If i have an a well ordered set in which every chain admits an upper bound then the maximal element is unique

Question

It is clear that the Zorn lemma guarantees the existence. I prove that the minimal element is unique, and obviously the set is totally ordered. So because the uniqueness of the successor of all elements of the set, in same sense it seems reasonable to me that we will find the uniqueness of the maximal element, but i don't know how to prove it

Answer 1

I think that i solved it.

Thm: Let X be a set, Let $\leq$ an a order relation, Let assumed that:

1. Every chain of $(X,\leq)$ admit an upper bound
2. $(X,\leq)$ is totally ordered

Then exist an unique maximal element.

Proof: Because 1) the Zorn lemma guarantee us that exist maximal elements. Without loss the generality assuming that we have two maximal elements, $x_1,x_2$. Assuming by absurd that $x_1\neq x_2$, then because the total ordering must be $x_1 \leq x_2 \vee x_2 \leq x_1 $. In any case one of them can't be a maximal element, so se found an absurd so $x_1 = x_2$ $\square$

So we can note that

Lemma: If $(X,\leq)$ is well-ordered then is also totally ordered.

So at the end we can obtain what i ask like a corollary of these theorem.