Complete Boolean algebra has dimension zero

It suffices to show that every \((-1)\)-connected object \(X\in\Shv(\Lambda)\) admits a global section. We construct, by transfinite induction, a strictly increasing family of elements \(U_\alpha\in\Lambda\), together with compatible sections \(s_\alpha\in X(U_\alpha)\), for as long as \(U_\alpha\neq\top\). Start with \(U_0=\bot\) and its unique section. At a limit ordinal \(\lambda\), set \(U_\lambda=\bigvee_{\alpha<\lambda}U_\alpha\). The compatible sections glue uniquely to a section over \(U_\lambda\) by the sheaf condition.Suppose that \(U_\alpha<\top\), and let \(V=U_\alpha^c\). Since \(X\to *\) is an effective epimorphism, there is a covering \(V=\bigvee_iV_i\) such that \(X(V_i)\) is nonempty for every \(i\). Choose a nonzero \(V_i\) and a section \(t_i\in X(V_i)\). The intersection \(U_\alpha\land V_i\) is initial, so \(s_\alpha\) and \(t_i\) agree there. They therefore glue to a section over \(U_{\alpha+1}:=U_\alpha\lor V_i\), and this element strictly contains \(U_\alpha\).If this process never reached \(\top\), it would define a strictly increasing map from the class of ordinals to the underlying set of \(\Lambda\), which is impossible. Hence some \(U_\alpha\) is \(\top\), and the corresponding section is a global section of \(X\).