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In the following we fix a topos. I'll speak of sets instead of objects and of subsets instead of subobjects.

Let $X$ be a set and assume $F$ is a set of subsets of $X$ that contains $\emptyset, X$, is closed under finite unions and arbitrary intersections. Thus you have a sort of topology given by closed sets. So let's define $U \subseteq X$ open if its complement is closed in $X$, i.e. is contained in $F$. Now clearly $\emptyset,X$ are open and since we have

$X \setminus \cup_{i \in I} U_i = \cap_{i \in I} X \setminus U_i$,

the open subsets are closed under unions. But we only have

$X \setminus U \cup X \setminus V \subseteq X \setminus (U \cap V)$,

right? In intuitionistic logic, we only have $(\neg \phi \vee \neg \psi) \rightarrow \neg (\phi \wedge \psi)$ and the converse does not hold in general. So it seems that we cannot prove that the intersection of two open subsets is open again. Wah!

Is it possible to correct this? Is there a construction or an assumption on the closure operator, which is, say, automatic in the topos of sets, and yields a honest topology?

share|cite|improve this question
@Martin: Is there any reason to believe that an honest topology will necessarily exist, say, for simplicial commutative rings? – Harry Gindi Aug 6 '10 at 12:43
Martin, have you looked at Toen-Vezzosi's HAG I? If I remember correctly, they give a very nice generalization of a lot of the commutative algebra relevant to algebraic geometry by looking at monoidal model categories, which might be relevant. – Harry Gindi Aug 6 '10 at 14:34

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