The Tarski-Lindenbaum theorem of the middle value In Ettore Casari, La Matematica della Verita' (2006), the following theorem (attributed to Tarski and Lindenbaum 1926), is stated, which Casari calls the theorem of the middle value of Tarski-Lindenbaum (MV theorem):

If $A \subseteq B \subseteq C$ and $A' \subseteq C'$ and $f : A \to A'$, $g : C \to C'$ are bijections, then there is a bijection $h : B \to B'$, for some $B'$ such that $A' \subseteq B' \subseteq C'$.

The MV theorem is stated without proof after proving the Knaster-Tarski
theorem theorem (KT) on monotone functions and the Cantor-Schroeder-Bernstein theorem (CSB).
The author claims that the MV theorem follows from what has been discussed before, presumably indicating that the MV theorem follows from either the KT theorem or from the CSB theorem. I have two questions:

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*Does the MV theorem follow from the KT theorem or from the CBS theorem?

*What is the significance of the MV theorem?

 A: Not a very satisfactory answer, but some considerations to the proof of the MV theorem:

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*One might think that analogously to the proof of CBS (see e.g. Joel David Hamkins answer how to use KT for that), one could define for some tricky set $X\subseteq A$:
$$h(x)=\begin{cases}f(x)&\text{if $x\in X$,}\\g(x)&\text{if $x\in B\setminus X$.}\end{cases}$$
This cannot work if $g(B\setminus X)\subseteq A'$ and $B$ has a strictly larger cardinality than $A$. This obstacle can at least happen if $B$ is finite, so it is strange that one has to special-case at least the finite case (if this approach works at all) which is not necessary for CSB.


*If $B$ is infinite and has a strictly larger cardinality than $A$, and one wants to proceed somewhat along the lines of 1, possibly with a different definition of $h$ on $B\setminus X$, it seems that one should at least apply the subtraction theorem which states that $B\setminus A$ has then the same cardinality as $B$ to exclude that the obstacle from 1 won't happen in this case. Unfortunately, as I learnt recently, this subtraction theorem is in ZF equivalent to the axiom of choice.


*If we do assume AC (or at least the existence of a well-order on $B$), one can define $h$ by transfinite induction, but this proof has of course no relation with KT or CSB (except that the historically earliest proofs of CSB were also based on transfinite induction, using AC.)
