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**eskimo343** Suppose $\displaystyle C$ is a well orderable set, $\displaystyle f: C \times C \rightarrow C$, $\displaystyle A \subseteq C$, and let $\displaystyle A_f =_{df} \cap \{ X \subseteq C | A \subseteq X \text{ }\& \text{ }f[X \times X] \subseteq X \}$ be the closure of $\displaystyle A$ under $\displaystyle f$. Define the sets$\displaystyle \{ A_n \}_{n \in \mathbb{N}}$ by the recursion $\displaystyle A_0=A$, $\displaystyle A_{n+1}=A_n \cup f[A_n \times A_n]$. Then we have that $\displaystyle A_f=\cup_{n \in \mathbb{N}} A_n$.

Show that if A is infinite, then $\displaystyle A_f =_c A$.

I am having trouble proving this part now [that if $\displaystyle A$ is infinite, then $\displaystyle A_f =_c A$]. I know that by the first part we have that $\displaystyle A_f=\cup_{n \in \mathbb{N}} A_n$. However, I do not see how to show that if $\displaystyle A$ is infinite, then $\displaystyle A_f =_c A$. I need help on this. Thanks.