Prove that for every function , there is a such that .
Hint: A fixed point theorem may be useful. We may us the Countable Principle of Choice, but not the Axiom of Choice.
Are you referring to Knaster–Tarski fixpoint theorem? Since it requires a monotonic function, maybe one can consider . Then is monotonic, so it has a fixpoint : , which implies . And the Countable Principle of Choice is used to show that is a complete lattice. I am not sure about this, but it may be a start...