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Consider these two axioms:

  1. Every partial order extends to a linear order.
  2. Every partial preorder (reflexive and transitive relation) extends to a linear preorder while preserving strict orderings: i.e., whenever $x<y$ (namely: $x\lesssim y$ but not $y\lesssim x$) 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 $x\sim y$ iff $x\lesssim y$ and $y\lesssim x$, 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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  • $\begingroup$ I have to check more carefully, but I am pretty sure that D is badly argued: it's not clear that $\preceq$ as I defined it is transitive. $\endgroup$ Sep 4 '20 at 14:16

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