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To prove that the function \( g: \mathbb{N} \to A \cup \{x\} \) defined by 

\[
g(n) = 
\begin{cases}
x, & \text{if } n = 1, \\
f(n-1), & \text{if } n \neq 1
\end{cases}
\]

is **bijective**, we need to show that it is both:

1. **Injective** (one-to-one): Each element in the domain \( \mathbb{N} \) maps to a unique element in the codomain \( A \cup \{x\} \).
2. **Surjective** (onto): Every element in the codomain \( A \cup \{x\} \) is mapped to by some element in the domain \( \mathbb{N} \).

### Step 1: Proving Injectivity

Let \( g(n_1) = g(n_2) \) for two arbitrary elements \( n_1, n_2 \in \mathbb{N} \). We need to show that \( n_1 = n_2 \).

- **Case 1: \( n_1 = 1 \) or \( n_2 = 1 \)**:
  If either \( n_1 = 1 \) or \( n_2 = 1 \), then \( g(n_1) = g(n_2) = x \). But the function \( g(n) = x \) only when \( n = 1 \), so if \( g(n_1) = g(n_2) = x \), we must have \( n_1 = n_2 = 1 \).
  
- **Case 2: \( n_1 \neq 1 \) and \( n_2 \neq 1 \)**:
  In this case, \( g(n_1) = f(n_1 - 1) \) and \( g(n_2) = f(n_2 - 1) \). Since \( g(n_1) = g(n_2) \), we have \( f(n_1 - 1) = f(n_2 - 1) \). But since \( f \) is injective (as given), we know that \( n_1 - 1 = n_2 - 1 \), which implies \( n_1 = n_2 \).

Thus, \( g \) is injective.

### Step 2: Proving Surjectivity

Let \( y \in A \cup \{x\} \). We need to show that there exists some \( n \in \mathbb{N} \) such that \( g(n) = y \).

- **Case 1: \( y = x \)**:
  By definition, \( g(1) = x \), so \( n = 1 \) satisfies \( g(n) = x \).
  
- **Case 2: \( y \in A \)**:
  Since \( f: \mathbb{N} \to A \) is surjective, there exists some \( m \in \mathbb{N} \) such that \( f(m) = y \). In this case, if \( n = m + 1 \), then \( g(n) = f(n - 1) = f(m) = y \).

Thus, \( g \) is surjective.

### Conclusion:

Since \( g \) is both injective and surjective, it is **bijective**.

---

Do you want further clarification on any steps? Here are five related questions you can explore:

1. How does the bijection between \( \mathbb{N} \) and \( \mathbb{N} \cup \{x\} \) relate to the cardinality of sets?
2. Can this proof be generalized to show that adding a finite number of elements to a countable set preserves countability?
3. What role does the injectivity of \( f \) play in the overall proof?
4. How would the proof change if we were working with an uncountable set instead of a countable set?
5. What would happen if \( x \in A \) in the proof? Would the function \( g \) still be bijective?

**Tip:** When proving injectivity, always check if the function behaves differently for special cases (e.g., specific values in the domain like \( n = 1 \)).

Learn how to prove that the function g: N → A ∪ {x}, where g(n) = { x if n=1, f(n-1) otherwise }, is bijective by showing it is injective and surjective. This problem covers key concepts in set theory and function properties.

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