Talk:Zorn's Lemma Implies Zermelo's Well-Ordering Theorem

This proof seems a bit confused. (considering $X \cup \set {x_0}$ when $x_0 \in X$, awkward notation for defining $\preceq'$, conclusion doesn't make sense to me) It's not quite up to standard yet, but I've cleaned up the confusion. To do:


 * Prove that the union over a chain is an upper bound for that chain, pretty easy but should be spelt out
 * Show that $\preceq'$ is a (total) order
 * Show that $\preceq'$ is a well-order (if $A \subseteq E \cup \set {x_0}$ is non-empty, then its $\preceq'$-least element is $x_0$ if $x_0 \in A$, or its $\preceq$-least element otherwise)

Caliburn (talk) 12:32, 15 March 2022 (UTC)


 * The first does not seem "pretty easy" and may be independent of ZF? Certainly it can't be true that we know that the union of well-orderable sets is well-orderable, otherwise we could just take the union of singletons and find that all sets are well-orderable. Nor can we know it's not true, since then Well-Ordering and hence Choice would fail. I don't know how the criteria that they're included in each-other changes anything. Seeing as I am very much out of my depth with this I'll post a health warning and maybe come back to it much later, would appreciate if someone could look at this. Caliburn (talk) 16:48, 18 March 2022 (UTC)


 * Indeed you struck upon a bad case of handwaving at least, and possibly an entirely wrong argument. I have added a note explaining how I would fix this proof, which is typical of applications of Zorn. Before we start fleshing it out, please read over my suggestion and let me know if it makes sense.
 * To give you an idea, in the argument made before we could take the chain $W_n = \set{ -n, \ldots, -1}$ with standard ordering and this would lead to all sorts of problems. &mdash; Lord_Farin (talk) 20:07, 18 March 2022 (UTC)


 * I've had a look at the (correct) proof along these lines and think I understand it. I would err on the side of this proof idea being wrong, since they are only considering the underlying sets and want to order $\WW$ by $\subseteq$ which does not account for the order. (where as you say it needs to be specifically chosen) I don't think there's any counterexamples as stated in ZF, since that would mean we have a set that isn't well-orderable, implying $\neg \textbf {AC}$, meaning something's gone wrong somewhere. (???) I would guess that it's not provable either, but I'm not really sure. It would also change the conclusion slightly, but the contradiction getting $E = X$ is basically the same it looks like. Caliburn (talk) 20:28, 18 March 2022 (UTC)


 * I agree. It btw suffices to disprove the conclusion of the argument (union is well-ordered), it is not necessary to disprove the theorem in itself. For this the $W_n$ I mentioned above are enough. Key point is that "well-ordered" is linked to Definition:Well-Orderable Set and the point of choosing the well-ordering is nontrivial in a non-choice environment. In conclusion I would suggest to write the correct proof and dispose of this handwaving failure. We should check Folland if he really makes this mistake; I can't imagine. &mdash; Lord_Farin (talk) 20:34, 18 March 2022 (UTC)


 * This is where I got a bit confused. Wouldn't $\bigcup_n W_n$ be well-ordered wrt $\ge$, since we're not caring about the order on $W_n$ or $\bigcup_n W_n$ and we're free to pick any? (more generally doesn't the well-orderability of $\N$ imply the well-orderability of any countable set?) I think this is the whole problem with only looking at underlying sets. Caliburn (talk) 20:48, 18 March 2022 (UTC)

The problem with "we're free to pick any" is that we are trying to prove there is any in the first place... Otherwise the union isn't in $\WW$ and the argument fails. And the only way to construct it is by means of the chain, which is inadequately prepared for that task. &mdash; Lord_Farin (talk) 22:14, 18 March 2022 (UTC)