It is known that the statement "Every set admits a total order" is independent of ZF. See here, for example. However, can it be proven in ZF that for every set $Y$, there exists a totally ordered set $X$ and a surjective map $X\to Y$?

  • 1
    $\begingroup$ It's definitely not provable in ZFA (since it's not provable there that "every linearly ordered set is well orderable" implies AC). $\endgroup$ May 15 at 13:48
  • 2
    $\begingroup$ This is indeed not provable in ZF, but it's tricky to show. I'll add an answer when I have time later today if nobody else beats me to it coughAsafcough! $\endgroup$ May 15 at 16:38
  • $\begingroup$ @Noah: I saw, I know what the answer is, but I am moving house today and everything is frantic. So by all means, take that one at your leisure... :) $\endgroup$
    – Asaf Karagila
    May 15 at 18:04
  • $\begingroup$ While we await the knowledge drop from Noah, also note that this would imply the partition principle implies every set can be linearly ordered, which is open $\endgroup$ May 15 at 19:26


You must log in to answer this question.

Browse other questions tagged .