Maximal element by vacuous truth Does the following argument hold in trying to find a poset in which every element is a maximal element. The set has to have at least two elements.
Define a relation $S$ on the set $P=\{1, 2\}$ by $S=\{(1, 1), (2, 2)\}$. This relation is a partial order relation as it is reflexive, anti-symmetric and transitive. Because an element $a\in P$ is said to be a maximal element if for all $b\in P$ we have that $a\preceq_S b\implies a=b$. For two distinct elements $a.b\in P$ the statement $a\preceq_S b\implies a=b$ is vacuously satisfied as $(a, b)\notin S$ and thus aren't comparable. This means that both $1,2$ are maximal elements of $P$.
Is this reasoning correct?
Edit: I don't want the solution to the problem, I would like to figure it out myself. This is just something I thought of when I was attempting the problem.
 A: Your solution works perfectly, and your reasoning is fine. Indeed, it is in a sense the only solution.
If every element in the poset $(P, \leq)$ is maximal, and you have two comparable elements $a$ and $b$, then we have both $a\leq b$ and $b\leq a$, so we must have $a=b$ by antisymmetry. This shows that distinct elements cannot be comparable, and $\leq$ is actually $=$.
A: You defined the relation $S$ as being exactly the equality relation $=$. Then with your reasoning, you have shown that $S$ was a partial order, and that with this partial order, every element of $P$ was a maximal element.
This reasoning is correct, and in fact works for any set, not just $P$.
A natural question that follows is: is the converse true? Do there exist other partial orders for which every element is a maximal element, or is the equality the only partial order with that propriety?
The answer is: yes, the converse is true. If a partial order $S$ on a set $P$ satisfies the property "every element is maximal", then $S$ is the equality relation. Since you say that you don't want the solution, I won't include the proof (although the other answer did include it).
