拓扑学 2026 秋助教讲稿 (1) [notes-topology-1]

讲稿
SimplicialCat

Finite topological spaces

Problem. Classify topologies on $\{a,b,c,d\}$ up to homeomorphism.

Solution. Finite topological spaces are equivalent to finite preorders. A topology on a finite set $X$ corresponds to the preorder $\leq$ defined as follows:

$x\leq y$ if and only if every neighborhood of $x$ contains $y$, or equivalently, $x$ lies in the closure of $\{y\}$.

Conversely, a preorder $\leq$ on $X$ defines a topology by specifying a basis

$\{\uparrow x\mid x\in X\}$, where $\uparrow x$ is the subset $\{y\in X\mid y\geq x\}$.

Remark. When $x\leq y$ we sometimes say that $x$ is a specialization of $y$ and that $y$ is a generalization of $x$, two terms from algebraic geometry. In algebraic geometry, one often consider generic points on varieties (e.g. an affine line) and talk about their specializations. The closure of a generic point contains all its specializations.

Moreover, under the above correspondence, finite $T_0$ spaces are equivalent to finite partial orders. Non-$T_0$ spaces (preorders) can be made into $T_0$ spaces (partial orders) by identifying indistinguishable pairs, reducing the classification problem.

To classify topologies on $\{a,b,c,d\}$, we first classify connected $T_0$ space, which correspond to connected partial orders.

We list all connected partial orders on finite sets of cardinality at most $4$ (which are relatively easy to enumerate):

Cardinality $1$

$\{a\}$

Cardinality $2$

$\{a < b\}$

Cardinality $3$

$\{a < b < c\}$
$\{a < b, a < c\}$
$\{a < c, b < c\}$

Cardinality $4$ (count: $10$)

$\{a < b < c < d\}$
$\{a < b < c, b < d\}$
$\{a < b, a < c, a < d\}$
$\{a < b, a < c < d\}$
$\{a < b < d, c < d\}$
$\{a < c, b < c < d\}$
$\{a < c, b < c, b < d\}$
$\{a < d, b < d , c < d\}$
$\{a < b, a < c, b < d, c < d\}$
$\{a < c, a < d, b < c, b < d\}$

Based on this result, we may add indistinguishable points to existing ones to obtain a classification of connected preorders on finite sets of cardinality at most $4$ (write $a \sim b$ for $a\leq b \land b\leq a$):

Cardinality $0$

$\varnothing$

Cardinality $1$

$\{a\}$

Cardinality $2$

$\{a < b\}$
$\{a \sim b\}$

Cardinality $3$ (count: $6$)

($3$ listed above)
$\{a < b \sim c\}$
$\{a \sim b < c\}$
$\{a \sim b \sim c\}$

Cardinality $4$ (count: $21$)

($10$ listed above)
$\{a < b < c \sim d\}$
$\{a < b, a < c \sim d\}$
$\{a < c, b < c \sim d\}$
$\{a < b \sim c < d\}$
$\{a \sim b < c, a < d\}$
$\{a \sim b < d, c < d\}$
$\{a \sim b < c < d\}$
$\{a < b \sim c \sim d\}$
$\{a \sim b < c \sim d\}$
$\{a \sim b \sim c < d\}$
$\{a \sim b \sim c \sim d\}$

A general topological space is the union of its connected components. Therefore the number of topologies on $\{a,b,c,d\}$ is

$$ \underbrace{21}_{4} + \underbrace{6}_{3+1} + \underbrace{3}_{2+2} + \underbrace{2}_{2+1+1} + \underbrace{1}_{1+1+1+1} = 33. $$