TITLE: Topos with enough points but not coherent
QUESTION [8 upvotes]: By Deligne's theorem, each coherent topos has enough points. What would be an example of a Grothendieck topos with enough points which is not coherent?

REPLY [16 votes]: Here are some examples :

For any topological space $X$, the topos of sheaf $\operatorname{Sh}(X)$ has enough points. In most cases this is not a coherent topos. If I remember correctly (for $X$ sober), this is only coherent if $X$ is a spectral space. In any case, spaces like $\mathbb{R}$ are definitely not coherent toposes.

For any topos $\mathcal{T}$, and any set of points $X$ of $\mathcal{T}$, you get geometric morphism $X \to \mathcal{T}$ (where by $X$ I mean the topos $\operatorname{Set}/X$), you can take its image factorisation $X \twoheadrightarrow I \hookrightarrow \mathcal{T}$ and $I$ is a subtopos of $\mathcal{T}$ with enough points. Generally, non coherent toposes don't have a lot of coherent subtoposes (though of course this can happen), so this will often gives example of non-coherent topos with enough topos. There is also a way to do this for $X$ the class of all points of $\mathcal{T}$ despite the size issues.

There is another completeness theorem like Deligne's which says that any "separable" Grothendieck topos has enough points. Separable essentially means that the topos can be defined by a site whose underlying category is countable and whose topology is generated by a countable family of basic covering Sieve. This is done in Makkai and Reyes's book "First order categorical logic" (theorem 6.2.4).


There are many separable toposes that are not coherent. In terms of classifying toposes, coherent means you only use finitary logic, while separable means you can use infinitary logic but the theory should have a countable signature and a countable set of axioms. So none of the two class is included in the other, but I would personally consider that in term of which classical topos they contains, separable toposes form a much larger class.