Hausdorff Maximal Principle

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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 1

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.

$\blacksquare$


Proof 2

Let $\preceq$ be an ordering on the set $\mathcal P$.

Let $X$ be a chain in $\left({\mathcal P, \preceq}\right)$.

By definition, a maximal chain in $\mathcal P$ that includes $X$ is a chain $Y$ in $\mathcal P$ such that $X \subseteq Y$ and there is no chain $Z$ in $\mathcal P$ with $X \subseteq Z$ and $Y \subsetneq Z$.

Let us define $\mathcal C$ as:

$\mathcal C = \left\{{Y : Y}\right.$ is a chain in $\mathcal P$ and $\left.{X \subseteq Y}\right\}$

Then:

  • a maximal chain in $\mathcal P$ that includes $X$ and
  • a maximal element of $\mathcal C$ under the partial order induced on $\mathcal C$ by inclusion

are one and the same.

It thus suffices to show that $\mathcal C$ contains such a maximal element.


According to Zorn's Lemma, $\mathcal C$ contains a maximal element if each chain in $\mathcal C$ has an upper bound in $\mathcal C$.

Let $W$ be a chain in $\mathcal C$, and let $Z = \bigcup W$.

Let $a, b \in Z$.

Then $\exists A, B \in W: a \in A, b \in B$.

Since $W$ is a chain in $\mathcal C$, one of $A$ and $B$ includes the other.

Suppose WLOG $A \subseteq B$.

Then $a, b \in B$.

Since $B$ is a chain in $\mathcal P$, either $a \preceq b$ or $b \preceq a$.

Therefore $Z$ is a chain in $\mathcal P$.

By definition of $Z$, we have that $\forall A \in W: A \subseteq Z$.

Thus $Z$ is an upper bound for $W$ under inclusion.

Finally we have that:

$\forall A \in W: X \subseteq A$

and so $X \subseteq Z$.

Thus $Z \in \mathcal C$

$\blacksquare$


Proof 3

Let $\left({\mathcal C, \subseteq}\right)$ be the set of all chains in $P$ ordered by inclusion.

By Set of Chains is Chain Complete for Inclusion, $\mathcal C$ is a chain complete ordered set.

Now define $f: \mathcal C \to \mathcal C$ as follows:

If $C$ is a maximal chain then $f \left({C}\right) = C$.
Otherwise $f$ chooses arbitrarily, using the axiom of choice, some chain $D$ which strictly contains $C$.

By construction, $f \left({C}\right) \supseteq C$.

By the above, $\mathcal C$ is chain complete.

Therefore the Bourbaki-Witt Fixed Point Theorem applies and $f$ has a fixed point $F \left({M}\right) = M$.

But by the above construction, the only fixed points of $f$ are maximal chains.

Therefore $M$ is a maximal chain.

$\blacksquare$


Axiom of Choice

This theorem depends on the Axiom of Choice.

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.


Sources