Exercise in well-ordered sets Definitions: Given a subset $A$ of a non-empty well-ordered set $X$ we define the supremum as follows:  $\text{sup(A)}:=\text{min}\bigg(\big\{\,y\in X \oplus\{ +\infty\}: \forall x\in A \;(x\le y)\, \big \} \bigg)$. 
If $x\in X$ we define the successor of $x$ by the formula: $\text{succ}(x):= \text{min}\big( ( x, +\infty ]\big)$
Exercise: If $x$ is an element of a well-ordered set $X$ show that exactly one of the following statements hold: 
1.- (Limit Case) $x=\text{sup} \big( \,[\text{min}(X),x)\, \big)$
2.- (Successor case) $x=\text{succ}(y)$ for some $y$. 
Hi everyone I have trouble with the above exercise. I'd like to put my attempt but  everything I've tried is wrong (Basically my idea is split the proof in two cases one to show that at most one case hold assuming by contradiction the opposite, and the other  which shows that at least one hold ). Also I don't have a real intuition about what the limit case really is, I mean,  with the successor case is easy only thinking in the natural numbers. But what about the limit case. 
(This is what I have so far)
Sketch proof: We have to show that at most (1) or (2) hold. Suppose for the sake of contradiction that both cases occurs simultaneously. Let $P(x)$ be the statement "$x$ is such that (1) and (2)  holds at the same time". We set $\{x\in X: \,P(x)\,  \}$, then by hypothesis is a non-empty subset of $X$ which is well-ordered, then it has a minimum element $x$. Thus for all $x'< x$ either (1) or (2) holds but not both. 
We already know that $x= \text{min} ((y, +\infty])$, then $x$ is the least such that $x>y$. Thus $y$ is either in (1) or (2) but not at the same time. 
If $y$ is in (1). This means that $y= \text{sup} ([\text{min}(X),y)$, i.e., $y$ is the least such as $y\ge w$ for all $w\in [\text{min}(X),y)$. Also we have that $y\in [\text{min}(X),x)$. 
We claim that $[\text{min}(X),x)=[\text{min}(X),y) \cup \{y\}$. Let $z\in [\text{min}(X),x)$, i.e., $\text{min}(X)\le z<x$, we have to check the case when $y\le z< x $ if were some $z\not= y$ contradicts that $x$ is the successor of $y$, hence $z=y$, the case when $z<y$ is trivially true. Hence $[\text{min}(X),x)\subset [\text{min}(X),y) \cup \{y\}$ the other inclusion is trivially true. Therefore $y$ is indeed an upper bound for  $[\text{min}(X),x)$ and also is the least, i.e., $y = \text{sup} ( \,[\text{min}(X),x)\,)$ and $y\not= x$ a contradiction.
If $y$ is in (2), $y= \text{succ}(z)$. Then $y\ge w$ for all $w\in [\text{min}(X),y)= [\text{min}(X),z\,]$. Then $y$ is the least which would imply $y=\text{sup} ( \,[\text{min}(X),y)\, )$ and contradicts the minimality of $x$.
For the other case, I'm not sure of how prove that at least one always hold? Any suggestion, please? Thanks
 A: I will write $y^+$ to mean the successor of $x$ to save on typing.
Only one can be true
Suppose $x = y^+$. Then $y \in [\min X\,..\,x)$.
Let $z \in [\min X\,..\,x)$. Then $z<x = y^+$. Either $z\le y$ or $z>y$. Suppose $z>y$. Then $y<z<y^+$, a contradiction. So $z\le y$. As this holds for all such $z$, $y$ is an upper bound of $[\min X\,..\,x)$. As it is an element of that set, it is in fact the maximum, and therefore the supremum. Since a set can have only one supremum, and $y \ne x$, $x$ is not the supremum of $[\min X\,..\,x)$.
At least one is true
Let $s$ be the supremum of $[\min X\,..\,x)$. Either $s\in[\min X\,..\,x)$ or $s$ is strictly greater than each element of $[\min X\,..\,x)$. I think you can probably finish this argument.
A: First we'd like to show that at most one hold. Suppose for the sake of contradiction that both cases occurs simultaneously. We have $x= \text{succ} (y)= \text{sup} ([\text{min}(X),x))$. Let $z$ be an arbitrary element of $([\text{min}(X),x))$, since $y\in [\text{min}(X),x)$ and the set is total, either $z>y$ or $z\le y$. For the former $y<z<x$ which contradicts that $x=\text{succ} (y)$. For the latter, since $z$ was arbitrary $y\ge z$, $\forall z\in[\text{min}(X),x)$, i.e., is an upper bound, it is the least because $y\in [\text{min}(X),x)$. Since $y\not=x$ we have a contradiction.
To conclude the proof we shall show that at least on of the cases always happen. Let $x\in X$ be arbitrary. Then the set $[\text{min}(X),x)$ is bounded above. Let $\mathcal{S}$ be the set of all the upper bounds of $[\text{min}(X),x)$, clearly is non-empty and therefore it has a least element that we call $s$. Then $s= \text{sup}([\text{min}(X),x))$ and either lies in the set or not. 
If were the case that $s\in [\text{min}(X),x)$, then is the maximal element and then we claim $x=\text{succ}(s)$. If were not the case, let $z=\text{succ}(s)$. Now since $z<x$, we have $z\in [\text{min}(X),x)$ but  by construction $s<z$, a contradiction. Hence $x=\text{succ}(s)$. 
On the other hand, if we have that $s\notin [\text{min}(X),x)$, then we claim that $s= x$. If were not the case then either $s<x$ or $s>x$. If $s<x$, then $s\in [\text{min}(X),x) $ a contradiction. If $s>x$, we already know that for all $w\in [\text{min}(X),x) $, $w<x$ then is an upper bound which is least than the least upper bound, a contradiction. Hence $x= \text{sup}([\text{min}(X),x))$.
