least-upper-bound and greatest-lower-bound properties

1. For a partial-ordered set, I was wondering if the least-upper-bound/greatest-lower-bound property means that any nonempty subset that has an upper/lower bound has a least-upper/greatest-lower bound, or any subset that has an upper/lower bound has a least-upper/greatest-lower bound? Is least-upper-bound property also called Dedekind completeness?
2. Why is the following statement true:

A partial ordered set has the least upper bound property if and only if it has the greatest lower bound property.

Thanks and regards!

"Least upper bound" property is that every nonempty set that is bounded above has a least upper bound; dually for "greatest lower bound", so it is only required that nonempty sets have the property.

(For example, the real numbers have the least upper bound property; if you also required the empty set to have a least upper bound, this would require the reals to have a least element).

Yes, Dedekind completeness is the same thing as the least upper bound property.

For 2: If $S$ is a nonempty set that is bounded below, let $B$ be the set of lower bounds of $S$. Show that $B$ is (i) nonempty; and (ii) bounded above. Conclude that $B$ has a least upper bound. Show that the least upper bound of $B$ is also the greatest lower bound of $S$. The converse is proven dually: the least upper bound of a nonempty set that is bounded above is equal to the greatest lower bound of the set of upper bounds.

• Thanks! (1) I can understand that every element in the set is an upper bound of the empty subset, so the least-upper bound of the empty subset is the least element of the set. I was wondering how to understand a lower bound of the empty subset and the greatest-lower bound of the empty subset? (2) is my part 2 still true if allowing both properties apply to any subset, empty or not? – Tim May 27 '11 at 5:28
• @Tim: Every element is a lower bound and upper bound for the empty set, because the statement that $y\leq x$ for all $x\in \varnothing$ (respectively, $x\leq y$) is vacuous because there are no $x\in \varnothing$. Thus, a greatest lower bound of $\varnothing$ (if one exists) is the greatest element of the entire set, i.e. the maximum element; similarly a least upper bound of $\varnothing$ will be a minimum element. – Zev Chonoles May 27 '11 at 5:32
• @Tim: the supremum of $X$ is the infimum of the set of upper bounds for $X$, when either of them exists. – Arturo Magidin May 27 '11 at 10:17

Arturo has answered your questions quite well; I would just like to mention that for a general poset, the property without the non-empty requirement is referred to as bounded completeness; the version with the non-empty requirement is called conditional completeness. Because a least upper bound for the empty set is necessarily a minimum element and a greatest lower bound for the empty set is a maximum element, that explains the claim in the article that

A conditionally complete lattice is either a complete lattice, or a complete lattice without its maximum element 1, its minimum element 0, or both.

• Thanks! (1) Is the property without the non-empty requirement referred to as completeness or bounded completeness? The article seems to call it as the latter. What do you mean by the former, i.e. "completeness"? (2) is it true that a partially ordered set is a complete lattice if and only if it is a conditionally complete lattice and has both the maximum and minimum elements? I am trying to understand your quote. – Tim May 27 '11 at 5:35
• @Tim: Apologies, you are correct. I've edited. And yes, your statement in (2) is correct; one direction is clear, and a conditionally complete lattice has a least upper bound and greatest lower bound for every non-empty subset, which combined with the minimum and maximum elements (l.u.b. and g.l.b. of $\varnothing$, respectively), means every subset has a l.u.b. and g.l.b., hence it is complete. – Zev Chonoles May 27 '11 at 5:41