# Why is the Axiom of Well-ordered Choice not strong enough to prove Zorn's Lemma?

This is based on this question: How strong is the axiom of well-ordered choice?

The "axiom of well-ordered choice" says that any transfinitely-indexed family of sets has a choice function. The individual sets within the family need not be well-ordered, but the family must be. We also assume the family is set-sized, so that the indices only go up to some ordinal.

I was going to give a direct example of where I'm stuck with Zorn's lemma, but it seems like it's easier to go with the Well-Ordering Principle (which indirectly uses Zorn's lemma anyway).

The usual proof sketch is, you begin with some set $$S$$ and select an initial element $$e_0$$. Then, via transfinite recursion you select an element $$e_\alpha$$ for any ordinal $$\alpha$$ from the set $$S \setminus \{e_n: n < \alpha\}$$. This process must "run out" of elements to select at some point, or else $$S$$ would have as many elements as ordinals, which it can't because it's a set. The resulting sequence of $$e_n$$'s is a well-ordering on $$S$$.

Of course, one cannot simply "select" elements from this family of sets without something like the axiom of choice. With AC, we can define a choice function on the power set $$P(S) \setminus \{\}$$, so that we know that we are always able to choose an element. This can be thought of as a choice function on a "partially ordered family of sets."

However, we don't really need a choice function on all of $$P(S) \setminus \{\}$$ for this. We only need a choice function on a particular transfinitely-indexed family of subsets of $$S$$ - each of which contains one less element than the last, or is equal to the intersection of all previous subsets at a limit ordinal. So rather than a partially ordered family of sets, we get a well-ordered family of sets. Since this is transfinitely-indexed, why can't we use the "well-ordered AC" here?

The same basic question applies to Zorn's lemma above, although I thought the above example was clearer. You begin with an initial element in some poset which has upper bounds for every chain, then transfinitely choose a sequence of larger elements until you run out. This is typically formalized by mapping a chain to the set of its larger elements, and then using AC to create a choice function selecting exactly one larger element from each set. This is equivalent to using AC on a partially ordered family of sets, however we only need to use it on one particular chain, which is a well-ordered family of sets. Why can't we use the well-ordered AC here?

Basically, why isn't the well-ordered AC equivalent to the usual AC?

• Well. Not every set can be well-ordered. – Asaf Karagila Mar 9 '19 at 19:46

To see the difference, let's follow your idea for a two-element set $$X=\{a,b\}$$. Our first "stage" is: pick an element $$e_0$$ from the set $$X$$. Our second stage is: pick an element $$e_1$$ from the set $$X\setminus\{e_1\}$$.
Incidentally, note that DC - which is basically "dependent choice for $$\omega$$-many stages" - is indeed strictly stronger than the axiom of countable choice, which is the "index-set-instead-of-stages" version. So there's a pattern here.