# nLab Hausdorff maximal principle

The Hausdorff maximal principle is a version of Zorn's lemma, equivalent to the usual version and thus (given excluded middle) equivalent to the axiom of choice.

## Statement and proofs

Given a poset (or proset) $S$, let a chain in $S$ be a subset $A$ of $S$ which, as a sub-proset, is totally ordered. A chain $A$ is maximal (as a chain) if the only chain that $A$ is contained in is $A$ itself.

###### Theorem (Hausdorff maximal principle)

Every chain in a proset is contained in a maximal chain.

###### Proof

We will use Zorn's lemma. Let $P$ be a proset and let $C\subseteq P$ be a chain. Consider the collection $𝒞$ of chains in $P$ that contain $C$, ordered by inclusion. If $\left\{{C}_{\alpha }{\right\}}_{\alpha \in A}\subseteq 𝒞$ is a family totally ordered by inclusion, then the union ${\bigcup }_{\alpha }{C}_{\alpha }$, with the order coming from $P$, is also totally ordered: any two elements $x\in {C}_{\alpha },y\in {C}_{\beta }$ are comparable in $\mathrm{max}\left({C}_{\alpha },{C}_{\beta }\right)$. The hypotheses for Zorn’s lemma therefore obtain on $𝒞$, and we conclude that $𝒞$ has a maximal element, which is clearly maximal in the collection of all chains.

###### Proof of converse

Conversely, suppose that the Hausdorff maximal principle holds; we will prove Zorn’s lemma. Suppose given a poset (or preorder) $P$ such that every chain in $P$ has an upper bound. Since $\varnothing$ is a chain, the Hausdorff maximal principle implies that $P$ contains a maximal chain $C$; let $x$ be an upper bound of $C$. Then $x$ is maximal: if $x\le y$, then $C=C\cup \left\{y\right\}$ by maximality of $C$; therefore $y\in C$ and hence $y\le x$ since $x$ is an upper bound of $C$.