Talk:Well-order

message: 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)
thread_title: Definition?

message: 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)
As far as I was aware, a poset is by definition is a "non-strict" relation. In my uni Discrete Math text (K.Rosen, Discrete Mathematics) it's simply any reflexive, antisymmetric, transitive relation on a set (thus "non-strict"). A total ordering is a poset with one extra requirement: (a,b) or (b,a) for all a,b in the set. Rosen doesn't even talk about "strict" posets. But the strictness or otherwise of the ordering doesn't effect the well-ordering condition anyway, since there is a least element whether or not the relation is strict or non-strict. So there is no need to talk about it? In fact I think doing so just add confusion. --Meef4H (talk) 08:25, 24 December 2023 (UTC)
thread_title:  Strict or nonstrict? 

message: 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)
thread_title:  intro para 

message: How does the sentence Every well-ordered set is uniquely order isomorphic to a unique ordinal number, called the order type of the well-ordered set. relate to the axiom of choice? Unless I am confusing something, I think "axiom-of-choice-deniers" don't accept this. Albmont (talk) 15:52, 10 December 2008 (UTC)
replies: text: You don't need choice to prove the above statement.  You do need the axiom of replacement to prove a certain formalization of it (namely the one that identifies ordinals with von Neumann ordinals.) --Trovatore (talk) 18:46, 10 December 2008 (UTC)
replies: text:  Ok, I saw that in article ordinal number, section "Von Neumann definition of ordinals", this point is made; maybe the reference to the AC should be made more explicit, since "intuition" yells that the AC is invoked whenever we talk about infinite sets :-) Albmont (talk) 10:40, 11 December 2008 (UTC)
replies: text: It does?  What intuition is that?  I would agree with this statement:  The way of thinking about infinite sets that we've found most useful, makes it intuitively clear that AC is true.  Is that what you meant? --Trovatore (talk) 20:14, 11 December 2008 (UTC)
replies: text:  Humour... it's a difficult concept... Of course, "intuition" and "set theory" are disjoint sets classes. Albmont (talk) 11:04, 12 December 2008 (UTC)

text:  BTW, I added a  link to the article about ordinal numbers; there is the natural place to explore in more details what constructions do and what constructions don't require AC. Albmont (talk) 11:05, 12 December 2008 (UTC)
replies: text: Intuition is absolutely central to set theory, as to any mathematics; you can't make progress without it.  The important (and difficult) thing is to get the right intuitions.  --Trovatore (talk) 17:07, 12 December 2008 (UTC)
thread_title:  Axiom of choice 

message: I'll briefly cite some:
"The natural numbers are a well-order."
"The set {1/n : n =1,2,3,...} has no least element and is therefore not a well-order."
How come these errors weren't caught by anyone? This is mis-information and should be deleted or amended. 212.149.212.108 (talk) 14:07, 4 February 2012 (UTC)
replies: text: Right above this list of examples it says A countably infinite subset of the reals may or may not be a well-order with the standard "≤". so these examples are using that relationship.  RJFJR (talk) 16:10, 4 February 2012 (UTC)
thread_title:  Many fundamental mistakes in the "Reals" subsection 

message: I think the previous paragraph in the Reals section about the countability of the well-ordered subsets of R was a little confusing. Any help in making the demonstration clearer is welcome, as you can probably see I'm not a native speaker. Odexios (talk) 14:39, 21 May 2012 (UTC)
thread_title:  Countability of well-ordered subsets of R 

message: JR has reverted Tobias's changes on the grounds that "well-founded only applies to strictly-less; total order only applies to less-or-equal".  Those may be the definitions in the linked articles but I do not think either restriction is really standard.  In general these things are understood from context, and the appropriate modifications made silently and probably without even noticing.  For example Wadge reducibility is nonstrict, and is wellfounded for as high as determinacy goes — no one bothers AFAIK to state this as "the strict part of Wadge reducibility".
To be sure, we can't be that sloppy in an article for general consumption, but interpreting whatever choice was made by whoever wrote the linked articles as somehow canonical is not the way to go either.  Maybe we could use an explanatory footnote or something? --Trovatore (talk) 04:34, 12 December 2012 (UTC)
replies: text: The existing lead already says "If ≤ is a (non-strict) well-ordering, then < is a strict well-ordering. A relation is a strict well-ordering if and only if it is a well-founded strict total order.". So there is no reason for Tobias's change unless it is more correct which it is not. JRSpriggs (talk) 04:56, 12 December 2012 (UTC)
replies: text: Sorry about that, I really wasn't aware of that convention. Anyway, I would like to restore my other changes to the lead section which I hope are correct and uncontroversial. I am trying to improve the flow of the lead section which seems a bit unconnected to my eyes. —Tobias Bergemann (talk) 08:23, 12 December 2012 (UTC)
replies: text: OK. JRSpriggs (talk) 09:28, 12 December 2012 (UTC)
thread_title:  Strict again 

