Finite topological spaces

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 X be a topological space. Then the specialisation preorder on (the underlying set of) X is the relation x \leq y if and only if x \in \overline{\{y\}}.

Note that it is indeed a preorder: clearly x \leq x, and if x \leq y and y \leq z, then \{y\} \subseteq \overline{\{z\}}, so x \in \overline{\{y\}} \subseteq \overline{\{z\}}, showing x \leq z. We denote this preorder by X^{\operatorname{sp}}.

Note that the relation x \leq y is usually denoted y \rightsquigarrow x in algebraic geometry, which is pronounced “y specialises to x”.

Definition. Given a preorder (X,\leq), the Alexandroff topology on X is the topology whose opens U \subseteq X are the cosieves, i.e. the upwards closed sets (meaning x \in U and x \leq y implies y \in U).

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 X_{\geq x} = \{y \in X\ |\ y \geq x\} for any x \in X. We denote the set X with its Alexandroff topology by X^{\operatorname{Alex}}.

Likewise, the closed sets in X are the sieves (or downwards closed sets); for instance the principal sieves X_{\leq x} = \{y \in X\ |\ y \leq x\}. The closure of S \subseteq X is the sieve X_{\leq S} = \bigcup_{s \in S} X_{\leq s} generated by S; for instance the closure of a singleton \{x\} is the principal sieve X_{\leq x}.

Theorem. Let F \colon \mathbf{PreOrd} \to \mathbf{Top} be the functor X \mapsto X^{\operatorname{Alex}}, and G \colon \mathbf{Top} \to \mathbf{PreOrd} the functor Y \mapsto Y^{\operatorname{sp}}.

  1. Let (X,\leq) be a preorder, Y a topological space, and f \colon X \to Y a function. Then f is a monotone function X \to Y^{\operatorname{sp}} if and only if f is a continuous function X^{\operatorname{Alex}} \to Y.
  2. The functors F and G are adjoint: F \dashv G.
  3. The composition GF \colon \mathbf{PreOrd} \to \mathbf{PreOrd} is equal (not just isomorphic!) to the identity functor.
  4. The restriction of FG \colon \mathbf{Top} \to \mathbf{Top} to the category \mathbf{Top}^{\operatorname{fin}} of finite topological spaces is equal to the identity functor.
  5. If Y is a topological space, then Y is T_0 if and only if Y^{\operatorname{sp}} is a poset.
  6. If (X,\leq) is a preorder, then X is a poset if and only if X^{\operatorname{Alex}} is T_0.
  7. The functors F and G give rise to adjoint equivalences

        \begin{align*}F\!:\mathbf{PreOrd}^{\operatorname{fin}} &\leftrightarrows \mathbf{Top}^{\operatorname{fin}}:\!G \\F\!:\mathbf{Pos}^{\operatorname{fin}} &\leftrightarrows \mathbf{Top}^{\operatorname{fin}}_{T_0}:\!G.\end{align*}

Proof. (1) Suppose f \colon X \to Y^{\operatorname{sp}} is monotone, let Z \subseteq Y be a closed subset, and let W = f^{-1}(Z). Suppose b \in W and a \leq b. Since f is monotone and Z is closed, we get f(a) \leq f(b), i.e. f(a) \in \overline{\{f(b)\}} \subseteq Z. We conclude that a \in W, so W is downward closed, hence closed in X^{\operatorname{Alex}}.

Conversely, suppose f \colon X^{\operatorname{Alex}} \to Y is continuous, and suppose a \leq b in X. Then a \in \overline{\{b\}}, so by continuity we get f(a) \in f\left(\overline{\{b\}}\right) \subseteq \overline{\{f(b)\}}, so f(a) \leq f(b).

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

    \begin{align*}\operatorname{Hom}_{\mathbf{PreOrd}}\big(X,G(Y)\big) &\stackrel\sim\to \operatorname{Hom}_{\mathbf{Top}}\big(F(X),Y\big) \\f &\mapsto f\end{align*}

is a bijection.

(3) Since \overline{\{x\}} = X_{\leq x}, we conclude that y \in \overline{\{x\}} if and only if y \leq x, so the specialisation preorder on X^{\operatorname{Alex}} is the original preorder on X.

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

    \[Z = \bigcup_{y \in Z} \overline{\{y\}}\]

of closed subsets of Y. Thus any closed subset of FG(Y) is closed in Y, so the topologies agree.

(5) The relations x \leq y and y \leq x mean x \in \overline{\{y\}} and y \in \overline{\{x\}}. This is equivalent to the statement that a closed subset Z \subseteq Y contains x if and only if it contains y. The result follows since a poset is a preorder where the first statement only happens if x = y, and a T_0 space is a space where the second statement only happens if x = y.

(6) Follows from (5) applied to Y = F(X) since X = G(Y) by (3).

(7) The equivalence \mathbf{PreOrd}^{\operatorname{fin}} \leftrightarrows \mathbf{Top}^{\operatorname{fin}} follows from (3) and (4), and the equivalence \mathbf{Pos}^{\operatorname{fin}} \leftrightarrows \mathbf{Top}^{\operatorname{fin}}_{T_0} then follows from (5) and (6). \qedsymbol

Example. The Alexandroff topology on the poset [1] = \{0 \leq 1\} is the Sierpiński space S = \{0,1\} with topology \{\varnothing, \{1\}, S\}. As explained in this post, continuous maps X \to S from a topological space X to S are in bijection with open subsets U \subseteq X, where f \colon X \to S is sent to f^{-1}(1) \subseteq X (and U \subseteq X to the indicator function \mathbf 1_U \colon X \to S).

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

    \[{\arraycolsep=-1em\begin{array}{ccccc} & & \{\varnothing,\{x\},\{y\},X\} & & \\ & \ \ / & & \backslash\ \  & \\ \{\varnothing,\{x\},X\} & & & & \{\varnothing,\{y\},X\} \\ & \ \ \backslash & & /\ \  & \\ & & \{\varnothing,X\}.\! & & \end{array}}\]

These correspond to 4 possible preorder relations \{(a,b)\ |\ a \leq b\} \subseteq X\times X, sitting in the following diagram (where vertical arrows indicate inclusion top to bottom):

    \[{\arraycolsep=-1.5em\begin{array}{ccccc} & & \{(x,x),(y,y)\} & & \\ & \ \ / & & \backslash\ \ & \\ \{(x,x),(x,y),(y,y)\} & & & & \{(x,x),(y,x),(y,y)\} \\ & \ \ \backslash & & /\ \ & \\ & & \{(x,x),(x,y),(y,x),(y,y)\}.\!\! & & \end{array}}\]

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 Y 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 Y is a Hausdorff space, then the specialisation preorder is just the equality relation \Delta_Y \subseteq Y \times Y, 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 X^{\operatorname{Alex}} \to Y.

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