TITLE: Boolean non-hypercomplete $(\infty,1)$-toposes
QUESTION [10 upvotes]: Let's say that an $(\infty,1)$-topos is Boolean if for every object $X$, the lattice $Sub(X)$ of subobjects (i.e. $(-1)$-truncated morphisms into $X$) is a Boolean algebra.  I think this is equivalent to asking that the subobject classifier be an internal Boolean algebra, and hence to asking that the underlying 1-topos of 0-truncated objects is Boolean in the classical sense.
In particular, the $(\infty,1)$-topos of sheaves on a topological space $X$ is Boolean if and only if the lattice of open sets in $X$ is a Boolean algebra, i.e. every open set is also closed.  Thus, Booleanness is a sort of "zero-dimensionality" condition.  Lurie shows in Higher Topos Theory that other sorts of "finite-dimensionality" conditions imply that an $(\infty,1)$-topos is hypercomplete.  At the moment, however, I don't see whether Booleanness implies any of these other conditions.
Thus my question is: can a Boolean $(\infty,1)$-topos fail to be hypercomplete?

REPLY [10 votes]: If $G$ is a profinite group, then the topos of sets with a continuous $G$-action is
Boolean, but the associated $\infty$-topos is usually not hypercomplete.