Zorn’s Lemma

Definition (Partial Order and Partially Ordered Set).
Let $\mathcal{P}$ be a set equipped with a binary relation $\leq$. We say that $\leq$ is a partial order if it satisfies:

  1. Reflexivity: $x \leq x$ for all $x \in \mathcal{P}$
  2. Antisymmetry: If $x \leq y$ and $y \leq x$, then $x = y$
  3. Transitivity: If $x \leq y$ and $y \leq z$, then $x \leq z$

A set $\mathcal{P}$ together with a partial order is called a partially ordered set (or poset).

A subset $\mathcal{C} \subset \mathcal{P}$ is called a chain if every pair of elements is comparable:
for all $x, y \in \mathcal{C}$, either $x \leq y$ or $y \leq x$.

An element $m \in \mathcal{P}$ is called maximal if there is no element $n \in \mathcal{P}$ such that $m < n$.

Zorn’s Lemma.
Let $\mathcal{P}$ be a partially ordered set in which every chain $\mathcal{C}\subset\mathcal{P}$ has an upper bound in $\mathcal{P}$. Then:

\[\mathcal{P} \text{ contains at least one maximal element.}\]

Zorn’s Lemma is logically equivalent to the Axiom of Choice, and it is a powerful tool for constructing maximal objects.

How to Use Zorn’s Lemma in Practice.
To apply Zorn’s Lemma, follow these steps:

  1. Define a set $\mathcal{P}$ of “candidate” structures.
  2. Introduce a partial order $\leq$ on $\mathcal{P}$.
  3. Prove that every chain has an upper bound in $\mathcal{P}$.
  4. Conclude that $\mathcal{P}$ has a maximal element.

Intuition: Why Zorn’s Lemma is Useful

Zorn’s Lemma gives us a way to ‘build’ something maximal without having to explicitly describe it.
It guarantees the existence of large mathematical objects (fields, ideals, bases) under mild closure properties.

For example:

  • Every nonzero ring has a maximal ideal.
  • Every vector space has a basis.
  • Every field has an algebraic closure.

Zorn’s Lemma formalizes the idea:

“If I can always extend, then I can extend as far as possible.”

Example: Existence of Maximal Ideals.
Let $R$ be a commutative ring with $1 \ne 0$. Then $R$ has a maximal ideal.

Proof Sketch.
Let

\[\mathcal{I} := \{ I \subset R : I \text{ is an ideal and } I \ne R \}\]

ordered by inclusion. Any chain $\mathcal{C} \subset \mathcal{I}$ has an upper bound:

\[I^* := \bigcup_{I \in \mathcal{C}} I\]

which is also a proper ideal. So by Zorn’s Lemma, $\mathcal{I}$ has a maximal element — a maximal ideal.