Can anyone please help me in this question:
$(X,\leq )$ is a well-ordered set. $\forall x\in X$, either $x$ is the largest element of $X$ or there exists $y\in X$ such that
(i) $x<y$ and
(ii) if $x_0\in X$ such that $x\leq x_0\leq y$, then either $x = x_0$ or $x_0 = y$.
Will this be sufficient to start the proof:
If $x$ is the largest element then there is nothing to prove. Suppose $x$ is not the largest element. $$A=\{ z\in X \mid x<z\}\neq \emptyset$$ How do I proceed?
On
You are almost done.
Suppose $x$ is not the largest element of $X$.
Then $Z = \{z \in X : x < z\}$ is nonempty. Since $(X, <)$ is a well-ordering, every nonempty set has a least element. Let $y \in Z$ be least.
Then $x < y$. Now suppose there exists $x_0$ such that $x < x_0 < y$. Then $x_0 \in Z$. But $x_0 < y$ implies that $y$ was not the least element of $Z$. Contradiction.
You've done all of the necessary creative wok already. All that is left is to apply the definition of "well-founded" and note that this gives you an $y$ with all of the properties you need for concluding (i) and (ii).