Talk:Well-ordering theorem

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I have always learned that the Well-Ordering Principle is simply that the Naturals are well-ordered, and that the Well-Ordering Principle is equivalent to the Principle of Mathematical Induction.  The Well-Ordering Axiom, on the other hand, states that any set can be well-ordered, and that is what this article is about.  The Well-Ordering Principle certainly doesn't deserve to be called an axiom, and although the Well-Ordering Theorem can be proved from AC, it can also receive axiom status indifferently - it is independent of ZF.
My point is, it doesn't make sense to say that the Well-Ordering Theorem is "not to be confused with" the Well-Ordering Axiom, when as far as I know, everyone who says "Well-Ordering Axiom" is referring to the Well-Ordering Theorem, not to the Well-Ordering Principle, which is not equivalent to AC.  I move that the Well-Ordering Axiom redirect be pointed here, and the phrasing in the parentheses be changed. GTBacchus 07:11, 21 Jun 2004 (UTC)
thread_title:  Principle/Axiom/Theorem confusion 

message: This theorem seems to imply that there is a well-ordering on the reals, albeit not yet conceived-of. Why doesn't this imply that the reals (or any set subject to this theorem for that matter) could be put into 1-1 correspondence with \mathbb{N} in accordance with that particular well-ordering, contradicting the fact that many sets such as the reals are uncountable? Btyner 04:22, 16 February 2006 (UTC)
replies: text: Because, well, it doesn't. There's no way to construct such a 1-1 correspondence. You are probably thinking of a particular way to construct one, but without knowing which, we can't tell you why it doesn't work. --Mellum 07:35, 16 February 2006 (UTC)
replies: text: The Well-order article gives the impression that a well-order is just a list such that a given element of the set will be found on the list eventually (?). The well-ordering theorem says such a list of the reals exists. It seems like the uncountability of the reals hinges on the fact that no one has yet conceived of such a list. Btyner 15:13, 16 February 2006 (UTC)
replies: text: I don't see where it gives this impression. It says that you can make a list and will always know what to take next, but it doesn't say nor imply that this will eventually get every element, except in the finite case. That clearly doesn't even hold for countable sets: If I take 2, 4, 6, ... I always know what to take next and never run out of elements, but I won't hit all elements. --Mellum 20:14, 16 February 2006 (UTC)
replies: text: Thank you; I think I understand now. By the theorem, there is an ordering on \mathbb{R} such that all subsets thereof have a least element under that ordering. Therefore, \mathbb{R} has a least element, call it x_1. Now, \mathbb{R}-\{x_1\} is a subset of \mathbb{R}, so it too has a least element; call it x_2, and so on. You are saying that the set \{x_1,x_2,\ldots\}, being countable, leaves out most of \mathbb{R}. Btyner 03:31, 17 February 2006 (UTC)
thread_title:  well-ordering and uncountable 

message: also, cocks
(but seriously guys.  i came here looking for a proof.  can't you at least give me a linque?)
oh yeah re: unsourced: if it had a proof, that would be, well, proof, and the article wouldn't be 'unsourced' because anyone (with the proper background/sufficient time) could understand the proof and know whether the article was correct or not.  —Preceding unsigned comment added by 216.15.119.166 (talk) 06:54, 31 January 2008 (UTC)
thread_title:  this article is worthless without proof 

message: There should be more about set theory. Adrionwells (talk) 15:55, 11 November 2009 (UTC)
replies: text: Split into a separate article Zermelo's theorem (game theory), although I doubt there's enough there to survive.  It does not belong in this article.  — Arthur Rubin  (talk) 17:10, 11 November 2009 (UTC)
thread_title:  why is there so much about game theory in this article?? 

message: That looks so ugly in this article. Can someone edit it for the better?  — Preceding unsigned comment added by 108.41.98.105 (talk) 21:46, 8 January 2023 (UTC)
thread_title:  "Axiom of choice & its Equivalents" 

message: Let me reproduce the proof first:
:The well-ordering theorem follows from Zorn's lemma. Take the set \mathbf{A} of all well-orderings of subsets of X: an element of \mathbf{A} is an ordered pair (a,b) where a is a subset of X and b is a well-ordering of a. \mathbf{A} can be partially ordered by continuation. That means, define E ≤ F if E is an initial segment of F and the ordering of the members of E is the same as their ordering in F. If \mathcal{E} is a chain in \mathbf{A}, then the union of the sets in \mathcal{E} can be ordered in a way that makes it a continuation of any set in \mathcal{E}; this ordering is a well-ordering, and therefore, an upper bound of \mathcal{E} in \mathbf{A} . We may therefore apply Zorn's Lemma to conclude that \mathbf{A} has a maximal element, say (M,R). The set M must be equal to X, for if X has an element x not in M, then the set M∪{x} has a well-ordering that restricts to R on M, and for which x is larger than all elements of M. This well ordered set is a continuation of (M,R), contradicting its maximality, therefore M = X. Now R is a well-ordering of X.
I take issue with this part: 
:If \mathcal{E} is a chain in \mathbf{A}, then the union of the sets in \mathcal{E} can be ordered in a way that makes it a continuation of any set in \mathcal{E}; this ordering is a well-ordering, and therefore, an upper bound of \mathcal{E} in \mathbf{A}
Contrary to the above, the union of \mathcal{E} need not be well-ordered. Let X = \mathbb{Z}. Let A_n = \{-n, -(n-1), \ldots, 1, 0\}, and let E_n = (A_n, \leq|_{A_n}), where \leq|_{A_n} is a standard ordering on integers restricted to A_n. Finally let \mathcal{E} = \{E_n: n \in \mathbb{N}\}. Then the union of chain \mathcal{E} is clearly not well ordered. Therefore, I'm removing the erroneous proof. Dodek (talk) 22:45, 24 January 2020 (UTC)
thread_title:  The proof of Zorn lemma => well ordering principle is incorrect 

message: That looks so ugly in this article. Can someone edit it for the better?  — Preceding unsigned comment added by 108.41.98.105 (talk) 21:48, 8 January 2023 (UTC)
thread_title:  "Axiom of choice & its Equivalents"