## § Cantor Schroder Bernstein via fixpoint

- Given two injections $f: S \to T$, $g: T \to S$, we want to create a bijection.
- Suppose we have $S = T = N$, and $f(n) = g(n) = n + 1$.
- If $f$ were surjective, we are done, for then $f$ is the bijection.
- In this case, $f$ is not surjective, because $T-f(S) = {0}$. So $0$ has no preimage under $f$.
- We will create a new function $f'$ by perturbing $f$, such that it does map some element in $X$ to $0$ [which is currently missed ].
- Start with $f' \equiv f$. This means that $f'$ misses $0$.
- We can "force" a pre-image for $0$. How? Consider $g(0) = 1$, and set $f'(g(0)) \equiv 0$, or $f'(1) \equiv 0$.
- Whoops, but we have now "lost" a preimage for $f(1) = 2$, as now $2$ is not in the image of $f'$.
- Let's repeat the same process and fix it the same way. $f'(g(2)) \equiv 2$, or $f'(3) \equiv 2$.
- Now we have lost a pre-image for $f(3)$. Well, we just repeat the construction. For how long?
- Well, this is where we invoke the glory of a fixpoint theorem!
- See that we definitely need to reverse the arrows for $(T-f(S))$. If we start with a set $Y \subseteq T$ that we will reverse the arrows to, we will then need to reverse the arrows for $Y \cup F(G(Y))$.
- Thus, the set that we need to fiddle in $f'$ is $Y \mapsto (T-f(S))\cup F(G(Y))$.