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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).


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up vote 2 down vote accepted

Let $P=\mathbb{N}\cup\{a,b\}$, where $a$ and $b$ are distinct elements not in $\mathbb{N}$. Give $\mathbb{N}$ the usual order $\le$, and set $n\le a$ and $n\le b$ for each $n\in\mathbb{N}$. Make $a$ and $b$ incomparable. Then $a$ and $b$ are both upper bounds for $\mathbb{N}$ with no smaller upper bound.

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A wonderfully $\mathbb N$atural example. Thanks! – Matt Jan 21 '12 at 18:01

Here is an easy example. Take two disjoint sets $A$ and $B$, each with at least two elements, and let the partial order have everybody in $A$ less than everybody in $B$, but no other order relations. In this case, every element of $B$ is an upper bound of $A$, and furthermore satisfies your criteria, since distinct elements of $B$ are incomparable. But $A$ will have no least upper bound, precisely because the upper bounds of $A$ are exactly the elements of $B$, and these are incomparable.

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