Exercise 1.25

Question

From the last two exercises deduce Stone's theorem, that every Boolean lattice is isomorphic to the lattice of open-and-closed subsets of some compact Hausdorff topological space.

Amit Awasthi · Accepted Answer

\begin{proof}
  Consider the lattice $\mathscr{L}$ of open-and-closed subsets of $\operatorname{Spec} (A(L))$ as a poset under inclusion of subsets $\subseteq$. By Exercise $1.23iv)$, $\operatorname{Spec}(A(L))$ is compact Hausdorff. We claim that the map $\varphi\colon L 	o \mathscr{L}$ sending $f \mapsto X_f$ is an isomorphism of lattices.
  \par We first show $\varphi$ is a bijection. By Exercise $1.23iii)$, $\varphi$ is surjective. $\varphi$ is injective since if $X_f = X_g$, then
  \begin{equation*}
    \emptyset = X_{1-f} \cap X_f = X_{1-f} \cap X_g = X_{(1-f)g}
  \end{equation*}
  using Exercises $1.17i)$ and $1.23i)$, hence by Exercise $1.17ii)$, $(1-f)g \in \mathfrak{N}$. So, $\left((1-f)gight)^m = 0$ for some $m$, and $(1-f)g = 0$ since $A(L)$ is Boolean. Thus, $g = fg$. We can show $f = fg$ using the same argument, and so $g \le f$ and $f \le g$ imply $f = g$.
  \par We now know the map $\varphi^{-1}$ mapping $X_f \mapsto f$ is well-defined, and so by \cite[Thm.~2.3]{BS81}, it suffices to show $\varphi,\varphi^{-1}$ are both order-preserving. Using Exercise $1.17i)$,
  \begin{equation*}
    f \le g \Leftrightarrow f = fg \Leftrightarrow X_f = X_f \cap X_g \Leftrightarrow X_f \subseteq X_g.\qedhere
  \end{equation*}
\end{proof}

Exercise 1.25

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Exercise 1.25

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