## Cantor Schroeder

Define ancestors as follows: Consider $a \in A$, if $g(B)$ then we call $g^{-1}(a)$ the first ancestor of $a$. If $g^{-1}(a) \in f(A)$ then we call $f^{-1}(g^{-1}(a))$ the second ancestor of $a$ and so on. Show that this divides $A$ into three disjoint subsets: $A_{\infty}, A_{e}, A_o$ where $A_{\infty}$ consists of the elements having an infinite number of ancestors. $A_o$ consist of elements having odd number of ancestor and $A_e$ consists of elements having an even number of ancestors.

So fixing an element of $a \in A$ we can find no ancestors, some ancestors, or an infinite number ancestors. If we fix $b \in B$ we can do the same. If $a \in A_{\infty}$ then $a \notin A_e$ and $a \notin A_o$ and so on.

And $f(A_{\infty}) = B$? How about $f(A_e)$?

Now define $F(a) = \begin{cases} f(a) \ \ \ \ a \in A_{\infty} \cup A_{e} \\ g^{-1}(a) \ \ \ \ a \in A_o \end{cases}$

To show $F$ is a one to one correspondence between $A$ and $B$ we just check $F(a) = F(a') \implies a = a'$?