Misunderstanding the Axiom of Choice and its equivalences

One of the equivalent statements of the Axiom of Choice is that

$\prod X_i\neq \emptyset\iff X_i\neq \emptyset\text{ for all }i.$

One can prove that this is equivalent to the statement that

Every chain in every partial order is contained in a maximal chain.

But why you have to assume AC for that? If you have a chain you can just add every element which is compareable to every element of the chain, to get a new chain, right?...

I guess this intuitive argument works only for countable sets, where you can really take new elements "step by step" and add them to your chain. But with AC we can make sure that its also working for arbitrary amount of sets i.e. with AC we can guarantee that the "thing" we will get after adding all the elements is indeed a chain. Am I right? It is still a bit confusing for me. I hope that someone can clarify a bit...

• If you have a chain, and you "just add every element which is comparable to every element of the chain", what makes you think you will get a chain? Say $a$ and $b$ are comparable to every element of the chain you started with, so you add them to the chain, but who says $a$ and $b$ are comparable to each other?
– bof
Commented Nov 19, 2016 at 12:28
• @bof First I add $a$ to the chain $C$ and get a new chain. Then I will add $b$ to the chain, if $b$ is compareable to $C\cup \{a\}$.
– Marc
Commented Nov 19, 2016 at 13:41
• Yes, that makes a lot more sense than what you said the first time. You might want to consider editing your question. So, how do you know which to start with, $a$ or $b$? The final chain will depend on what order you take the elements in, right? So how do you do that without the axiom of choice?
– bof
Commented Nov 19, 2016 at 14:08
• @bof why it does mater which one of them I take first? The statement is that there is a maximal chain. So we don't care about uniqueness?
– Marc
Commented Nov 19, 2016 at 14:20
• Suppose that $a>b$ and $a>c$ and that $b,c$ are incomparable . Consider the chain $D=\{a\}.$ If you add every element that is comparable to every element of $D$ you must add $b$ and $c .$ And then you don't get a chain. Commented Nov 21, 2016 at 23:07

1. Usually there is no canonical choice as to what element you're adding to your chain, and this may affect your future choices as well (if you add $x$, then you cannot add any element incomparable with $x$ later on).