Hausdorff's Maximal Principle implies Axiom of Choice/Proof

Theorem

Let Hausdorff's Maximal Principle be accepted.

Then the Axiom of Choice holds.

Proof

Let Hausdorff's Maximal Principle be accepted.

Let $S$ be a non-empty set.

Let $\Sigma$ be the set of all choice functions of all non-empty subsets of $S$.

From Countable Set has Choice Function every finite subset of $S$ has a choice function.

Hence $\Sigma$ is a non-empty set.

Lemma

$\Sigma$ is closed under chain unions.

$\Box$

By Hausdorff's Maximal Principle it follows that $\Sigma$ has a maximal element $f$.

Hence $f$ is seen to be a choice function for $S$.

This needs considerable tedious hard slog to complete it. In particular: Fill in the details To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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: Exercise $5.1$