Proposition:
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.
Proof:
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 $x$.
Question:
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?
The proof is so confusing that it is preferable create ons's own proof.
Since every chain has an upper bound, A has a maximal element m by Zorn's lemma. As m is maximal, it cannot have a successor.
A least element is not needed; only the existence of an element.
To avoid the use of Zorn's lemma and AxC:
Assume every element x has an immediate successor f(x).
Pick an element a.
Define by transfinite induction
f(0) = a. f(k + 1) = f(f(k)) and
when k is a limit ordinal, f(k) = sup { f(x) : x < k }.
By Hartog's lemma, there is an ordinal exceeding the cardinality of A.
So the domain of f is limited to an initial segment of ordinals,
{ x ordinal : x < z } for some ordinal z.
If z is a successor ordinal, then there is s with z = s + 1.
f(f(s)) = f(s + 1) = f(z), a contradiction.
If z is limit ordinal, then f(z) = sup { f(x) : x < z },
another contradiction.
Consequently, somebody cannot have a successor.