Poset with atoms that isn't atomic

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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.

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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.