TITLE: Poset with atoms that isn't atomic
QUESTION [5 upvotes]: An atomic set is defined as:
A partially ordered set with a least element 0 is atomic if every
element b > 0 has an atom a below it, that is, there is some a such 
that b ≥ a :> 0.

Isn't this equal to saying: "a poset has atoms"? If a poset has atoms, isn't it automatically atomic? Of course the same applies to coatoms, and coatomic.

REPLY [5 votes]: The answer is "no".  It is possible that a poset has some atoms, and also some elements that do not have atoms below them.
For example, consider the ground set $\mathbb{Q}^{\ge 0}\cup \{i\}$, with the usual order on $\mathbb{Q}^{\ge 0}$, and the extra relation $i>0$ and no others (i.e. $i$ is incomparable to all nonzero elements of $\mathbb{Q}^{\ge 0}$).  Now, $i$ is an atom, but there are no other atoms, and the poset is not atomic.

A few extra thoughts:
The example above satisfies the "has atoms" criterion but not the "atomic"  criterion.  It is also possible to go the other way.  Consider the poset containing only $0$ and nothing else.  It satisfies the "atomic" criterion vacuously, but does not satisfy the "has atoms" criterion.