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 Axiom of choice § Equivalents).^[1][2] Ernst Zermelo introduced the axiom of choice as an "unobjectionable logical principle" to prove the well-ordering theorem.^[3] 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.^[3] 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".^[4] However, it is considered difficult or even impossible to visualize a well-ordering of ${\displaystyle \mathbb {R} }$; such a visualization would have to incorporate the axiom of choice.^[5] 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.^[6] 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.^[7]

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?^[8]

Proof from axiom of choice

The well-ordering theorem follows from the axiom of choice as follows.^[9]

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

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

An essential point of this proof is that it involves only a single arbitrary choice, that of ${\displaystyle R}$; applying the well-ordering theorem to each member ${\displaystyle S}$ of ${\displaystyle E}$ separately would not work, since the theorem only asserts the existence of a well-ordering, and choosing for each ${\displaystyle S}$ a well-ordering would require just as many choices as simply choosing an element from each ${\displaystyle S}$. Particularly, if ${\displaystyle E}$ 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