TITLE: Are all sets totally ordered ?
QUESTION [32 upvotes]: The question is the title. 
Working in ZF, is it true that: for every nonempty set X, there exists a total order on X ? 
If it is false, do we have an example of a nonempty set that has no total order?
Thanks

REPLY [2 votes]: I'm just a student, so the deep logical principles behind François G. Dorais's answer are a drop over my head. However, I was able to find another Ultrafilter Lemma/Boolean Prime Ideal Theorem equivalent that seems a natural fit to the problem: the restricted Tukey-Teichmüller Theorem. Specifically, let $\mathscr A$ be the set of all relations on $X$ whose transitive closures are strict (partial) orderings. It's not hard to show that $\mathscr A$ satisfies the premises of rTT, where $(x,y)'$ is defined as $(y,x)$ for each $(x,y) \in S \times S$. Then rTT effectively shows that some element of $\mathscr A$ is connected, and thus its transitive closure is a strict total ordering of $S$ (in fact it is itself transitive, but this is not important). I've written up a more detailed outline on ProofWiki.