# How to understand the proof of the below statement similar to Zorn's lemma?

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?

• This "proof" is obviously just the start of a proof and not a complete proof of the proposition. Are you sure that's the whole proof you have? – Eric Wofsey Apr 27 at 19:02
• @EricWofsey, I am reading the book "A Book of Set Theory" by Charles C. Pinter. The proof I mentioned above is there, not more than that. If you know the complete proof, please post it. – krishna Apr 27 at 19:16
• It's not the whole proof: the definitions and lemmas that are after the last sentence you posted are parts of proof too. – mihaild Apr 27 at 19:42
• They did not show all elements have an immediate successor. That was as assumption, which was proven wrong. – William Elliot Apr 27 at 23:51

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 },
• If by "AxC" you mean the axiom of choice, you have not avoided it. How is $f(x)$ defined if $f$ has two or more immediate successors? – bof Apr 28 at 5:55