message: In the "Equivalent Formulations" section, the article states that the Axiom of Dependent Choice is required to prove that a well-ordered set  cannot have an infinite decreasing sequence. However, isn't the fact that the set is well-ordered precisely what the Axiom of Dependent Choice is there to prove? Since we are assuming the set is well-ordered, it is already order-isomorphic to a unique ordinal, and thus has an infinitely decreasing sequence if and only if the ordinal does as well. However, it is easy to prove that any decreasing sequence of ordinals (and thus decreasing sequence of elements of an ordinal) is finite by transfinite induction without any form of the Axiom of Choice. 
If every ordinal smaller than alpha has no infinitely decreasing subset, then alpha has no infinitely decreasing subset, for its first element must be some alpha_0 < alpha. After this element, every other element is in alpha_0, so the remainder of the sequence must be finite. However, adding a term to a finite sequence does not yield an infinite sequence.
Am I missing something?  — Preceding unsigned comment added by 140.180.243.190 (talk) 22:10, 25 March 2013 (UTC)
replies: text: You are looking at the wrong direction of implication. In well-order#Equivalent formulations, #3 can be deduced from the other versions without DC as you see. Where DC is needed is to go from #3 to the other versions. Without some version of the axiom of choice, there could be a non-well-ordered linear ordering which lacks any infinite decreasing sequences. If x0>x1>x2>...xn is a finite sequence of elements above the end of the well-ordered initial segment of the ordering, then there is an xn+1<xn which is also above the end. But we need DC to make an infinite sequence of such choices. JRSpriggs (talk) 08:37, 26 March 2013 (UTC)
thread_title:  Axiom of Dependent Choice and "Every Strictly Descending Sequence is Finite" 

message: "Every well-ordered set is uniquely order isomorphic to a unique ordinal number, called the order type of the well-ordered set. "
This is a consequence of ZF, but is independent of Zermelo Set Theory (and if I recall correctly is the biggest reason for preferring ZF over Z set theory.) Is it fair to mention this? Given that Z is a common axiomitization I feel it's a valid thing to mention, unlike for example something that's a consequence of the axiom of infinity.  — Preceding unsigned comment added by James Waddington (talk • contribs) 21:23, 15 July 2014 (UTC)
replies: text: I'm not opposed to mentioning after the fact that you need Replacement to prove this property.  But we should still state the property without waffling.  --Trovatore (talk) 21:33, 15 July 2014 (UTC)
thread_title:  Choice of Set theory and independence result. 

message: Substituted at 02:41, 5 May 2016 (UTC)
thread_title: Assessment comment

message: > In mathematics, a well-order (or well-ordering or well-order relation) on a set S is a total order on S with the property that every non-empty subset of S has a least element in this ordering. The set S together with the well-order relation is then called a well-ordered set.
Is there such a thing as a well-order relation? Will not any total-ordering relation suffice, with the well-ordering arising from the nature set only? --Meef4H (talk) 09:18, 23 December 2023 (UTC)
replies: text: Not really sure what you're getting at here.  An ordering is a kind of binary relation, but not all total orderings are wellorderings. --Trovatore (talk) 18:05, 23 December 2023 (UTC)
replies: text: I was getting at that there is no such thing as a well-order relation, only a well-ordering (a set plus a relation). I.e. there is no property of a relation taken alone that distinguishes it as a "well-order" relation as distinct from a total ordering. AFAICT anyway. Meef4H (talk) 07:47, 24 December 2023 (UTC)

text: I.e. I want to change the above quoted text to remove the first instance of "well-order relation" and replace the second one with just "relation". Saying "well-order relation" make it sound like it's some special type of relation stand alone. Meef4H (talk) 07:54, 24 December 2023 (UTC)
replies: text:   I echo the words of  above: I too am not really sure what you're getting at. Firstly, it is not true that "there is no such thing as a well-order relation"; the expression "well-order relation" exists, is commonly used by mathematicians, and has the meaning attributed to it in this Wikipedia article. Secondly, "Saying 'well-order relation' make it sound like it's some special type of relation stand alone" doesn't have a clear meaning; it is indeed a special type of relation, but what are the words "stand alone" intended to convey? The only way I can make any sense out of your remarks is to assume that you have a misunderstanding of what the word "relation" means in mathematics. A relation is a set of ordered pairs from a set. For example, we can define a relation called "is less than" on the set of integers. That relation then includes, amongst others, the pairs (2, 87), (-5, -1), and so on. We can also define another relation, also called "is less than" on the set of positive real numbers. That will include the pair (2, 87) but not (-5, -1); it will also include (2.7, π). Note that these are two different relations. In a context where it is clear what particular relation we are referring to, we can just use a name for it such as "is less than" or "<", but in any context where order relations on different sets are used, we have to use a notation which distinguishes between them. I suspect from what you have said above that you had not realised that the standard interpretation of "relation" in mathematics is a specific relation on a specific set. JBW (talk) 15:35, 4 January 2024 (UTC)
replies: text: Consider a so called "well-order relation" on the natural numbers. The natural numbers are a subset of the real numbers. Now consider the same "well-order relation" on the reals. It's not longer well-order relation. The well-ordered-ness comes from the set not from the relation. Meef4H (talk) 23:54, 4 January 2024 (UTC)
replies: text:  I don't know what you mean by "consider the same well-order relation on the reals". The relation you have referred to is a relation on the natural numbers; it is not a relation on the real numbers, so "the same well-order relation on the reals" has no meaning. A relation is a relation on a particular set, not some abstract concept divorced from any set; a relation on a different set is a different relation. JBW (talk) 22:19, 5 January 2024 (UTC)
replies: text: "The relation you have referred to is a relation on the natural numbers ..." It's defined for all natural numbers, it's not necessarily a function over the natural numbers. "A relation is a relation on a particular set ..". No. When we say 'relation on a set' we typically mean a binary relation from/to the same set, but that's not true in general. Relations can also be n-ary. "not some abstract concept divorced from any set ...". A relation itself can be completely represented by a set of tuples. I think you have some fundamental misunderstanding about what a relation is :) Meef4H (talk) 02:40, 6 January 2024 (UTC)
thread_title: Definition? :: well-order relation