Zorn Lemma: (Halmos' book) what are the advantages of working with collections of chains?

Question

This is a follow up to Explanation of Proof of Zorn's lemma in Halmos's Book

Zorn's lemma. If X is a partially ordered set such that every chain in X has an upper bound, then X contains a maximal element.

Proof sketch.

step 1:

Let be collection of weak initial segments of elements of X, ordered by set inclusion; we show that if has a maximal set, then X has a maximal element

step 2:

Let be the collection of all chains in X, ordered by set inclusion; show that if has a maximal set, then has a maximal set

The passage from to cannot introduce any new maximal elements.

Question

My question concerns the following comment by Halmos in Naive Set Theory:

One advantage of the collection is the slightly more specific form that the chain hypothesis assumes; instead of saying that each chain has some upper bound in , we can say explicitly that the union of the sets of , which is clearly an upper bound of , is an element of the collection .

I presume that Halmos is referring to an advantage of over .

But this advantage is not clear to me.

Wouldn't the exact same property, claimed to be an advantage of , apply to ?

Assume that we define as a chain in .

Why can't we say explicitly that the union of the sets of , (which also is clearly an upper bound of ?), is an element of the collection ?

In summary: I don't understand the first advantage of over , claimed by Halmos.

Answer 1

The fact that the union of a chain is an upper bound of that chain does not hold for $\mathcal{S}$. Recall that a weakly initial segment is a set of the form $\{a: a\le b\}$ for some fixed $b$; that is, every weakly initial segment has (in fact, is determined by) a top element. To quote Halmos:

More precisely, we consider, for each element $x\in\mathcal{X}$, the weak initial segment $\overline{s}(x)$ consisting of $x$ and all its predecessors.

This means that if I have a chain in $\mathcal{S}$ with no greatest element, the union of this chain will not be an element of $\mathcal{S}$!

For example, take my poset to be $\mathbb{N}\cup\{\infty\}$ with the usual ordering, and let $[n]=\{1, 2, . . . , n\}$. Then $\{[n]: n\in\mathbb{N}\}$ is a chain in $\mathcal{S}$, but its union is $\mathbb{N}$, which is not an element of $\mathcal{S}$.

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Halmos also lists a second advantage: that a subset of an element of $\mathcal{X}$ is an element of $\mathcal{X}$. Note that this again fails for $\mathcal{S}$: e.g. in the example above, the set $\{1, 12\}$ is a chain - so an element of $\mathcal{X}$ - but is not a weak initial segment - that is, not an element of $\mathcal{S}$. But $\{1, 12\}$ is a subset of $\{1, 2, 3, . . . , 12\}=[12]$, which is an element of $\mathcal{S}$.