Every partial order with well ordered chains is well founded is equivalent to the axiom of choice

Question

The theorem that every partial order with well ordered chains must be well founded can be proved with the Hausdorff Maximal Principle which is equivalent to the axiom of choice. Does this theorem imply the axiom of choice? The proof I have thought was that given a set of sets $A$ we define on it's union a partial order where $a< b$ if a is an element of the transitive closure of b. This can be verified to be a partial order since transitivity follows from the definition of transitive closure and irreflexivity follows from the foundation axiom. We also have that it's chains are also well ordered because of the foundation axiom. By the theorem for each $b\in A$ it will be a subset of its union therefore we can find a minimal element allowing us to choose an element for each b. Is this true and is the proof correct?

Answer 1

Every partial order has well-ordered chains, for example, singletons, or the empty chain.

Presumably you're asking about maximal well-ordered chains. And the answer to that is indeed positive.

It is easy to prove Zorn's lemma from this assumption: Assume that every partial order has maximal well-ordered chains, and let $(P,\leq)$ be a partial order in which every chain has an upper bound. If $C$ is a maximal well-ordered chain, let $x$ be an upper bound for it; if $x\notin C$, show that $C\cup\{x\}$ is still a well-ordered chain, so this is impossible, and therefore $x\in C$. From this follows that $x$ must be maximal, otherwise there would be an upper bound of $C$ which is not in $C$, which we established is impossible.

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If what you're suggesting is "Every partial order whose chains are all well-ordered is well-founded", this is not enough to prove the axiom of choice.

To see that, first we need to prove another auxiliary choice principle: Dependent Choice, which is weaker than the axiom of choice. Dependent Choice can be stated as the principle: If $(P,\leq)$ is a partial order such that there is no infinite decreasing sequence, then $P$ is well-founded.

Now, assume Dependent Choice holds. If a partial order is not well-founded, then it has a decreasing sequence, which is a chain that is by definition not well-ordered, and therefore not all chains are well-ordered.

In the other direction, suppose that every partial order whose chains are well-ordered is well-founded, and let $(P,\leq)$ be a partial order without any decreasing sequences. Then consider $(Q,\prec)$ where $Q$ is the set of all decreasing sequences in $P$, ordered by reverse end-extensions (i.e., $\vec b\prec\vec a$ if $\vec a$ is an initial segment of $\vec b$).

It is not hard to verify that every chain in $Q$ is well-ordered, and therefore $Q$ is well-founded. Now suppose that $A$ is any subset of $P$, look at all the decreasing sequences in $A$, this is a subset of $Q$, so it has a minimal element, which is a decreasing sequence in $A$ that cannot be extended any further inside $A$, but those are finite sequences, and therefore this minimal sequence has an element which is minimal in $A$.