Infinite totally ordered set with all non-empty subsets containing both endpoints? I'm wondering if there are any infinite sets which can be totally ordered in such a way that all non-empty subsets contain a maximal and minimal element.

It is clear to me that this is impossible in discrete orderings and in the standard bounded dense orderings like $[0,1]$, but is it known that this is impossible in general?
This is indeed impossible. This is because any linear order $L$ has either an infinite increasing sequence or an infinite decreasing sequence (or both), and either provides a counterexample.



Technically, the observation above requires choice (think about a Dedekind-finite infinite set of reals). However, we do not in fact need choice for this problem:

Suppose $L$ is a linear order. Let 

$$D=\{x\in L: \text{ there are finitely many points } &lt;x\}$$

If $D$ has a maximal element $y$, then $\{z\in L: y&lt;z\}$ is infinite (since $L$ is infinite) and has no least element, so $D$ can't have a maximal element. If $D$ is nonempty, then $D$ is a counterexample to the desired property.

So $D=\emptyset$. But then $L$ doesn't have a minimal element, so $L$ itself is a counterexample.

Concisely:


  Suppose $L$ is a linear order. If $L$ has a least element, then the set of elements with finitely many smaller elements has no maximum. 


Similarly, if $L$ has a greatest element, then the set of elements with finitely many bigger elements has no minimum.



Incidentally, going back to the observation at the top of the answer: while this isn't the best way to solve this problem since it requires choice, it leads to a really interesting question. Namely, the set $\{\omega, \omega^*\}$ (where $L^*$ denotes the reverse of $L$) is a basis for the infinite linear orders: a linear order is infinite iff it contains a copy of $\omega$ or of $\omega^*$. Moreover, this is clearly the smallest basis for the infinite linear orderings. So we can ask:


  What about, say, uncountable linear orderings?


This turns out to be a very deep problem in set theory; the biggest positive result is due to Justin Moore:


  Suppose the Proper Forcing Axiom (PFA) holds. Then $\{\omega_1,\omega_1^*,S, C,C^*\}$ is a basis for the uncountable linear orders whenever $C$ is an arbitrary Countryman line and $S$ is an $\aleph_1$-dense set of real numbers.


(Here a Countryman line is a linear order of cardinality $\aleph_1$ whose square is a countable union of chains; the existence of such an object is not immediate, but provable in ZFC alone.)

Interestingly, the choice of Countryman line is meaningful - different Countryman lines can be non-isomorphic - but the choice of $S$ turns out not to be: assuming PFA, any two $\aleph_1$-dense sets of reals without endpoints are isomorphic (this follows from work of Baumgartner, I'm not sure where it was first explicitly stated).
Prove that a partial order with a top and bottom, for which
every not empty subset contains the top and the bottom, has
exactly one point.