Show that every ordered set with the well ordering has the least upper bound property

Question

Here is a proof attempt:

Let $S_a =\{x\in A:x \leq a\}$ (also known as a section of $A$). We firstly prove that $\forall a \in A$, $S_a$ has a supremum in $A$.

Clearly, every $S_a$ is bounded above by $a$. We consider $T =A \setminus S_a$. Now, $T$ being non empty [unless $a = \max\{A\}$], it has a minimal element $m$.

Now, $a$ is an element of $S_a$ with these properties:

1. $a$ is an upper bound of $S_a$ and $a \in S_a$ (not strict partial order relation).
2. Any element less than $a$ is not an upper bound of $S_a$.
3. $a\leq m$ and $m\notin S_a$.

Clearly, $\sup S_a =a$.

We generalise this to the case of any bounded above set $K$. Let $\max K =M$. The section corresponding to the element $M$, i.e. $S_M$ contains $K$. So, $K \subset S_M$ and $\max S_M = \max K= M$. Now, $M$ is such an element that:

1. $M$ is an upper bound of both $S_M$ and $K$.

2. Any element greater than $M$ is not in the two sets.

3. No element less than $M$ can be an upper bound of either of the sets.

So, $\sup S_M = \sup K = M$

I don't know if this is a valid proof. A thorough verification would be of immense help.

Thank You.

Answer 1

Let $S$ be a well ordered set.
If $A$ is a not empty subset of $S$ with an upper bound $b$.
Then $B$, the set of upper bounds of $A$ is not empty.
So $B$ has a least element $d$.
If $d$ is a successor, then the predecessor is the maximum element of $A$.
If $d$ is not a successor, then $d$ is the supremum of A.