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[parent] Zorn's lemma and the well-ordering theorem equivalence of Hausdorff's maximum principle (Proof)

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

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

According to Zorn's lemma, $\A$ 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

$\displaystyle \phi(a) = \begin{cases} \{a\} &\text{if }\{a\}\cup\bigcup_{b<a}\p... ...is totally ordered under }\preceq.\ \emptyset &\text{otherwise}. \end{cases}.$
It follows that $\bigcup_{x\in X}\phi(x)$ is a maximal totally ordered subset of $X$ as required.




"Zorn's lemma and the well-ordering theorem equivalence of Hausdorff's maximum principle" is owned by mathcam. [ full author list (2) | owner history (1) ]
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See Also: Zorn's lemma, axiom of choice, Zermelo's well-ordering theorem, Hausdorff's maximum principle

Other names:  proof ofZorn's lemma, proof of Hausdorff's maximum principle, proof of the maximum principle

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Cross-references: transfinite recursion, function, members, contains, postulate, Zorn's lemma, union, transitivity, symmetry, partial order, easy to see, iff, relation, well-ordering, collection, maximal element, contained, totally ordered set, element, subset, totally ordered, maximum principle, upper bound, chain, partially ordered set
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This is version 6 of Zorn's lemma and the well-ordering theorem equivalence of Hausdorff's maximum principle, born on 2002-09-29, modified 2007-06-24.
Object id is 3493, canonical name is ZornsLemmaAndTheWellOrderingTheoremEquivalenceOfHaudorffsMaximumPrinciple.
Accessed 13146 times total.

Classification:
AMS MSC03E25 (Mathematical logic and foundations :: Set theory :: Axiom of choice and related propositions)

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