Math Problem Statement
how to prove that g is bijective
how to prove that g is bijective
Solution
To prove that the function defined by
\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$$).Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Set Theory
Denumerability
Functions
Injectivity
Surjectivity
Bijection
Formulas
g(n) = { x, if n = 1; f(n - 1), if n ≠ 1 }
Theorems
Theorem 7.14 (Denumerability of A ∪ {x})
Suitable Grade Level
Undergraduate Mathematics
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