MathOverflow Asked by Alexander Pruss on September 6, 2020

Consider these two axioms:

- Every partial order extends to a linear order.
- Every partial preorder (reflexive and transitive relation) extends to a linear preorder while preserving strict orderings: i.e., whenever $x<y$ (namely: $xlesssim y$ but not $ylesssim x$) holds in the original order, it holds in the extended order.

**Question:** Does 2 imply 1 in ZF?

**Notes:**

A. In ZF, Boolean Prime Ideal implies 1 (but not conversely if ZF is consistent), and 1 implies 2 (take the quotient of the preordered set under $sim$, where $xsim y$ iff $xlesssim y$ and $ylesssim x$, and apply 1).

B. Also, 2+(every set has a linear order) implies 1. (Use 2 to extend the partial order $le$ to a total preorder $lesssim$ preserving strict orderings. We now need to turn $lesssim$ into a linear order. To do that, we just need to linearly order within each equivalence class under $sim$. Do that by taking a linear order on the whole set and using that to induce the order in each equivalence class—though not between them.)

C. Claim 2 implies Banach-Tarski and thus the existence of nonmeasurable sets (Pawlikowski’s proof of Banach-Tarski from Hahn-Banach can be adapted), and hence 2 is not provable in ZF (if ZF is consistent).

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