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?

  • 2
    $\begingroup$ 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? $\endgroup$ – Eric Wofsey Apr 27 at 19:02
  • $\begingroup$ @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. $\endgroup$ – krishna Apr 27 at 19:16
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    $\begingroup$ It's not the whole proof: the definitions and lemmas that are after the last sentence you posted are parts of proof too. $\endgroup$ – mihaild Apr 27 at 19:42
  • $\begingroup$ They did not show all elements have an immediate successor. That was as assumption, which was proven wrong. $\endgroup$ – 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 },
another contradiction.

Consequently, somebody cannot have a successor.

  • $\begingroup$ 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? $\endgroup$ – bof Apr 28 at 5:55

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