well-ordering theorem

{{short description|Theoretic principle in mathematics stating every set can be well-ordered.}}

{{redirect|Zermelo's theorem|Zermelo's theorem in game theory|Zermelo's theorem (game theory)}}

In mathematics, the well-ordering theorem, also known as Zermelo's theorem, states that every set can be well-ordered. A set X is well-ordered by a strict total order if every non-empty subset of X has a least element under the ordering. The well-ordering theorem together with Zorn's lemma are the most important mathematical statements that are equivalent to the axiom of choice (often called AC, see also {{section link|Axiom of choice|Equivalents}}).{{cite book |url=https://books.google.com/books?id=rqqvbKOC4c8C&pg=PA14 |title=An introduction to the theory of functional equations and inequalities |page=14 |location=Berlin |publisher=Springer |isbn=978-3-7643-8748-8 |first=Marek |last=Kuczma |year=2009 |authorlink=Marek Kuczma}}{{cite book |url=https://books.google.com/books?id=ewIaZqqm46oC&pg=PA458 |title=Encyclopaedia of Mathematics: Supplement |first=Michiel |last=Hazewinkel |year=2001 |authorlink=Michiel Hazewinkel |page=458 |location=Berlin |publisher=Springer |isbn=1-4020-0198-3 }} Ernst Zermelo introduced the axiom of choice as an "unobjectionable logical principle" to prove the well-ordering theorem.{{cite book |url=https://books.google.com/books?id=RkepDgAAQBAJ&pg=PA23 |title=Handbook of Mathematics |first=Vialar |last=Thierry |year=1945 |page=23 |location=Norderstedt |publisher=Springer |isbn=978-2-95-519901-5 }} One can conclude from the well-ordering theorem that every set is susceptible to transfinite induction, which is considered by mathematicians to be a powerful technique. One famous consequence of the theorem is the Banach–Tarski paradox.

History

Georg Cantor considered the well-ordering theorem to be a "fundamental principle of thought".Georg Cantor (1883), “Ueber unendliche, lineare Punktmannichfaltigkeiten”, Mathematische Annalen 21, pp. 545–591. However, it is considered difficult or even impossible to visualize a well-ordering of $\mathbb{R}$ , the set of all real numbers; such a visualization would have to incorporate the axiom of choice.{{cite book |url=https://books.google.com/books?id=RXzsAwAAQBAJ&pg=PA174 |title=The Logic of Infinity |page=174 |publisher=Cambridge University Press |isbn=978-1-1070-5831-6 |first=Barnaby |last=Sheppard |year=2014 }} In 1904, Gyula Kőnig claimed to have proven that such a well-ordering cannot exist. A few weeks later, Felix Hausdorff found a mistake in the proof.{{citation|title=Hausdorff on Ordered Sets|volume=25|series=History of Mathematics|first=J. M.|last=Plotkin|publisher=American Mathematical Society|isbn=9780821890516|year=2005|contribution=Introduction to "The Concept of Power in Set Theory"|pages=23–30|url=https://books.google.com/books?id=M_skkA3r-QAC&pg=PA23}} It turned out, though, that in first-order logic the well-ordering theorem is equivalent to the axiom of choice, in the sense that the Zermelo–Fraenkel axioms with the axiom of choice included are sufficient to prove the well-ordering theorem, and conversely, the Zermelo–Fraenkel axioms without the axiom of choice but with the well-ordering theorem included are sufficient to prove the axiom of choice. (The same applies to Zorn's lemma.) In second-order logic, however, the well-ordering theorem is strictly stronger than the axiom of choice: from the well-ordering theorem one may deduce the axiom of choice, but from the axiom of choice one cannot deduce the well-ordering theorem.{{cite book |authorlink=Stewart Shapiro |first=Stewart |last=Shapiro |year=1991 |title=Foundations Without Foundationalism: A Case for Second-Order Logic |location=New York |publisher=Oxford University Press |isbn=0-19-853391-8 }}

There is a well-known joke about the three statements, and their relative amenability to intuition:

The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?{{Citation|last=Krantz|first=Steven G.|chapter=The Axiom of Choice|date=2002|pages=121–126|editor-last=Krantz|editor-first=Steven G.|publisher=Birkhäuser Boston|language=en|doi=10.1007/978-1-4612-0115-1_9|isbn=9781461201151|title=Handbook of Logic and Proof Techniques for Computer Science}}

Proof from axiom of choice

The well-ordering theorem follows from the axiom of choice as follows.{{Cite book |last=Jech |first=Thomas |title=Set Theory |publisher=Springer |year=2002 |isbn=978-3-540-44085-7 |pages=48|edition=Third Millennium }}

Let the set we are trying to well-order be $A$ , and let $f$ be a choice function for the family of non-empty subsets of $A$ . For every ordinal $\alpha$ , define an element $a_\alpha$ that is in $A$ by setting $a_\alpha\ =\ f(A\smallsetminus\{a_\xi\mid\xi<\alpha\})$ if this complement $A\smallsetminus\{a_\xi\mid\xi<\alpha\}$ is nonempty, or leaves $a_\alpha$ undefined if it is. That is, $a_\alpha$ is chosen from the set of elements of $A$ that have not yet been assigned a place in the ordering (or undefined if the entirety of $A$ has been successfully enumerated). Then the order $<$ on $A$ defined by $a_\alpha < a_\beta$ if and only if $\alpha<\beta$ (in the usual well-order of the ordinals) is a well-order of $A$ as desired, of order type $\sup\{\alpha \mid a_\alpha\text{ is defined}\}+1$ .

Proof of axiom of choice

The axiom of choice can be proven from the well-ordering theorem as follows.

:To make a choice function for a collection of non-empty sets, $E$ , take the union of the sets in $E$ and call it $X$ . There exists a well-ordering of $X$ ; let $R$ be such an ordering. The function that to each set $S$ of $E$ associates the smallest element of $S$ , as ordered by (the restriction to $S$ of) $R$ , is a choice function for the collection $E$ .

An essential point of this proof is that it involves only a single arbitrary choice, that of $R$ ; applying the well-ordering theorem to each member $S$ of $E$ separately would not work, since the theorem only asserts the existence of a well-ordering, and choosing for each $S$ a well-ordering would require just as many choices as simply choosing an element from each $S$ . Particularly, if $E$ contains uncountably many sets, making all uncountably many choices is not allowed under the axioms of Zermelo-Fraenkel set theory without the axiom of choice.

Notes

External links

Mizar system proof: http://mizar.org/version/current/html/wellord2.html

Category:Axiom of choice

Category:Theorems in the foundations of mathematics

Category:Axioms of set theory