Talk:Well-order

From Wikipedia, the free encyclopedia

WikiProject Mathematics
Mathematics Portal
This article is within the scope of WikiProject Mathematics, which collaborates on articles related to mathematics.
Mathematics rating: B Class High Priority  Field: Foundations, logic, and set theory

I think this would be clearer with an example. Does a well-order require the definitions of mathematical sets?

Contents

[edit] my example stinks

I added an example of how to well order the integers (all the integers), but my reference book is at home, I didn't write it well and it needs a proof it well orders the integers. Somebody please improve it. (marked for cleanup) RJFJR 21:46, Feb 7, 2005 (UTC)

[edit] definition?

The definition of a well ordered set says that every subset of a set has a least element. However, since the relational operator must be anti-symmetric, I'm having a hard time imagining a totally ordered set which itself has a least element but has a subset without a least element.

Can anyone give an example of why this definition is necessary? It might be something nice to include on the page. 129.110.240.1 05:18, 25 Mar 2005 (UTC)


Easy. The set of real numbers [0,1]. It has a least number, 0. What is the least element of the subset consisting of the interval (0,1] ? It is the smallest number greater than zero. But what is the smallest real number greater than zero? [Actually, if the Axiom of Choice is true then there is a way to well order the reals, but the proof, in addition to requiring AC, is non constructive so no one knows how to well order the reals. Unlike well ordering integers where there is a way to well order the negative numbers). RJFJR 15:22, Mar 25, 2005 (UTC)

I removed the cleanup template from here since it has been cleanedup. RJFJR 13:46, Jun 24, 2005 (UTC)

[edit] New version

The new version looks great! RJFJR 13:43, Jun 24, 2005 (UTC)


[edit] Strict or nonstrict?

The article is not very specific about whether it's discussing strict or nonstrict partial orders. The link to total order uses the nonstrict notion, but the examples given seem to be strict. Among adepts this is one of those things you don't worry too much about because it's usually clear from context, but in an article like this maybe we should be a little more careful. --Trovatore 8 July 2005 00:04 (UTC)

[edit] intro para

The intro para was stated in such a way that the linearity of the ordering was a consequence of the proposed definition (a partial order in which every subset has a least element). The alternative way of defining a well ordering is as a linear order which is well founded in the sense that every nonempty subset has a minimal element. The third possibility, namely a poset in which every nonempty subset has a minimal element, is not the same. To prevent confusion between least and minimal, I think it is better to be explicit that a well ordering is defined to be a well founded linear order. CMummert 13:10, 1 July 2006 (UTC)

[edit] unhyphenated spelling

I have removed the {{fact}} tag from the spelling note; of course this is not the sort of thing for which citations are likely to be available. It's the kind of thing you'd find in a dictionary, but no one writes dictionaries of contemporary set-theoretic usage. If you want a citation in which the spelling is used, I can find that for you once I get home (I'd put it on the talk page, not in the article). --Trovatore 23:22, 30 April 2007 (UTC)


OK, here we go.

  • Moschovakis, Yiannis N. (1980). Descriptive Set Theory. North Holland, p. 104. ISBN 0-444-70199-0. 

The relevant quote is

There are various names attached to relations that satisfy some of these conditions and we put them down here for the record.
[...]
(2) \preceq is a wellordering if it is a wellfounded ordering.

Hope this helps, --Trovatore 05:17, 1 May 2007 (UTC)

[edit] well-ordering of reals under ZF

Is it known whether {\mathbb R} can be proven to have a well-ordering using only the ZF axioms (without AC)? I guess that means using some conventional construction of R, with Dedekind cuts or whatever. Thanks. 75.62.4.229 (talk) 11:32, 13 December 2007 (UTC)

ZF does not prove that R can be wellordered. In fact the usual proof (or at least the first proof that I learned) that ZF does not prove AC, goes through the fact that ZF does not prove there's a wellordering of the reals.
However it is also consistent with ZF that the reals can be wellordered, but some larger set (say, the powerset of the reals) cannot be. --Trovatore (talk) 18:33, 13 December 2007 (UTC)
However, if V=L holds, then there is a specific (but very complicated) formula which well-orders the reals. JRSpriggs (talk) 05:51, 14 December 2007 (UTC)
I don't really see how that relates to the question. --Trovatore (talk) 07:06, 14 December 2007 (UTC)
To Trovatore: You are right that I was not addressing the specific question that was asked. But if his question was motivated by wanting to avoid having to postulate the existence of a set without having a way of constructing it, then I was pointing out that there is a way to construct it provided one is willing to discard all sets which are not in the constructible universe. JRSpriggs (talk) 08:06, 14 December 2007 (UTC)
Actually, it is sufficient to discard the reals which are not constructible. That is, because the constructible universe is well-ordered by a specific formula, its intersection with the reals is also well-ordered by that formula. In other words, you can use the construction of the constructible real numbers themselves to construct a well-ordering of the set of constructible real numbers. JRSpriggs (talk) 05:29, 15 December 2007 (UTC)