Hausdorff Maximal Principle/Proof 1

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Theorem

Let $\left({\mathcal P, \preceq}\right)$ be a partially ordered set.

Then there exists a maximal chain in $\mathcal P$.


Proof

Let $S$ be the set of all chains of $\mathcal P$.

$S \ne \varnothing$ since the empty set is an element of $S$.

From Subset Relation is Ordering, we have that $\left({S, \subseteq}\right)$ is partially ordered by inclusion.

Let $C$ be a totally ordered subset of $\left({S, \subseteq}\right)$.

Then $\bigcup C$ is a chain in $C$ by Set of Chains is Chain Complete for Inclusion.

This shows that $S$, ordered by set inclusion, is an inductive ordered subset.

By applying Zorn's Lemma, the result follows.

$\blacksquare$


Axiom of Choice

This proof depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.


Source of Name

This entry was named for Felix Hausdorff.