If is the topology on a topological space , and if a map that preserves finite meets and arbitrary joins is considered an instance of “seeing a point ”, then is “sober” if every point we see is really there (i.e., is induced from a point = continuous map ), and if we never see double.
A topological space has enough points if and only if its quotient is sober. Spaces with enough points are also reflective, and a topological space is iff this reflection is sober.
Examples (and non-examples)
Any nontrivial indiscrete space is not sober, since it is not . More interestingly, the space with the Zariski topology is but not sober, since every subvariety is an irreducible closed set which is not the closure of a point. Its soberification is, unsurprisingly, the scheme , which contains “generic points” whose closures are the subvarieties.
The last (non-)example shows that sobriety is not a hereditary separation property, i.e., subspaces of sober spaces need not be sober.
The Alexandroff topology on a poset is also not, in general, sober. For instance, if is the infinite binary tree (the set of finite -words (lists) with the “extends” preorder), then the soberification of its Alexandroff topology is Wilson space?, the space of finite or infinite -words (streams).