Stone representation for Boolean algebras

Derive the dichotomy. Let $U$ be an ultrafilter and $a\mathrm{\notin }U$. If $u\wedge a\mathrm{\ne }0$ for every $u\in U$, then $V=\{b:(u\wedge a)\le b\text{ for some }u\in U\}$ is a proper filter containing $U$ and $a$. It contains 1 and is upward closed; if two witnesses are $u,v$, their meet has witness $u\wedge v\in U$. It omits 0 by the assumed nonzero meets. Thus it strictly extends $U$, contradicting maximality. Hence some $u\in U$ satisfies $u\wedge a=0$. Distributivity gives $u=u\wedge (a\vee \mathrm{\neg }a)=(u\wedge a)\vee (u\wedge \mathrm{\neg }a)=u\wedge \mathrm{\neg }a$, so $u\le \mathrm{\neg }a$ and $\mathrm{\neg }a\in U$. Both $a$ and $\mathrm{\neg }a$ cannot belong to a proper filter, since their meet is 0. This proves exactly one belongs, without attributing the claim to the extension interface.

Injectivity. If $a\mathrm{\ne }b$, [F2] supplies an ultrafilter in exactly one of $[a],[b]$, so these subsets differ. Thus $b\mapsto [b]$ is injective.

Boolean operations and basis. Meet closure and upward closure give $[a\wedge b]=[a]\cap [b]$. If $a\vee b\in U$ but neither summand belongs to $U$, step 1.1 puts both complements in $U$; their meet has zero meet with $a\vee b$ by distributivity, contradicting properness. Conversely either summand in $U$ puts their join in $U$. Thus $[a\vee b]=[a]\cup [b]$. Also $[\mathrm{\neg }a]=X\setminus [a]$, $[0]=\mathrm{\varnothing }$ and $[1]=X$. Hence the map preserves all Boolean operations and every $[a]$ is clopen. The generating family contains $X$ and is closed under finite intersections, so it is a basis for the generated topology.

Hausdorff separation. Distinct ultrafilters cannot be properly included in each other, by their maximality among proper filters. Thus for $U\mathrm{\ne }V$ some $b\in U\setminus V$ exists. Step 1.1 gives $\mathrm{\neg }b\in V$, so the disjoint open sets $[b]$ and $[\mathrm{\neg }b]$ separate them.

Compactness for basic covers. Suppose $\{[b_i]:i\in I\}$ covers $X$ but has no finite subcover. For every finite $J\subseteq I$ there exists an ultrafilter outside its union, so $c_J={\bigwedge }_{i\in J}\mathrm{\neg }{b}_i$ lies in a proper filter and is nonzero, using step 1.1; for $J=\mathrm{\varnothing }$ this uses $c_J=1$ and the fact that failure of the empty subcover implies $X\mathrm{\ne }\mathrm{\varnothing }$. The family $F=\{b:c_J\le b\text{ for some finite }J\subseteq I\}$ is a proper filter: it contains 1, is upward closed, is meet-closed using $J\cup J^{\mathrm{\prime }}$, and omits 0 because every $c_J\mathrm{\ne }0$. By [F2] extend it to an ultrafilter $U$. It contains every $\mathrm{\neg }{b}_i$, hence no $b_i$, contradicting the cover. No simultaneous choice of witnessing ultrafilters for the finite J is made.

General covers and clopens. Any open cover has the refinement of all basic sets contained in one of its members. This covers by step 2.1, so step 3.2 yields finitely many basic sets. For these finitely many sets choose containing original members, yielding a finite original subcover (finite choice, available under AC). Thus $X$ is compact. If $C\subseteq X$ is clopen, [F3] makes $C$ compact. The basic sets contained in $C$ cover it, so finitely many suffice and $C=[b_1]\cup \cdots \cup [b_r]=[b_1\vee \cdots \vee b_r]$ by step 2.1. When $C=\mathrm{\varnothing }$ take the empty finite join, namely 0. Thus the Boolean homomorphism is onto all clopens.

Conclusion and trivial case. By steps 1.2, 2.1 and 4.1 the map is a bijective Boolean homomorphism; its inverse preserves the operations by applying injectivity to each homomorphism identity. By steps 2.1, 3.1 and 4.1 the space is compact Hausdorff with a clopen basis. If $0=1$ no proper filter can both contain 1 and omit 0, so $X=\mathrm{\varnothing }$ and its only clopen is $\mathrm{\varnothing }$; the asserted isomorphism is the unique map between one-element Boolean algebras. Every cover of the empty space has the empty finite subcover. The extension supplier uses AC/Zorn, and this proof uses no product-Tychonoff argument.