Boolean Algebras, Stone Duality, and the Prime Ideal Theorem: Examples and Counterexamples

The nonzero elements are the nonempty subsets. Each singleton $\{s\}$ is minimal nonzero; a subset with two distinct elements has a nonempty proper singleton subset, so is not an atom. Thus the atoms are exactly the singletons. F1 now lists all ultrafilters as $U_s$, uniquely for $s\in S$.

For $A\subseteq S$, $U_s\in [A]$ exactly when $A\in U_s$, exactly when $s\in A$. Therefore $[A]=\{U_s:s\in A\}$ and $[\{s\}]=\{U_s\}$. Every singleton in the Stone space is open, so every subset is open. For the explicit instance $S=\{0,1\}$, the four algebra elements give respectively $[\mathrm{\varnothing }]=\mathrm{\varnothing }$, $[\{0\}]=\{U_0\}$, $[\{1\}]=\{U_1\}$ and $[S]=\{U_0,U_1\}$.

If $S=\mathrm{\varnothing }$, the algebra has one element and no atoms; F1 gives no ultrafilters, and the empty map identifies the two empty spaces. If $S$ has one point, step 2.1 gives a one-point discrete Stone space with exactly two clopens. This verifies the claimed identification in every finite case. QED.

The family $I$ contains zero, is downward closed and is closed under union; it omits $\{0,1,2\}$, so is proper. The condition $A\mathrm{△}{A}^{\mathrm{\prime }}\subseteq \{2\}$ is equivalent to $A\cap \{0,1\}=A^{\mathrm{\prime }}\cap \{0,1\}$: membership at $0,1$ must agree, while membership at $2$ is unrestricted. The four classes are $\{\mathrm{\varnothing },\{2\}\}$, $\{\{0\},\{0,2\}\}$, $\{\{1\},\{1,2\}\}$ and $\{\{0,1\},\{0,1,2\}\}$.

Define $r(A)=A\cap \{0,1\}$. It preserves unions and intersections, sends the two bounds to $\mathrm{\varnothing }$ and $\{0,1\}$, and satisfies $r(\{0,1,2\}\setminus A)=\{0,1\}\setminus r(A)$. Its zero fibre is $I$. F2 factors it through $B\mathrm{/}I$; step 1.1 shows the factor sends the four classes bijectively to $\mathrm{\varnothing },\{0\},\{1\},\{0,1\}$. Thus it is the asserted Boolean isomorphism.

The atoms of each powerset algebra are its singletons, so F3 identifies its ultrafilters with point filters. For $i=0,1$, the inverse image of the point filter $V_i$ in $\mathcal{P}(\{0,1\})$ is $r^{-1}[V_i]=\{A:i\in r(A)\}=\{A:i\in A\}=U_i$ in $B$. Its image contains exactly $U_0,U_1$, omitting $U_2$. By F4 the preimage of any basic clopen $[A]$ is the set of those $V_i$ with $i\in A\cap \{0,1\}$, explicitly the clopen $[r(A)]$. This is the stated two-point inclusion with its discrete topology. QED.

If $G$ is a forcing filter, nonemptiness and upward closure put $S\in G$. For $A,B\in G$, directedness supplies a nonempty $C\in G$ with $C\subseteq A\cap B$. Thus $A\cap B$ is a condition, and upward closure puts it in $G$. This proves Boolean filter closure, while $\mathrm{\varnothing }\mathrm{\notin }G$ holds because $G\subseteq P$. Conversely a proper Boolean filter contains $S$, hence is nonempty, and its meet $A\cap B\in G$ is a nonempty common stronger condition. The two notions therefore agree here.

For $S=\{0,1\}$ the conditions are $a=\{0\}$, $b=\{1\}$ and $t=\{0,1\}$, with $a,b\subseteq t$. A filter must contain $t$. It cannot contain both $a,b$, since $a\cap b=\mathrm{\varnothing }$ and there is no common stronger condition. The three possibilities are consequently $\{t\}$, $\{a,t\}$ and $\{b,t\}$, and each satisfies F1. The last two are maximal; adding either singleton to $\{t\}$ gives one of them.

The maximal families decide each of the four complementary pairs in the Boolean algebra, as F2 also predicts. The empty family fails F1, although its upward and pairwise-directed conditions alone would be vacuous. If $S$ is a singleton there is only the condition $t$ and the filter $\{t\}$. If $S=\mathrm{\varnothing }$, $P$ is empty and is excluded by the definition of forcing preorder; there is also no proper Boolean filter on $\mathcal{P}(S)$. Thus the correspondence spends no choice and supplies no nonexistent meets on arbitrary preorders. QED.