The well-known Zermelio's theorem states that every set can be well-ordered. Since arbitrary well-ordering is a linear ordering, from this theorem it follows the following corollary:
(A) An arbitrary set can be linearly ordered.
It is well-known that Zermelio's theorem is equivalent to the axiom of choice.
Question: Can Corollary (A) be proven without axiom of choice?
Dear Peter Breuer!
Thank You for answer.
1) It seems to me that in the in the phrase in quotes some misprint is enable ("with the axiom of choice..." should be replaced by "without... ..." :)).
2) Hence, in such models the set of real numbers is one of examples of linearly ordered set that is not well-orderable, but Statement (A) in my question asserts that "an arbitrary set can be linearly ordered". So, your answer says nothing yet, because it remains unclear, whether is Statement (A) being valid in such models?
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if you can make sense of the following thread :
https://math.stackexchange.com/questions/311246/does-linear-ordering-need-the-axiom-of-choice
(i'm having hard time with it ... but thanks for the question ; had me review a pile of long-forgotten concepts !)
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Dear Fabrice Clerot!
Thank You for nice answer.
The paper doi:10.2307/2275540, which I had found in your link, gives the complete answer to my question.
I think there is a misunderstanding about well-ordering. Well-ordering does not require linear ordering although, for finite sets of finite objects, one can be imposed if there is any kind of canonical form for expressing the objects. And for distinguishable members of finite sets, AC is pretty much in the bag.
My example of a very successful non-strict well-ordering is any Boolean Algebra on the set of all subsets of a finite set with more than two members. Those subsets are enumerable, and an enumeration can be taken as a strict ordering.
I recently delved into that for a different purpose. The draft about it is at https://github.com/orcmid/miser/blob/master/oMiser/boole.txt
Thanks Paul Marcoux
. There you have it Yaroslav Ivanovich Grushka . First, well-orderings do not impose linear ordering, and, in fact, it appears that not all sets are well-ordered, and one must deal with the apparent independence of the Axiom of Choice.
Dear Dennis Hamilton!
I think, You are confused with well-ordered sets and well-founded sets.
For well-ordering relation see https://www.encyclopediaofmath.org/index.php/Well-ordered_set. For well-founded relation see https://en.wikipedia.org/wiki/Well-founded_relation or DOI:10.1016/S0049-237X(08)70781-X, Definition 3.1.
Ah yes, Yaroslav Ivanovich Grushka , I think you are right. I mean well-founded, which applies, for example, to Boolean Algebras. I must clean up the nomenclature in my work.
Is it not still the case that the Zermelo theorem depends on the Axiom of Choice?
Let S be an arbitrary subset of R^(N) (the space of all sequences of real numbers). On R^(N) consider the lexicographic order, which seems to be total. Then the restriction of this order relation to SXS makes S a totally ordered set. Examples of Banach spaces X contained in R^(N), for which a natural Schauder base is easy to be pointed out are: X=l_p, 1