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
(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
(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):
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
.