Let X be well-ordered. Let S be a bounded subset of X. Then clearly by well-ordering there exists a sup of S (the set of upper bounds of S is a subset of X and so by well-ordering it has a minimal element). I feel like the sup must belong to S itself (it will if X is the naturals) but I can neither think of a proof nor a counterexample.
Thanks!
No, the supremum need not be included in $S$. Consider any infinite well-order that isn't $\omega$ - a good example is $\omega + \omega$, a copy of the naturals with another copy of the naturals tacked onto the end. This is a well-ordered set. But if we let $S$ be just that first copy of the naturals, $S$ clearly has no greatest element - that would be a greatest natural number!