For sets $x, y$ we write $x\leq y$, if there is an injection $\iota: x \to y$, and we write $x \leq^\* y$ if either $x = \emptyset$ or there is a surjection $s: y \to x$. In ${\sf (ZF)}$ we have that $x \leq y$ implies $y \leq^\* y$.


Consider the following statements:



> 
> Partition principle (PP): For all sets $x, y$ we have that $x \leq^\* y$ implies $x\leq y$.
> 
> 
> 



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> Dual Cantor-Bernstein (CB)\*: For all sets $x,y$, if $x\leq^\* y$ and $y \leq^\* x$, then there is a bijection $\varphi: x\to y$.
> 
> 
> 


Via the "normal" [Cantor-Bernstein theorem](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem), which a theorem of ${\sf (ZF)}$, we can show that (PP) imples (CB)\* in ${\sf (ZF)}$.


It seems to be [open](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjR2u3Dk5H3AhWU8LsIHWS9C7IQFnoECAMQAQ&url=https%3A%2F%2Fprojecteuclid.org%2Fjournals%2Fnotre-dame-journal-of-formal-logic%2Fvolume-31%2Fissue-3%2FThe-dual-Cantor-Bernstein-theorem-and-the-partition-principle%2F10.1305%2Fndjfl%2F1093635502.full&usg=AOvVaw3lP42qtQ0HrR-MxttLfcuW) whether (CB)\* implies the [Axiom of Dependent Choice (DC)](https://en.wikipedia.org/wiki/Axiom_of_dependent_choice). Since (PP) is stronger than (CB)\*, this begs the question:


Does (PP) imply (DC) in ${\sf (ZF)}$?



Does the partition principle imply (DC)?
Yes. This is a combination of facts.


1. $\sf PP$ implies that if a set $X$ can be mapped onto an ordinal $\alpha$, then $\alpha$ injects into $X$. In other words, it implies that $\aleph^\*(X)=\aleph(X)$ for any set $X$.
2. $\sf AC\_{WO}$, that is the axiom of choice from families of sets indexed by an ordinal, is equivalent to the statement "For every $X$, $\aleph^\*(X)=\aleph(X)$".
3. $\sf AC\_{WO}$ implies $\sf DC$. This is due to the fact that if $T$ is a tree of height $\omega$ without cofinal branches, we can consider the various well-orderable subtrees of $T$, define a rank function on those and use that to define a rank function on $T$. Then, using $\sf AC\_{WO}$ we can show that it is impossible that $T$ itself is not truly well-founded, which would mean that it has a maximal element.


