This article is rated B-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics High‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles High This article has been rated as High-priority on the project's priority scale.

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

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)

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)

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

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)[reply]

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)[reply]

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)[reply]

OK, here we go.

• Moschovakis, Yiannis N. (1980). Descriptive Set Theory. North Holland. pp. p. 104. ISBN 0-444-70199-0. ((cite book)): |pages= has extra text (help)

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) ${\displaystyle \preceq }$ is a wellordering if it is a wellfounded ordering.

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

Is it known whether ${\displaystyle {\mathbb {R} ))$ 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)[reply]

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)[reply]

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)[reply]

I don't really see how that relates to the question. --Trovatore (talk) 07:06, 14 December 2007 (UTC)[reply]

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)[reply]

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)[reply]

Can the wellorder of the reals as proved to exist under ZF+AC be explicitly stated, or described in any detail? I'm trying to wrestle with the implications of the wellordering theorem. I can conceptualize wellorders for all finite and countable sets. It would be nice to have an idea of what a wellorder of an uncountable set looks like or how it acts. Thanks! - UC Berkeley math student. —Preceding unsigned comment added by 98.210.233.159 (talk) 01:13, 19 April 2010 (UTC)[reply]

In general, the well-ordering of the reals may not be definable in any reasonable way. For example, in a model of ZFC + projective determinacy, every projective set is Lebesgue measurable, so there cannot be a projective well-ordering of the reals, because this would lead to the existence of a projective Vitali set. On the other hand, if V = L then there is a ${\displaystyle \Delta _{2}^{1))$ well ordering of the reals. So although AC implies that there is a well-ordering of the reals, there may or may not be a nice formula that defines this well-ordering. — Carl (CBM · talk) 01:38, 19 April 2010 (UTC)[reply]

There is however a limit to how bad the simplest formula can be, if there's a formula at all. That's because there's a canonical, definable, wellorder of the ordinal-definable reals. If there is any definable wellorder of the reals, then every real is ordinal-definable, and therefore the canonical wellorder of the OD reals is a wellorder of all the reals.

Woodin has also placed a limit on the best lower bound on the complexity you can get just from large cardinals of the sort that we know about (the ones that are preserved under small forcing). As I understand it, he showed that small forcing suffices to add a ${\displaystyle \Delta _{2}^{2))$ wellorder of the reals; therefore, no known large cardinals can refute the existence of such a wellorder. I don't know whether that's lightface or boldface. --Trovatore (talk) 08:01, 19 April 2010 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Humour... it's a difficult concept... Of course, "intuition" and "set theory" are disjoint sets classes. Albmont (talk) 11:04, 12 December 2008 (UTC)[reply]

BTW, I added a ((main)) 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)[reply]

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)[reply]

How is it equivalent to say if a set is totally ordered then it is well-ordered? The way wolfram math world defines this makes this statement incorrect. http://mathworld.wolfram.com/WellOrderedSet.html There it uses the example of ${\displaystyle \mathbb {Z} }$, which is totally ordered but is not well-ordered. Am I misreading this part of the article, or is this part incorrect? If it's correct, I think there needs to be a better explanation. Katachresis (talk) 04:50, 18 June 2010 (UTC)[reply]

I think that you are misreading it. It does not say that totally ordered and well ordered are the same.

None the less, I added a few words which I hope will make it clearer to you. JRSpriggs (talk) 08:14, 18 June 2010 (UTC)[reply]

Ok, it makes more sense to me, thanks. Katachresis (talk) 23:15, 20 June 2010 (UTC)[reply]

What is the reason for the insistence in the lead on the total order involved being strict? Specifying a total order, or a strict total order, is essentially the same thing, and this insistence is distracting. David Olivier (talk) 09:22, 11 May 2011 (UTC)[reply]

"Well-founded", in the definition in the lede, also presumes strict, at least in its formal definition. If you look closely, if x R x, then {x} does not have an R-minimal element. And, von Neumann ordinals are always strictly ordered by the element relation.

