# Sober Topological Space and its associated site

I was reading recently for Grothendieck topologies on an arbitrary category $\mathcal{C}$ (i.e. sites) and although did understand a couple of things I read something on Wiki article https://en.wikipedia.org/wiki/Grothendieck_topology for Grothendieck topologies which didn't get.

If $X$ is a sober topological space, says, that we can recover the topological space again from its associated site. What does this sentence mean? How do we recover $X$ and under which construction does this happen?

Thank you!

Given a topological space $T$, we can form the poset (seen as a category) $\mathcal O(T)$ of its open subsets. This can be made into a site by making the covers of an object $U$ (that is, an open subset of $T$) precisely those collections of subsets of $U$ whose union is $U$. (This is spelled out on the Wikipedia page in the sections "Motivation" and "Sieve".) However, the site structure on this category does not even matter.

Recovering in this case means that if we start with a sober space $T$, produce the category $\mathcal O(T)$, and then forget where this category came from, we can reconstruct (a space homeomorphic to) $T$. Abstractly, this just means that the "mapping" $T \mapsto \mathcal O(T)$ is "injective", but more specifically in this case we have a construction to find (a space homeomorphic to) $T$.

Sobriety states that every irreducible closed set (cannot nontrivially be written as the union of two closed sets) has a unique generic point; that is, a unique point whose closure is that closed set. Clearly, each point is the generic point of its own closure. Hence, the mapping that takes a point to its closure provides a bijection between the points of a space and the irreducible closed subsets.

Now, the complement of an irreducible closed subset is precisely an open set $U \subsetneq T$ such that there are no open subsets $V,W$, both distinct from $U$, such that $V = V \cap W$. Note that we can recognize this condition just from the poset structure of $\mathcal O(T)$, even if we forget they are subsets of a topological space: $V \cap W$ is the greatest lower bound of $V$ and $W$. The greatest lower bound is also called the "meet", which is why in any poset which has all meets, these objects are called meet-irreducible. Thus, given a category $\mathcal C$, which we know is the category of open subsets of a sober space, we form a new set $T$ whose points are in one-to-one correspondence with the meet-irreducible objects of $\mathcal C$. Then, we form a topology on the set $T$ as follows: we make an open subset for each object of $\mathcal C$, whose points are those corresponding to the meet-irreducible objects which are not above (in the partial order sense) that object.

(Originally, this answer incorrectly claimed that irreducible closed sets in $T$ correspond to open subsets $U$ of $T$ for which, if $V$ is open and $U \subseteq V \subsetneq T$ then $U = V$. In fact, these are generally only some of the irreducible closed sets. Thanks to Alex Kruckman in the comments for pointing out the mistake.)

Then $T$ is homeomorphic to the sober space of which $\mathcal C$ was the category of open subsets. The homeomorphism takes a point of $T$ to the generic point of the complement of its open set.

• Excellent Mees, +1 for your beautiful answer! Thank you! – user321268 Aug 26 '16 at 14:59
• @mayer_vietoris This is a nice answer, except that the description of the space associated to $O(T)$ is not correct. The complement of an irreducible closed set is not necessarily almost maximal. Otherwise, there could never be a containment between irreducible closed sets - think of the Zariski topology, in which the whole space is closed irreducible, but is complement (the empty set) is properly contained in lots of open sets that aren't the whole space. Instead, the points of $T$ are in bijection with the meet irreducible opens - the open sets which can be expressed as an intersection of – Alex Kruckman Aug 29 '16 at 22:17
• ... two strictly larger open sets. – Alex Kruckman Aug 29 '16 at 22:18
• @AlexKruckman, thank you for pointing out the mistake. My bad -- I got cocky thinking I remembered it correctly without reconsidering the details. The answer has been corrected. – Mees de Vries Aug 30 '16 at 15:07
• "both distinct from $U$, such that $V = V \cap W$" Do you mean $U = V \cap W$ here? – B. Mehta 2 days ago