# equivalence of Hausdorff’s maximum principle, Zorn’s lemma and the well-ordering theorem

## Hausdorff’s maximum principle implies Zorn’s lemma.

Consider
a partially ordered set^{} $X$, where every chain has an upper bound. According to the maximum principle
there exists a maximal totally ordered^{} subset $Y\subseteq X$. This then has an upper bound, $x$. If
$x$ is not the largest element in $Y$ then $\{x\}\cup Y$ would be a totally ordered set in which $Y$
would be properly contained, contradicting the definition. Thus $x$ is a maximal element^{} in $X$.

## Zorn’s lemma implies the well-ordering theorem.

Let $X$ be any non-empty set, and let $\mathcal{A}$ be the collection^{} of pairs $(A,\le )$, where $A\subseteq X$
and $\le $ is a well-ordering on $A$. Define a relation^{} $\u2aaf$, on $\mathcal{A}$ so that for all $x,y\in \mathcal{A}:x\u2aafy$
iff $x$ equals an initial of $y$. It is easy to see that this defines a partial order^{} relation on $\mathcal{A}$
(it inherits reflexibility, anti symmetry^{} and transitivity from one set being an initial and thus a subset of
the other).

For each chain $C\subseteq \mathcal{A}$, define ${C}^{\prime}=(R,{\le}^{\prime})$ where R is the union of all the sets $A$ for all $(A,\le )\in C$, and ${\le}^{\prime}$ is the union of all the relations $\le $ for all $(A,\le )\in C$. It follows that ${C}^{\prime}$ is an upper bound for $C$ in $\mathcal{A}$.

According to Zorn’s lemma, $\mathcal{A}$ now has a maximal element, $(M,{\le}_{M})$. We postulate^{} that $M$ contains all
members of $X$, for if this were not true we could for any $a\in X-M$ construct $({M}_{*},{\le}_{*})$ where
${M}_{*}=M\cup \{a\}$ and ${\le}_{*}$ is extended so ${S}_{a}({M}_{*})=M$. Clearly ${\le}_{*}$ then defines a well-order on
${M}_{*}$, and $({M}_{*},{\le}_{*})$ would be larger than $(M,{\le}_{M})$ contrary to the definition.

Since $M$ contains all the members of $X$ and ${\le}_{M}$ is a well-ordering of $M$, it is also a well-ordering on $X$ as required.

## The well-ordering theorem implies Hausdorff’s maximum principle.

Let $(X,\u2aaf)$ be a partially ordered set, and let $\le $ be a well-ordering on $X$. We define the function $\varphi $ by transfinite recursion over $(X,\le )$ so that

$$ |

It follows that ${\bigcup}_{x\in X}\varphi (x)$ is a maximal totally ordered subset of $X$ as required.

Title | equivalence of Hausdorff’s maximum principle, Zorn’s lemma and the well-ordering theorem |

Canonical name | EquivalenceOfHausdorffsMaximumPrincipleZornsLemmaAndTheWellorderingTheorem |

Date of creation | 2013-03-22 13:04:45 |

Last modified on | 2013-03-22 13:04:45 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 9 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 03E25 |

Synonym | proof ofZorn’s lemma |

Synonym | proof of Hausdorff’s maximum principle |

Synonym | proof of the maximum principle |

Related topic | ZornsLemma |

Related topic | AxiomOfChoice |

Related topic | ZermelosWellOrderingTheorem |

Related topic | HaudorffsMaximumPrinciple |