Is there a poset which has an element which does not have immediate succesor and is not maximal as well?

Question

Here's my question: If $A$ is a partially ordered set and suppose some element $a$ of $A$ does not have any immediate successor then is $a$ maximal in $A$? The converse is obviously true because if $a$ were maximal then $a$ cannot have a immediate successor because that would contradict the definition. Now, if $A$ was well ordered and if I assume that $a$ was not maximal then the set $\{ x\in A : a<x \}$ would be nonempty and so the least element of this set would be the immediate successor. What about in the case where $A$ is not so special? I tried to think about it but could not come up with proof or a counterexample.

Answer 1

Let $A = (-1,1] \cup (2,3)$ in the order inherited from $\Bbb R$ (so a linear order) has element $1$ that is not maximal (e.g. $2\frac12 \in A$ is larger) but has no immediate successor.

Also in $\Bbb R$ and $\Bbb Q$ all elements are not maximal and none of them have immediate successors.

Answer 2

Order natural numbers so that 1 is less than any other number, and the rest is reversed:

1 < ...... < 7 < 6 < 5 < 4 < 3 < 2.

Number 1 in this ordering satisfies your requrements - it's not maximal but it has no immediate successor.