Talk:Well-ordering principle

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Z vs N[edit]

In the algebraic definition, I think it's usual to talk about all of Z rather than just N, because it is Z which is an integral domain, and Z which amalgamates the relevant algebraic properties. The natural numbers themselves are well-ordered in a set-theoretic (and order-theoretic) sense, of course; but the property of interest is algebraic here, no?

Will this be understood as I've left it? --VKokielov 03:42, 11 September 2006 (UTC)

If I understand you correctly, you (or perhaps Birkhoff-MacLane, which I doubt) define the phrase
R is a well-ordered integral domain
as an abbreviation for
R contains a well-ordered subset, called the natural numbers, of which any subset always contains a least element.
This does not make sense to me. First, "always contains a least element" just repeats "well-ordered". Second, I assume that by "called the natural numbers" you mean "isomorphic to the natural numbers". Third: the rational numbers also have this property (namely, that they contain the well-ordered subset of the natural numbers).
You may mean the following:
The ring of integers is characterized among the integral domains by the fact that it can be linearly ordered in such a way that
  1. the order agrees with the operations (i.e., addition and multiplication are monotone)
  2. the positive elements are well-ordered.
Aleph4 11:38, 11 September 2006 (UTC)
Right -- so I've caught myself and fixed it. --VKokielov 22:57, 11 September 2006 (UTC)
Hmm, I don't get it (the "fixed" part that is). Every version of this article since 11 September 2006 says "every well-ordered integral domain is isomorphic to the integers" which is silly, since the integers are not well ordered (and no integral domain can therefore be well-ordered, prime fields being not ordered at all). I'm throwing out the phrase; you may put something back in that makes sense. Marc van Leeuwen (talk) 14:55, 4 March 2011 (UTC)

Comment on well-ordering principle[edit]

I am writing concerning the Well-ordering principle. It is stated, at the address <>, that the proposition:

Any nonempty subset of the Natural must contain a least element.

May have to be an axiom. I do not see why the following proof would not work in most any framework, i.e. why there is a need for an axiom?

Let S be a nonempty subset of the Naturals. Then S contains some element, say n. Consider the set {1,2,3,…,n} If 1 is in S, then it must be the least element of S, and we are done. Otherwise, if 2 is in S, then we are done. Continue in this manner, until the smallest element of S is found. Note: If none of the elements 1,2,3,…,n-1 is in S, then n must be the smallest element of S and we arrived at this conclusion in a finite number of steps.

I do recognize that I am making an assumption:

Given any nonempty set, an element of the set may be chosen.

But, this assumption (axiom) is much more benign than the Axiom of Choice.


Hajdupe 19:41, 8 May 2007 (UTC)

request for a layman example[edit]

this is to enable us laymen to understand this prionciple Sanjiv swarup (talk) 13:27, 23 September 2008 (UTC)

Does every math wikipedia article start with a shit first sentence?[edit]

Well-ordering is not a property of ordered sets. The reals are ordered, but not well-ordered.

It is indeed a property of ordered sets; the reals simply do not have this property. If you dislike the wording, you are always welcome to change it. — Carl (CBM · talk) 13:55, 21 December 2011 (UTC)