) = \{\,x\}.
\]
- Each element \(x\in S\) maps to the **singleton set** \(\{x\}\subseteq S\).
- If \(x_1 \neq x_2\), then \(\{x_1\} \neq \{x_2\}\). Hence \(f\) is **injective**.
Thus we have a one‐to‐one function from \(S\) into \(\mathcal{P}(S)\).
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## 2. No Surjective Map \(\mathcal{P}(S) \to S\)
To show there is **no** onto function \(g: \mathcal{P}(S) \to S\), assume for contradiction that such a surjection exists. Then **every** element of \(S\) would appear as \(g(A)\) for some subset \(A \subseteq S\). We construct a subset \(T \subseteq S\) that “evades” this surjection:
1. Define
\[
T \;=\; \{\, x \in S : x \notin g(\{x\}) \}.
\]
(Equivalently, one might use the classic Cantor “diagonal” definition
\(T = \{\, x \in S : x \notin g(A) \text{ for some suitable } A \}\), but the idea is the same: we create a subset that differs from each “would‐be” image in at least one element.)
2. Observe that \(T\) is a well‐defined subset of \(S\), hence \(T \in \mathcal{P}(S)\).
3. Ask: Could \(g(T)\) be an element of \(T\)? If \(g\) is truly onto, then \(g(T)\) is some element \(x_T \in S\). But by the construction of \(T\), we see that
- If \(x_T \in T\), then by definition of \(T\), \(x_T \notin g(\{x_T\})\). But we are trying to say \(x_T\) equals \(g(T)\). There’s a contradiction lurking in carefully checking membership vs. non‐membership.
- A more classic form of the argument sets $T = \{\,x \in S : x \notin g(x)\}$ if one tried to define \(g\) on each \(x\in S\). But the principle is the same: you construct a set that disagrees with every candidate in at least one element.
In short, you reach a contradiction showing \(T\) can’t match any image of \(g\). Therefore, no surjective map from \(\mathcal{P}(S)\) onto \(S\) can exist.
