Talk:Well-ordering theorem

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Is this technically speaking a theorem (in that it follows from the axiom of choice); or is it a principle (it is equivalent to the axiom of choice)? Chas zzz brown 03:02 Jan 14, 2003 (UTC)

It is a theorem of ZFC, in that it is logically entailed by the axioms of ZFC. One could take it to be an axiom, in which case the proposition conventionally called the "axiom of choice" would be a theorem. But that is not conventionally done. Lots of mathematicians know and use the axiom of choice every day without ever in their lives thinking about well-orderings, so I think that way of doing things is appropriate. Michael Hardy 03:06 Jan 14, 2003 (UTC)

Placing it as part of the default ZFC makes sense to me (I'm pro-Choice :)). I suppose the same goes for Zorn's lemma. BTW, it is helpful when adding new mathematical articles if you could place a link on the List of mathematical topics, so that these articles can be easily tracked when changes are made. Cheers Chas zzz brown 03:15 Jan 14, 2003 (UTC)

[edit] Principle/Axiom/Theorem confusion

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)

When I hear "Well ordering axiom/principle/theorem", I always think of the more general one first, and the case restricted to the natural numbers is just a special case. These could actually fit on the same page, with a section about the specialisation to the naturals. (Which obviously has special importance.)

-- Cale Gibbard 18:48, 05 Oct 2007 (UTC)

[edit] well-ordering and uncountable

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)

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)

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)

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)

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)

[edit] this article is worthless without proof

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)