One of my favourite bits of point set topology is messing around with easy topological spaces. What could be easier than *finite* topological spaces? The main result (below) is that the category of finite topological spaces is *equivalent* to the category of finite preorders.

Recall (e.g. from algebraic geometry) the following definition:

**Definition.** Let be a topological space. Then the *specialisation preorder* on (the underlying set of) is the relation if and only if .

Note that it is indeed a preorder: clearly , and if and , then , so , showing . We denote this preorder by .

Note that the relation is usually denoted in algebraic geometry, which is pronounced “ specialises to ”.

**Definition.** Given a preorder , the *Alexandroff topology* on is the topology whose opens are the cosieves, i.e. the upwards closed sets (meaning and implies ).

To see that this defines a topology, note that an *arbitrary* (possibly empty) union or intersection of cosieves is a cosieve. A subbase for the topology is given by the *principal cosieves* for any . We denote the set with its Alexandroff topology by .

Likewise, the closed sets in are the sieves (or downwards closed sets); for instance the *principal sieves* . The closure of is the sieve generated by ; for instance the closure of a singleton is the principal sieve .

**Theorem.** *Let be the functor , and the functor .*

*Let be a preorder, a topological space, and**a function. Then is a monotone function if and only if is a continuous function .**The functors and are adjoint: .**The composition is*equal*(not just isomorphic!) to the identity functor.**The restriction of to the category of finite topological spaces is*equal*to the identity functor*.*If is a topological space, then is if and only if is a poset.**If is a preorder, then is a poset if and only if is .**The functors and give rise to adjoint equivalences*

*Proof.* (1) Suppose is monotone, let be a closed subset, and let . Suppose and . Since is monotone and is closed, we get , i.e. . We conclude that , so is downward closed, hence closed in .

Conversely, suppose is continuous, and suppose in . Then , so by continuity we get , so .

(2) This is a restatement of (1): the map

is a bijection.

(3) Since , we conclude that if and only if , so the specialisation preorder on is the original preorder on .

(4) In general, the counit is a continuous map on the same underlying space, so is finer than . Conversely, suppose is closed, i.e. is a sieve for the specialisation preorder on . This means that if , then implies ; in other words . If and therefore is finite, there are finitely many such , so is the finite union

of closed subsets of . Thus any closed subset of is closed in , so the topologies agree.

(5) The relations and mean and . This is equivalent to the statement that a closed subset contains if and only if it contains . The result follows since a poset is a preorder where the first statement only happens if , and a space is a space where the second statement only happens if .

(6) Follows from (5) applied to since by (3).

(7) The equivalence follows from (3) and (4), and the equivalence then follows from (5) and (6).

**Example.** The Alexandroff topology on the poset is the Sierpiński space with topology . As explained in this post, continuous maps from a topological space to are in bijection with open subsets , where is sent to (and to the indicator function ).

**Example.** Let be a set with two elements. There are 4 possible topologies on , sitting in the following diagram (where vertical arrows indicate inclusion bottom to top):

These correspond to 4 possible preorder relations , sitting in the following diagram (where vertical arrows indicate inclusion top to bottom):

We see that finer topologies (more opens) have stronger relations (fewer inequalities).

**Example.** The statement in (4) is false for infinite topological spaces. For instance, if is the Zariski topology on a curve, then any set of closed points is downwards closed, but it is only closed if it’s finite. Or if is a Hausdorff space, then the specialisation preorder is just the equality relation , whose Alexandroff topology is the discrete topology.

I find the examples useful for remembering which way the adjunction goes: topological spaces generally have fewer opens than Alexandroff topologies on posets, so the continuous map should go .

**Remark.** On any topological space , we can define the *naive constructible topology *as the topology with a base given by locally closed sets for open and closed. In the Alexandroff topology, a base for this topology is given by the locally closed sets : indeed these sets are clearly naive constructible, and any set of the form for upward closed and downward closed has the property .

Thus, if is the Alexandroff topology on a preorder, we see that the naive constructible topology is discrete if and only if the preorder is a poset, i.e. if and only if is .