supremum of directed set is contained or not? I have seen definitions which state that a directed set contains its supremum, for example: 
http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Directed_set.html "for any a and b in A there must exist a c in A with a ≤ c and b ≤ c." But I have also seen sets such as $[o,a)$ used as examples of directed sets that do not contain their supremum. My sense is that finite directed sets must, and non-finite ones may or may not. What is the truth of the matter?
 A: I think I have found the answer. 
First, a finite directed set must contain the upper bound of any pair of its elements as per the standard definitions such as, "A directed set is a partially ordered set $A$,  $(A,\leq)$ such that whenever $a,b\in A$ there is an $x\in A$ such that $a\leq x$ and $b\leq x$." http://planetmath.org/directedset
Then if the upper bound of any pair of elements of a finite set is contained in the set, it is either the least upper bound of the pair or the least upper bound must be less than it but still $\geq$ than the pair, and hence also contained in the set. If the finite directed set is a chain it is clear that the least upper bound of the chain must then be contained in the set. If it is not a chain I am not certain, but I did find this: "It is well known that, for a partial order, having a least upper bound for each of its directed subsets of cardinality $\omega$, and having a least upper bound for each of its chains of cardinality $\omega$ are equivalent notions. (http://eprints.soton.ac.uk/261872/1/w-ind-compl.pdfI)"
However, it seems this is not actually the full form of the definition. I have found two references that state that a set is directed if it contains an upper bound for any of its finite subsets. 
"A non-empty subset D of a partial order P is directed if any pair of elements in D has an upper bound in D. Equivalently, D is directed if it contains an upper bound of any of its finite subsets." http://eprints.soton.ac.uk/261872/1/w-ind-compl.pdf
"A subset X of an ordered set L is directed if every finite subset of X has an upper bound in X. In particular, the void subset of a directed set has an upper bound, so directed sets are nonvoid. An ordered set L is upwards complete if every directed $X\subseteq L$ has a supremum in L." http://arxiv.org/pdf/0810.4705.pdf
Thus in the case of an infinite set every finite subset could contain a least upper bound and this would suffice to make the infinite set directed. However, since there is no greatest finite subset of an infinite set that does not contain its upper limit, such as $[0,a)$, the infinite set as a whole, although directed by virtue of all of its finite subsets being directed, does not contain its least upper bound.
