Let $A$ be a partially ordered set such that every chain (total ordered subset) of A has a supremum in A; assume that A has a least element p. Show that there exists an element $m ∈ A$ such that $m$ has no immediate successor.
In order to show this, we will suppose that every element $x ∈ A$ has an immediate successor; this assumption will lead to a contradiction.
If every element of $A$ has an immediate successor, then we can define a function$ f : A → A $ such that for each $ x ∈ A, f (x) $ is an immediate successor of $x$. Indeed, let $T$ be the set of all the immediate successors of $x$; by the Axiom of Choice, there exists a choice function $g$ such that $g(T ) ∈ T $ . We define $f$ by letting $f(x) = g(T );$ clearly, $f(x)$ is an immediate successor of $x$.
To show that an element has no immediate successor, they are using contradiction and finally showed all elements has immediate successors how is the proof related to the proposition?