I have to prove this: Show that in a well-ordered set, every element except the largest(if one exists) has an immediate successor.
Here is what I have: Let A be a well-ordered set.
let a belong to the set A.
Consider the set S = {b belongs to set A |a < b}
We want to use the fact that every nonempty subset of a well-ordered set has the smallest element.
If a is not the largest element of the set, then (a,b) = {x | a < x < b) = nonempty set, and has the smallest element b. Then b is the immediate successor of a if a<b and if there is no x such that a < x < b.
Therefore, in a well-ordered set, every element except the largest (if one exists) has an immediate successor.
I am correct?
No, that does not work: you pulled $b$ out of thin air. At the very least you need to say that if $a$ is not the largest element of $A$, then there is some $b\in A$ such that $a<b$; then you can consider the interval $(a,b)$. However, that interval may be empty, in which case it does not have a smallest element. And even if it is non-empty and therefore does have one, you cannot call that element $b$: you’ve already used that name for something else.
But in fact there’s no reason to look at such an interval in the first place. The point is that if $a$ is not the largest element of $A$, then $S\ne\varnothing$, and therefore $S$ has a least element $s$. Now show that $s$ is an immediate successor of $a$ by showing that there is no $b$ such that $a<b<s$.