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 .