Suppose I have a poset $(P,\leq)$, and am trying to prove that it is complete. If, for a general subset $S \subseteq P$, I've come up with a candidate $x$ for its supremum and am trying to prove it is such, am I correct in saying that it is (in general) invalid to suppose for contradiction that there as another upper bound $y$ of $S$ such that $ y < x$?
If $P$ is not a total order (or, more precisely, if the set of upper bounds of $S$ is not a chain), it may be possible to prove that $x$ is the least element from the subset of upper bounds of $S$ which are comparable with $x$ whilst there are other upper bounds not comparable with $x$ that satisfy the same criteria. Is there an example of this? I don't really care if there isn't an obvious one, I'd ultimately like verification that my reasoning here is correct.
This thought occured to me when I was trying to prove a certain poset was complete, and I proceeded in the fashion that I was used to in real analysis (where $\mathbb R$ with its usual order is obviously a total order).