On the other hand, requiring the relation to be reflexive allows identification of <<X, R>> with R, as otherwise the empty relation is an exemplar of both the order types 0 and 1, making the proof that Hartog's function H satisfies ${\displaystyle H(X)\leq ^{*}2^{X^{2))}$or ${\displaystyle 2^{H(X)}\leq 2^{2^{X^{2))))$ more difficult. Also, all the examples in this article are reflexive. Perhaps we should rewrite, after all. — Arthur Rubin (talk) 13:20, 11 May 2011 (UTC)[reply]

I'm not an expert and certainly don't understand all that is at stake (I have no idea what that Groundhog's function is :P). But it does seem akward to me that the first sentence defines the well-order relation as strict, but then does nothing with this strictness: the concept of a least element is in substance the same, whether the order is strict or not (and is actually easier to define with a reflexive order). True, using a reflexive order might make the second sentence a bit more complicated to state; but the readability of the first sentence seems more important to me. David Olivier (talk) 15:40, 11 May 2011 (UTC)[reply]

This is a slightly annoying point — I think I brought it up some years ago but never followed through on it. In actual usage, I agree, both strict and nonstrict orders may be described as wellorders, and no one worries about it much. In an article, we probably have to worry about it at least a little.

We could: (A) State that some authors take wellorders to be strict and others to be reflexive, and give the two corresponding definitions, (B) say that there are different definitions depending on context, or (C) give a definition that covers both cases (e.g. the irreflexive part has no infinite descending sequence). For (C) we need a source; is there any source that does that? (A) and (B) are more honest but harder to figure out how we deal with the sourcing. --Trovatore (talk) 19:45, 11 May 2011 (UTC)[reply]

My mother, in Set Theory for the Mathematician, used "Hartogs' function" as a representation of what we call Hartogs number. Sorry about the confusion. I can't help with sourcing, though. — Arthur Rubin (talk) 00:46, 12 May 2011 (UTC)[reply]

First, {x} always has a minimal element, whether you take the order to be strict or not. You just have to define "minimal" properly.

Second, we should not insist on the order being strict, certainly not in the first paragraph of the article. There is no need to mention "strict" at all, as long as you use symbols like < and ≤ which make it clear whether you are talking about a reflexive or areflexive relation. Of course, in the AC-characterization we have to say "no strictly decreasing sequence" (which we do already).

Third: Personally, I prefer strict orders when talking about well-orders, because then I can use the symbol ∈ to denote the order relation on ordinals.

--Aleph4 (talk) 13:50, 12 May 2011 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

OK. JRSpriggs (talk) 09:28, 12 December 2012 (UTC)[reply]

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)[reply]

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>...x_n is a finite sequence of elements above the end of the well-ordered initial segment of the ordering, then there is an x_n+1<x_n 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)[reply]

"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)[reply]

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)[reply]

In English, a hyphen is not generally used to connect an adverb, like "well" to a verb like "ordered". However, when an adverb not ending in "ly" modifies a verb acting as participle and together they modify a subsequent noun then the hyphen is used, as in "well-ordered set". This article, including its title, confuses this distinction and is using too many hyphens. 𝕃eegrc (talk) 18:29, 29 February 2016 (UTC)[reply]

The thing is, "well" is not actually an adverb here. If it were, then the noun form would be "good order" (ordinarily, an adverb cannot modify a noun).

The term "well order", "well-order", or (my preferred version) "wellorder", are better thought of as technical non-analyzable terms in which the word "well" has no independent meaning.

That's one reason I like "wellorder", as it seems to make this especially clear. But in any case, all three forms are used. --Trovatore (talk) 20:03, 29 February 2016 (UTC)[reply]

In the Joining section of the Hyphen article, the ways that hyphens are used to join words are listed. The compound modifier, as in "well-ordered set" is given. Also given are combinations of nouns. When speaking of a "well ordering" neither of these is applicable (unless "well" is noun, such as a place where we get water). That leaves the Other compounds section. Is it your thinking that a "well-ordering" is like a "lily-of-the-valley", which distinguishes a specific kind of lily from an arbitrary lily growing in the valley? Is it common enough for someone to use "a well ordering" to be an ordering that was done well, but which is not necessarily "a well ordering" in the sense of the article? If not, but you think that there are other uses for hyphens (such as "as technical non-analyzable terms in which the [connected] word has no independent meaning") can you give a citation, Wikipedia article, or whatever to support that? 𝕃eegrc (talk) 21:10, 29 February 2016 (UTC)[reply]

I would prefer no hyphen, but also no space. A wellordering is very specifically not an ordering done well. There's nothing "better" about a wellordering than any other ordering (though, certainly, sometimes it is more useful).

