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 $(A,\leq)$, where $A\subseteq X$ and $\leq$ is a well-ordering on $A$. Define a relation   $\preceq$, on $\mathcal{A}$ so that for all $x,y\in\mathcal{A}:x\preceq y$ 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 $C^{\prime}=(R,\leq^{\prime})$ where R is the union of all the sets $A$ for all $(A,\leq)\in C$, and $\leq^{\prime}$ is the union of all the relations $\leq$ for all $(A,\leq)\in C$. 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, $(M,\leq_{M})$. We postulate  that $M$ contains all members of $X$, for if this were not true we could for any $a\in X-M$ construct $(M_{*},\leq_{*})$ where $M_{*}=M\cup\{a\}$ and $\leq_{*}$ is extended so $S_{a}(M_{*})=M$. Clearly $\leq_{*}$ then defines a well-order on $M_{*}$, and $(M_{*},\leq_{*})$ would be larger than $(M,\leq_{M})$ contrary to the definition.

Since $M$ contains all the members of $X$ and $\leq_{M}$ 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 $(X,\preceq)$ be a partially ordered set, and let $\leq$ be a well-ordering on $X$. We define the function $\phi$ by transfinite recursion over $(X,\leq)$ so that

 $\phi(a)=\begin{cases}\{a\}&\text{if }\{a\}\cup\bigcup_{b

It follows that $\bigcup_{x\in X}\phi(x)$ 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