Boolean Prime Ideal Theorem
there exists a prime ideal $P$ in $S$ such that $I \subseteq P$ and $I \cap F = \varnothing$.
Proof from the Axiom of Choice
Let $T$ be the set of ideals in $S$ that contain $I$ and are disjoint from $F$, ordered by inclusion.
Let $N$ be a chain in $T$.
Then $U=\bigcup N$ is clearly disjoint from $F$ and contains $I$. It is also an ideal:
- Suppose $x\in U$ and $y \le x$.
- Then for some $A \in N$, $x \in A$.
- By the definition of union, $x \in U$.
- Suppose $x \in U$ and $y \in U$.
- Then for some $A,\,B \in N$, $x \in A$ and $y\in B$.
- By the definition of a chain, $A \subseteq B$ or $B \subseteq A$.
Suppose without loss of generality that $A \subseteq B$.
- Then $x \in B$. Since $y$ is also in $B$, and $B$ is an ideal,
- $x \vee y \in U$.
By Zorn's Lemma, $T$ has a maximal element, $M$.
It remains to show that $M$ is a prime ideal:
Every Boolean algebra is a distributive lattice, so by Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice, $M$ is a prime ideal.
Axiom of Choice
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.
Proof from the Ultrafilter Lemma
The Boolean Prime Ideal Theorem is weaker than the Axiom of Choice, but is similarly independent of ZF theory. It is sufficient to prove a number of important theorems, although such proofs are often more involved than ones relying on the Axiom of Choice.