Boolean Prime Ideal Theorem/Proof 2

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Theorem

Let $\struct {S, \le}$ be a Boolean lattice.

Let $I$ be an ideal in $S$.

Let $F$ be a filter on $S$.

Let $I \cap F = \O$.


Then there exists a prime ideal $P$ in $S$ such that:

$I \subseteq P$

and:

$P \cap F = \O$


Proof

Let $Q$ be the set of all ideals of $S$ that are disjoint from $F$.

For each $x \in S$, define:

$C_x = \set {q \in Q : x \in q}$


We want to construct a filter $\FF$ on $Q$ such that:


We can satisfy the first property by letting $\FF = \set {C_x : x \in I}$, but the second may not be satisfied since we could have $C_x \notin \UU$ and also $C_{\neg x} \notin \UU$.

To remedy this, we additionally include $C_x \cup C_{\neg x} \in \FF$ for every $x \in S$.

For then, given any $x \in S$ such that $C_x \notin \UU$, we can intersect its complement with $C_x \cup C_{\neg x}$ to deduce $C_{\neg x} \in \UU$.


The full proof in detail is below.

Propositions

We will first demonstrate several properties of $C_x$.

First, let $x, y \in S$ be arbitrary, with $x \le y$.

For any $q \in C_y$, we have by definition of $C_y$ that:

$y \in q$

But then, by ideal axiom $\paren 1$:

$x \in q$

and so:

$q \in C_x$

Thus:

$\paren 1 \quad x \le y \implies C_y \subseteq C_x$.


Now, let $x, y \in S$ once again be arbitrary.

By definition of the join, we have:

$x \le x \vee y$
$y \le x \vee y$

and so by $\paren 1$:

$C_{x \vee y} \subseteq C_x \cap C_y$

Conversely, let $q \in C_x \cap C_y$ be arbitrary.

Then, $x \in q$ and $y \in q$, so by Ideal is Closed under Join:

$x \vee y \in q$

and so:

$C_x \cap C_y \subseteq C_{x \vee y}$

From the above:

$\paren 2 \quad C_{x \vee y} = C_x \cap C_y$.

Finite Intersection Property

Define:

$\BB = \set {C_x : x \in I} \cup \set {C_x \cup C_{\neg x} : x \in S}$

which will be a sub-basis of $\FF$.

For that to be the case, $\BB$ must have the finite intersection property.


Let:

$G = \set {C_{x_i} : 1 \le i \le n} \cup \set {C_{y_j} \cup C_{\neg y_j} : 1 \le j \le m}$

be an arbitrary finite subset of $\BB$.

Define $\ds X = \bigvee_{i = 1}^n x_i$

By repeatedly applying $\paren 2$ above, we deduce that:

$\ds \bigcap_{i = 1}^n C_{x_i} = C_X$

By Ideal is Closed under Join:

$X \in I$

and hence:

$X \notin F$


Now, we consider the $y_j$.

For an arbitrary $k \in \set {1, \dots, m}$ and $a \notin F$, we can apply the Element Extension Lemma to find some $z_k$ such that:

either $z_k = y_k$ or $z_k = \neg y_k$

and:

$a \vee z_k \notin F$

In particular, we will choose:

$\ds a = X \vee \bigvee_{j = 1}^{k - 1} z_j$

where the $z_j$ have already been selected inductively.


After this process is completed, we will have selected $\sequence {z_j}_{1 \le j \le m}$ such that:

$\forall j: z_j \in \set {y_j, \neg y_j}$
$\ds X \bigvee_{j = 1}^m z_j \notin F$

Define:

$\ds Z = X \bigvee_{j = 1}^m z_j$

so then by Join Succeeds Operands:

$x_i \le X \le Z$
$z_j \le Z$


Let $Z^\le$ denote the lower closure of $Z$.

By Lower Closure of Element is Ideal, we then have:

$Z^\le \in Q$

which by the above remarks satisfies:

$x_i \in Z^\le$
$z_j \in Z^\le$

Therefore:

$Z^\le \in C_{x_i}$
$Z^\le \in C_{z_j} \subseteq C_{y_j} \cup C_{\neg y_j}$

for all $i, j$.

Hence:

$\ds Z^\le \in \bigcap G$

and so the intersection is indeed non-empty.


As $G$ was arbitrary, $\BB$ has the finite intersection property.

$\Box$

Ultrafilter

We can now apply the corollary to the Ultrafilter Lemma to obtain an ultrafilter $\UU$ on $Q$ such that:

$\BB \subseteq \UU$

Define:

$P = \set {x \in S : C_x \in \UU}$

Before proving that $P$ is a prime ideal, we will first demonstrate the other conditions of the theorem.


By definition of $\BB$, for every $x \in I$:

$C_x \in \BB \subseteq \UU$

and so:

$x \in P$

Thus, $I \subseteq P$ as required.


Additionally, since for each $x \in F$:

$C_x = \O$

we cannot have $C_x \in \UU$ be definition of an ultrafilter.

Hence, $P \cap F = \O$.

$\Box$

Prime Ideal

It only remains to show that $P$ is a prime ideal.

We will first show that $P$ is an ideal.

Since $P \supseteq I \ne \O$, it follows that:

$P \ne \O$

which is required by the definition of an ideal.


Let $x, y \in S$ such that $y \le x$.

If:

$x \in P$

we have by $\paren 1$ above that:

$C_y \supseteq C_x \in \UU$

and hence by filter on set axiom $\paren {\text F 4}$:

$y \in P$

which is ideal axiom $\paren 1$.


Let $x, y \in P$.

Then, $C_x, C_y \in \UU$.

By filter on set axiom $\paren {\text F 3}$, it follows that:

$C_x \cap C_y \in \UU$

But, by $\paren 2$ above:

$C_{x \vee y} = C_x \cap C_y \in \UU$

and hence:

$x \vee y \in P$

which satisfies ideal axiom $\paren 2$.

Hence, $P$ is an ideal.


Moreover, by definition of filter:

$F \ne \O$

so since:

$P \cap F = \O$

it follows that:

$P \ne S$

Therefore, $P$ is a proper ideal.


Lastly, we will show that, for each $x \in S$, either:

$x \in P$

or:

$\neg x \in P$

If $x \in S \setminus P$ is given, it follows that:

$C_x \notin \UU$

and so by definition of an ultrafilter:

$Q \setminus C_x \in \UU$


We also have:

$C_x \cup C_{\neg x} \in \BB \subseteq \UU$

so we can apply filter on set axiom $\paren {\text F 3}$ to find:

$\paren {Q \setminus C_x} \cap \paren {C_x \cup C_{\neg x}} \in \UU$

or equivalently:

$C_{\neg x} \setminus C_x \in \UU$


But for no ideal $q \in Q$ can we have:

$\set {\neg x, x} \subseteq q$

for by Ideal is Closed under Join and the definition of complement:

$\top \in q$

which contradicts $q \cap F = \O$ by Top in Filter.


Therefore:

$C_{\neg x} = C_{\neg x} \setminus C_x \in \UU$

and thus:

$\neg x \in P$

As $x \in S$ was arbitrary, we can apply Proper Ideal is Prime iff Contains Element or Complement, from which the theorem follows.

$\blacksquare$

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