I am trying to prove the following in Enderton's text 'Elements of Set Theory' (exercise 8 of chapter 4):
Let $f$ be a one-to-one function from $A$ into $A$, and assume that $c \in A - ran \ f$. Define $h: \omega \rightarrow A$ by recursion: $h(0) = c, h(n^{+}) = f(h(n))$. We show that $h$ is one-to-one.
So far, I have the following attempt but I am not sure that I am approaching this correctly.
Let $f$ be a one-to-one function from $A$ into $A$, and assume that $c \in A - ran \ f$. Define $h: \omega \rightarrow A$ by recursion: $h(0) = c, h(n^{+}) = f(h(n))$. We show that $h$ is one-to-one.
Let $S = \{n \in \omega \ |$ for every $m \in \omega$ different from $n$, $h(m) \neq h(n) \}$. We show that $S$ is inductive. As our base case, we have $h(0) \neq h(m)$ for any $m \neq 0$. This is because of the following. $h(0) = c$ and $\forall n \in \omega, (n = 0) \vee (\exists p \in \omega$ s.t. $n = p^{+}$) [Theorem 4C - p. 68]. Since for all $n^{+} \in \omega$, we have $h(n^{+}) = f(h(n)) = f(h(p^{+}))$, $c \notin ran \ f$ and so $ f(h(p^{+})) \neq c$.
Assuming $k \in S$, consider $h(k^{+}) = f(h(k))$. Since $k \in S$, we know that
$\forall n,k \in \omega, (n \neq k) \ \implies \ (h(k) \neq h(n))$. So we have $h(k^{+}) = f(h(k)) \neq f(h(n)) = h(n^{+})$, for some $n^{+} \in \omega$, because $h(k) \neq h(n)$ and the fact that $f$ is a one-to-one function. So for any $n^{+}, k^{+} \in \omega$, ($n^{+} \neq k^{+}) \implies (h(k^{+}) \neq h(n^{+}))$ and we have $k^{+} \in S$ making $S$ inductive.
An easier solution might be to use the Well Ordering Principle.
Define $A$ as the complement of $S$ as follows: $A=\{n\in \omega \mid \exists m \in w, n\ne m \wedge h(n)=h(m)\}$. To show that $h$ is one-one, it suffices to show that $A$ is empty.
As you have shown, $0\not \in A$. Assume $A$ is non-empty, then by the Well Ordering Principle, $A$ has a non-zero smallest element $m^+$. By definition of $A$, there is a non-zero $k^+\ne m^+$, such that $h(m^+) = h(k^+)$. Since $f$ is one-one, we get $h(m) = h(k)$ where $m\ne k$, hence $m \in A$. But this contradicts the minimality of $m^+$ and so $A$ is empty.