Answer
The map
\[
A \;\mapsto\; \chi_A
\] is a **bijection** between \(\mathcal{P}(S)\) and the set of all functions \(S \to \{0,1\}\). Consequently, they have the **same cardinality**.
Work Step by Step
A well‐known way to show that the power set \(\mathcal{P}(S)\) (the set of all subsets of \(S\)) has the same cardinality as \(T\) (the set of all functions \(S \to \{0,1\}\)) is to exhibit a natural **bijection** between them:
---
## Defining the Bijection
1. **Given a subset \(A \subseteq S\)**, define its **characteristic function** \(\chi_A\colon S \to \{0,1\}\) by
\[
\chi_A(x)
\;=\;
\begin{cases}
1, & x \in A,\\
0, & x \notin A.
\end{cases}
\]
Clearly, \(\chi_A\) is a function from \(S\) into \(\{0,1\}\).
2. **Conversely, given a function \(f\colon S \to \{0,1\}\)**, define the subset
\[
A_f \;=\;\{\,x \in S : f(x)=1\}.
\]
That is, you pick out exactly those elements of \(S\) that map to \(1\).
---
## Injectivity (One‐to‐One)
- If \(A \neq B\), then there is at least one element \(s \in S\) that belongs to exactly one of the two subsets.
- Hence \(\chi_A(s)\neq \chi_B(s)\), so the characteristic functions differ.
Therefore, distinct subsets produce distinct characteristic functions.
---
## Surjectivity (Onto)
- Every function \(f\colon S \to \{0,1\}\) is exactly the characteristic function of the subset \(A_f\).
So for **any** function \(f\) in \(T\), there is a subset \(A_f\in \mathcal{P}(S)\) whose characteristic function is \(f\).