It's just technical jargon, and the question is not whether it conforms to broader English usage (which clearly it doesn't, since an adverb can't modify a noun). "A well order" is pretty much ungrammatical, again, as you say, unless you mean an order inside a well, or perhaps an order that isn't sick. "A well-order" at least gets you over the WTF-adverb-before-noun-what-is-this? reaction. --Trovatore (talk) 21:18, 29 February 2016 (UTC)[reply]

Both Jech's book and Kunen's books use "well-order", with a hyphen, as in "R is a well-ordering of A" and "A is well-ordered". This is very common in the field of set theory, which this article is about. Given that the article was already established using the hyphenated form, I am in favor of maintaining it, personal preferences aside. — Carl (CBM · talk) 23:10, 29 February 2016 (UTC)[reply]

Oh, I agree. I wasn't actually suggesting moving the article to wellorder. --Trovatore (talk) 23:15, 29 February 2016 (UTC)[reply]

If these authors deliberately-wrote textbooks that clearly-use "well-order" as a verb then that counts as well-written by me. 𝕃eegrc (talk) 13:00, 1 March 2016 (UTC)[reply]

This discussion seems predicated on the notion that "well" is an adverb. But, in the context, it seems equally likely that it is a noun: as in, every well has a bottom. More precisely, this is a noun adjunct. This post suggests that noun adjuncts may or may not be hyphenated. My impression is that in English, such things are generally governed by practice rather than fixed rules. Sławomir
Biały 14:10, 1 March 2016 (UTC)[reply]

It never occurred to me that it might be "well" in the sense of a space with a bottom! I have always assumed that "well" means that the ordering is good (in some special sense). Does anyone have a reference to resolve this? 𝕃eegrc (talk) 14:29, 1 March 2016 (UTC)[reply]

Unlikely. Mathematicians rarely write about their terminology in that way, so finding references that talk about it is very unlikely. This is true for many aspects of mathematical terminology and notation, not just here. Of course, we also don't say anything about this in the article, which is where a reference would be required. (For the record, as someone who has worked with these orders professionally for years, I have never thought of the "well" as referring to a hole in the ground. ) — Carl (CBM · talk) 15:28, 1 March 2016 (UTC)[reply]

That, and the fact that well-order was very likely an English translation of the German wohlordnung. But it was worth a try :-) Anyway, the word "wohl" in German does have two different meanings, the relevant sense of which presumably does not correspond exactly to the English word well. So perhaps Torvatore's perspective is most natural, that one really ought to consider "well-order" as an indivisible unit. Sławomir
Biały 16:22, 1 March 2016 (UTC)[reply]

Leegrc, I didn't mean that "good ordering" wasn't the intent, etymologically. I think it was. I was saying that it can't be read that way, now.

Usually, when someone defines "good" in a proof, it's a way of saying, "this is just a notion I want to define for convenience in this particular proof. I understand that, of course, you can't really make 'good' a permanent technical term, but I don't expect this notion to ever be useful outside the current argument."

But what I think is definitive, did I mention that adverbs can't modify nouns? --Trovatore (talk) 17:41, 1 March 2016 (UTC)[reply]

Although we usually use the hyphen in the article, we are not entirely consistent. Probably that should be fixed. Also, perhaps we should have a short section listing the alternatives, or at least a sentence somewhere near the "wellordering" sentence, to indicate that sometimes the separated words do not have a hyphen between them — if for no other reason than to prevent someone who knows the English rules for hyphens from fixing the current article. (FWIW, I don't agree that "mathematicians rarely write about their terminology in that way", but it appears to be moot, so I won't argue it.) 𝕃eegrc (talk) 17:48, 1 March 2016 (UTC)[reply]

Those suggestions make sense, I think. I would note though that the usual English rules also forbid "a well order" (or even "a well ordering").

