Hausdorff's Maximal Principle implies Kuratowski's Lemma/Proof 1

Theorem

Let Hausdorff's Maximal Principle be accepted as true.

Then Kuratowski's Lemma holds.

Proof

Recall Hausdorff's Maximal Principle:

Let $A$ be a non-empty set of sets.

Let $S$ be the set of all chain of sets of $A$ (ordered under the subset relation).

Then every element of $S$ is a subset of a maximal element of $S$ under the subset relation.

$\Box$

Recall Kuratowski's Lemma:

Let $S$ be a set of sets which is closed under chain unions.

Then every element of $S$ is a subset of a maximal element of $S$ under the subset relation.

$\Box$

So, let us assume Hausdorff's Maximal Principle.

Let $S$ be a non-empty set of sets which is closed under chain unions.

$\set b$ is trivially a chain.

Hence by Hausdorff's Maximal Principle $\set b$ is a subset of a maximal chain of elements of $S$.

Hence $b$ is an element of a maximal chain $C$ of elements of $S$.

We have by hypothesis that $S$ is closed under chain unions.

Hence $\bigcup C \in S$.

Then $b \subseteq \bigcup C$.

If $\bigcup C$ were a proper subset of some element $x \in S$, $C$ would be a proper subset of the chain $C \cup \set x$.

But $C$ is maximal so that cannot happen.

Hence $\bigcup C$ is maximal .

Thus it is seen that Kuratowski's Lemma holds.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.10$