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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?

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Is there a good reference for the result that the ($\infty$, 1)-topos of sheaves on $X$ is Boolean iff $X$ satisfies "open=closed"? – Noah Schweber Jan 31 '13 at 3:43

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.

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Thanks! Can you point me to an argument for why it is not hypercomplete? – Mike Shulman Jan 31 '13 at 3:15
Ben Wieland showed me an argument that $B\mathbf{Z}_p$ is not hypercomplete; a sketch of his argument (which uses the Sullivan conjecture) is reproduced as Warning in "Higher Topos Theory". And $\mathbf{Z}_p$ is about as tame as profinite groups get (the topos even has finite cohomological dimension). – Jacob Lurie Jan 31 '13 at 5:47
Can you say any more about why the morphism $\alpha$ in is $\infty$-connective? Why does it follow from the displayed map being a homotopy equivalence, and why is that map a homotopy equivalence? (Also, is there maybe a typo in the codomain of that map?) – Mike Shulman Jun 20 '14 at 6:59

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