Does ZF prove that the union of a $\subseteq$-chain of well-orderable sets is again well-orderable? I'm interested in this mis-transcription of Folland: https://proofwiki.org/w/index.php?title=Zorn%27s_Lemma_Implies_Well_Ordering_Theorem&oldid=356679. It is dubiously stated that the union of a $\subseteq$-chain of well-orderable subsets of a set $X$ is well-orderable. Clearly we cannot provide a counter-example in ZF, since this would contradict the Well-Ordering Theorem giving a set that is not well-orderable. My guess would be that it is not provable either, and would somehow imply the well-ordering principle (thereby making the proof circular) or a weak version of it. So my question is - is this statement independent of ZF (I would guess so), and if so is this hard to show?
 A: Suppose, for example, that $R$ is a set of socks, namely, it can be partitioned into countably many sets, $P_n$, each of size exactly $2$, and no infinitely many of them admit a choice function. The existence of such sets is consistent with $\sf ZF$ as was shown by Cohen (based on the work of Fraenkel before him).
Note, simply, that $R_n=\bigcup_{k<n}P_k$ is a finite set, and so it is certainly well-orderable, and $R=\bigcup R_n$ is certainly not well-orderable. This means that the chain $\{R_n\mid n<\omega\}$ is a chain of finite sets whose union is not well-orderable.
This can be easily repeated whenever we have a set that cannot be well-ordered, but can be written as an increasing union of well-orderable sets. For example $\Bbb R$ may be a countable union of countable sets, which implies it cannot be well-ordered, and in some cases it can be written as a union of $\omega_1$-type chain of sets of size $\aleph_1$ (but still not be well-orderable).
Okay, so maybe we can do this works for any set that cannot be well-ordered? Well, not quite. We say that $A$ is amorphous if it cannot be written as a disjoint union of two infinite sets. Note that this includes all the finite sets, but we'll usually discard them implicitly.
One quirky property of amorphous sets is that every function into any linearly ordered set must have a finite range. In particular, any chain of subsets must be finite. But since all the well-orderable subsets of an amorphous set are finite, that means that any chain of well-orderable sets is a finite set of finite subsets, so its union is finite, and thus well-orderable.
So, the final question remaining, does the statement "the union of a chain of well-orderable sets is well-orderable" imply the axiom of choice after all? I'd reckon the answer is again negative, but I don't have a proof off-hand.
A: To make the proof more rigorous, we should consider instead the set $\mathcal W$ of pairs $(W,\le_W)$ where $W\subset X$ and $\le_W$ is a well-ordering on $W$. We define an "inclusion" order on $\mathcal W$ as follows
$$(W,\le_W) \le (W',\le_{W'}) ~\text{ iff }~ W \subset W' \text{ and } \le_{W'} \text{ extends } \le_W $$ Given a chain $(W_\alpha, \le_{W_\alpha})$ of pairs in $\mathcal W$, let $W = \cup_\alpha W_\alpha$ and define $\le_W$ by
$$x \le_{W} y ~\text{ iff }~ \exists \alpha : x,y\in W_\alpha \text{ and } x \le_{W_\alpha} y$$ In other words, we are taking the union of the graphs of the $\le_{W_\alpha}$. As $(W_\alpha,\le_{W_\alpha})$ is a chain, it is easy to see that $\le_W$ is a well-ordering on $W$. Thus $(W,\le_{W})$ is an upper bound of the chain in $\mathcal W$.