I think we also ought to make it clear that "well-order" in the sense of the article title is a noun, not a verb (though it can also be used as a verb, our article titles are supposed to be nouns; see WP:NOUN.) So the first sentence should start with just "[a] well-order", not "[a] well-order relation". --Trovatore (talk) 17:58, 1 March 2016 (UTC)[reply]

The fact that the adverb "well" remains "well" as an adjective instead of becoming "good" is common when irregularly transforming words are used in novel ways. Compare "Today, the bird flies; yesterday, the bird flew" with "This inning, the batter flies out; last inning, the batter flied out". I don't find it alarming that "well" fails to be irregular when the adverb-verb phrase "well order" transforms to the adjective-noun phrase "well order" or "well ordering".

Also note that there are no hyphens in "fly out", "flied out", "linear order", "total order", or "simple order". Just saying. 𝕃eegrc (talk) 19:54, 1 March 2016 (UTC)[reply]

Consider the infinitive form. You would write "to well-order a set", not "to order a set well". If "well" were acting as an adverb, the first form would be a split infinitive. Sławomir
Biały 11:49, 6 March 2016 (UTC)[reply]

The comment(s) below were originally left at Talk:Well-order/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Last edited at 18:32, 11 May 2007 (UTC). Substituted at 02:41, 5 May 2016 (UTC)

While reading this: Nik Weaver Predicativity Beyond Γ₀ (2009) I see that he has written:

the classical concept of well-ordering has a variety of formulations which are not predicatively equivalent (see §1.4 and §2.4). In fact, previous discussions of predicativism have tended to ignore this distinction, and this will emerge as a crucial source of confusion (see §1.4).

It seems that this article ignores these distinctions as well, and perhaps that could be remedied? 67.198.37.16 (talk) 21:45, 8 July 2016 (UTC)[reply]

Weaver's perspective is not a common one in mathematics. Because of the concept of undue weight, we should generally stick to the usual sources here, unless there are more mainstream sources that also worry about predicative definitions of well orderings. Analogously, the same holds for mathematical constructivism - few Wikipedia math articles spend time on what would happen if their topics were studied in constructive mathematics, because that is not a very common perspective in the references. It would give a distorted perspective if all our math articles were written from the POV of constructivism or predicativism. — Carl (CBM · talk) 22:55, 8 July 2016 (UTC)[reply]

This article contains the section Well-order#Equivalent formulations which gives different formulations of well-ordering and asserts that they are (provably) equivalent. Nik Weaver's definition seems to be the one referred to in that section as "Transfinite induction works for the entire ordered set". Can you give a reason to doubt that that definition is equivalent to the others listed there? JRSpriggs (talk) 05:26, 9 July 2016 (UTC)[reply]

I was simply reading around, and was interested in alternative formulations and discussions. Weaver's quote seemed notable. Regarding "undue weight", that's just ridiculous: every single math article on WP gives undue weight to one or another topic, and is grossly deficient in a large variety of ways. WP is a nice place to look up various terms and concepts and get a quick back-grounder; but once one starts reading about a topic in depth (i.e. from a book, journal articles) it becomes clear that WP is one giant ball of undue-weight. To turn that around and use it as an argument to NOT extend an article with useful information seems to be a preposterous absurdity. I suppose that sounds harsh, but really, just take a step back and take a look at what's going on in WP math and physics. Its a fair train-wreck. 67.198.37.16 (talk) 19:16, 9 July 2016 (UTC)[reply]

Although I doubt that Weaver would accept this characterization, it seems to me that he is implicitly modeling mathematics as

${\displaystyle L_{W_{t))}$

where t is time, W_t is the least limit ordinal for which Weaver has not yet chosen a unique preferred (and predicatively acceptable) fundamental sequence at time t, and L_α is the constructible sets below level α. Notice that

${\displaystyle \Gamma _{0}\leq W_{t}<\omega _{1}^{CK}\,.}$

What do you think? JRSpriggs (talk) 21:17, 11 July 2016 (UTC)[reply]

> 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)[reply]

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)[reply]

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)[reply]

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)[reply]

@Meef4H: I echo the words of Trovatore 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)[reply]

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)[reply]

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)[reply]

"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)[reply]