Tukey's Lemma implies Zorn's Lemma

Theorem

Let Tukey's Lemma be accepted as true.

Then Zorn's Lemma holds.

Proof

Recall Tukey's Lemma:

Let $S$ be a non-empty set of finite character.

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

$\Box$

Recall Zorn's Lemma:

Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set.

Let $T \subseteq \powerset S$ be the set of subsets of $S$ that are totally ordered by $\preceq$.

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

$\Box$

So, let us assume Tukey's Lemma.

Let $S$ be a non-empty ordered set, with $T$ as defined.

From Property of being Totally Ordered is of Finite Character:

$T$ is of finite character.

From Ordering on Singleton is Total Ordering it follows trivially that $T$ is non-empty.

Then by Tukey's Lemma:

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

Thus it is seen that Zorn's Lemma likewise 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 $5.7$