Spectral space

Let X be a topological space and let K^{${\displaystyle \circ }$}(X) be the set of all compact open subsets of X. Then X is said to be spectral if it satisfies all of the following conditions:

• X is compact.
• K^{${\displaystyle \circ }$}(X) is a basis of open subsets of X.
• K^{${\displaystyle \circ }$}(X) is closed under finite intersections.
• X is sober, i.e., every nonempty irreducible closed subset of X has a unique generic point.

From that X is sober it follows that X is T₀. Indeed the definition of a spectral space can be equivalently reformulated through explicitly assuming that X is T₀ and weaking the assumption that X is sober to only require it to be quasi-sober, i.e. every irreducible closed subspace possesses a (not nececssarily unique) generic point. This is the way the definition is formulated in Hochster's 1967 thesis.

Let X be a topological space. Each of the following properties are equivalent to the property of X being spectral:

1. X is homeomorphic to a projective limit of finite T₀ spaces.
2. X is homeomorphic to the spectrum of a bounded distributive lattice L. In this case, L is isomorphic (as a bounded lattice) to the lattice K^{${\displaystyle \circ }$}(X) (this is called Stone representation of distributive lattices).
3. X is homeomorphic to the spectrum of a commutative ring.
4. X is the topological space determined by a Priestley space.
5. X is a T₀ space whose locale of open sets is coherent (and every coherent locale comes from a unique spectral space in this way).

Let X be a spectral space and let K^{${\displaystyle \circ }$}(X) be as before. Then:

• K^{${\displaystyle \circ }$}(X) is a bounded sublattice of subsets of X.
• Every closed subspace of X is spectral.
• An arbitrary intersection of compact and open subsets of X (hence of elements from K^{${\displaystyle \circ }$}(X)) is again spectral.
• X is T₀ by definition, but in general not T₁.^[1] In fact a spectral space is T₁ if and only if it is Hausdorff (i.e. T₂) if and only if it is a boolean space if and only if K^{${\displaystyle \circ }$}(X) is a boolean algebra.
• X can be seen as a pairwise Stone space.^[2]

A spectral map f: X → Y between spectral spaces X and Y is a continuous map such that the preimage of every open and compact subset of Y under f is again compact.

The category of spectral spaces, which has spectral maps as morphisms, is dually equivalent to the category of bounded distributive lattices (together with homomorphisms of such lattices).^[3] In this anti-equivalence, a spectral space X corresponds to the lattice K^{${\displaystyle \circ }$}(X).