Maximal Element in Toset is Unique and Greatest

Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $M$ be a maximal element of $\left({S, \preceq}\right)$.

Then:
 * $(1): \quad M$ is the greatest element of $\left({S, \preceq}\right)$.
 * $(2): \quad M$ is the only maximal element of $\left({S, \preceq}\right)$.

Proof
By definition of maximal element:
 * $\forall y \in S: M \preceq y \implies M = y$

As $\left({S, \preceq}\right)$ is a totally ordered set, by definition $\preceq$ is a connected.

That is:
 * $\forall x, y \in S: y \preceq x \lor x \preceq y$

It follows that:
 * $\forall y \in S: M = y \lor y \preceq M$

But as $M = y \implies y \preceq M$ by definition of $\preceq$, it follows that:
 * $\forall y \in S: y \preceq M$

which is precisely the definition of greatest element.

Hence $(1)$ holds.

Suppose $M_1$ and $M_2$ are both maximal elements of $S$.

By $(1)$ it follows that both are greatest elements.

It follows from Greatest Element is Unique that $M_1 = M_2$.

That is, $(2)$ holds.

Also see

 * Minimal Element in Toset is Unique and Smallest


 * Maximal Element need not be Greatest Element