Question regarding a proof of Zorn's Lemma

Question

I am trying to understand this proof of Zorn's lemma.

I think the notation is clear to me and I can follow the proof until the point where the author defines conforming subsets although I am not entirely sure if I get the part using the axiom of choice.

Now my questions are:

1) Why do we need the axiom of choice? Until now I have viewed the axiom of choice as "we can create a new set from other sets even if we have uncountably many sets". Now wikipedia states it somewhat differently, i.e. that the axiom of choice gurantess a choice function, but I can see this formalizes my intuition since if such a function exists then we can use this function to pick elements from each set in a collection of sets. However, I am not entirely sure why the argument in the proof of Zorn's lemma requires a choice function. Given the assumptions we know that every chain in $X$ has a strict upper bound, so what's the use of the axiom of choice now?

2) Why does the author define the conforming property for arbitrary subsets? In the definition he uses an initial segment $P(A,x)$, but in the notation part he defined an initial segment only for chains. It also makes sense to define it only for chains since if a subset does not have a total order, then some elements might simpy not be in $P(A,x)$ because there is no relation between $x$ and those other elements.

I am quiet new to such deep set theory arguments and only want to understand this since it is relevant for a lemma on extensions of solutions to ODEs.

Thanks for any help and suggestions!

Answer 1

1) You know that for each chain in $X$ there exists an upper bound, but there might be infinitely many upper bounds for each chain, and there might be infinitely many chains. Humans don't really have experience of this: when did you last (consciously) make infinitely many decisions, each with infinitely many choices of outcome? The Axiom of Choice is precisely the assumption (taken to be self-evident) that this is always an allowable procedure in set-theoretical reasoning (no matter how "big" the "infinities" involved are). Saying "there exists a choice function such that..." is just a way of formalizing this notion.

2) Recall that a chain is a totally ordered subset. Condition (a) of being "conforming" is that the set $A$ is well ordered. In particular, note that well ordered implies totally ordered. So you are correct that a conforming set will always be a chain. The author just wrote things that way because they thought it would be easier to read than the alternative; restricting the definition to chains might seem weird to experienced readers, for two reasons:

• the definition makes complete sense in the more general context that $A$ is an arbitrary subset of $X$; and
• part of the definition means making another restriction to well ordered subsets anyway, so you don't gain anything by restricting the context at the outset.

Answer 2

Just because a chain has an upper bound, it does not mean it has exactly one upper bound, or even a canonically chosen upper bound. In fact, in the generic case, it won't have one.

So that means that we need to choose one. Well, fine, doing it once is easy. But when you need to choose them in succession, in a conforming way, then you already find yourself in a pickle after doing so for every natural number, as you've essentially had to make infinitely many arbitrary choices. And god forbid your partial order is even larger than that...

As for the definition of a conforming subset, note that by requiring the order induced by $\leq$ to be a well-ordering you've already required that the set is a chain since it is linearly ordered by $\leq$.