Can the Knaster-Tarski theorem be proved using the Schroeder-Bernstein theorem?

Question

The reverse can be done easily and the proof is well known I am wondering if the exact same argument can be used to prove reverse as well.

Answer 1

The Cantor-Schroeder-Bernstein theorem admits many proofs of various natures, which have been extended in diverse mathematical contexts to show that the phenomenon holds in many other parts of mathematics. So the question of whether the CSB property holds is an interesting mathematical question in many mathematical contexts (and it is particularly interesting in contexts where it fails).

And the proof of CSB from the Knaster-Tarski theorem that you have in mind proceeds as follows: suppose that $f:A\to B$ and $g:B\to A$ are both injective. Define $\varphi:P(A)\to P(A)$ on the power set of $A$ by $\varphi(X)=A-g[B-f[X]]$. It is easy to see by applying the functions and taking complements (twice) that $X\subset Y\to \varphi(X)\subseteq \varphi(Y)$, and so $\varphi$ is a monotone (order-preserving) operation on the power set of $A$, a complete lattice. Thus, by KT there is a fixed point $\varphi(X)=X$. From this, it follows that the function $h=(f|X)\cup(g^{-1}|(A-X))$ is a bijection between $A$ and $B$, and I leave all the details as a fun exercise.

Meanwhile, the proof of KT itself has a very short direct proof, which it would seem difficult to improve upon by using CSB. Namely, if $\varphi:L\to L$ is a order-preserving function on a complete lattice $L$, then let $d=\wedge\{e \mathrel{|} \varphi(e)\leq e\}$. Note that $\varphi(e)\leq e$ implies $d\leq e$ which implies $\varphi(d)\leq \varphi(e)\leq e$ and so $\varphi(d)\leq d$ and consequently $\varphi(\varphi(d))\leq \varphi(d)$, and so $\varphi(d)$ is one of the $e$'s, and so $d\leq \varphi(d)$ and hence $d=\varphi(d)$, as desired.

Note that the function $\varphi$ arising in the proof of CSB from KT is not merely monotone, but also continuous, since if $X=\bigcup_i X_i$, then $f[X]=\bigcup_i f[X_i]$ and so $B-f[X]=\bigcap_i B-f[X_i]$ and thus $g[B-f[X]]=\bigcap_i g[B-f[X_i]]$, because $g$ is injective, and hence $\varphi(X)=A-g[B-f[X]]=A-\bigcap_i g[B-f[X_i]]=\bigcup_i (A-g[B-f[X_i]])=\bigcup_i \varphi(X_i)$. In short, $\varphi(\bigcup_i X_i)=\bigcup_i\varphi(X_i)$, which means that $\varphi$ is continuous.

Thus, the standard argument shows that CSB follows not only from KT, but from the special case of continuous-KT, that is, KT restricted to continuous monotone functions. But the full KT is true for arbitrary monotone $\varphi$, even when they are not continuous, by the simple argument above. I take this as suggesting that the ``exact same argument,'' as you asked in your question, does not establish KT from CSB. Not every monotone $\varphi$ arises as the $\varphi$ used in the proof, since not every monotone map is continuous.

Meanwhile, I also observe that the direct proof of KT is so simple, it would seem difficult to find a substantially simpler proof of it by using any other principle, including CSB.