Topologies on Set with 3 Elements

Theorem
Let $S = \set {a, b, c}$ be a set with $3$ elements.

Then there exist $29$ possible different topologies on $S$:

The numbering is arbitrary.

Proof
The power set of $S$ is the set:


 * $\powerset S = \set {\O, \set a, \set b, \set c, \set {a, b}, \set {a, c}, \set {b, c}, \set {a, b, c} }$

A topology on $S$ is a subset of $\powerset S$.

Thus the set of all topologies on $S$ is a subset of the power set of $\powerset S$.

From Cardinality of Power Set of Finite Set:
 * $\card {\powerset {\powerset S} } = 256$

Half of the subsets of $\powerset {\powerset S}$ do not contain $\O$ and so are not topologies on $S$.

Half of the remaining subsets of $\powerset {\powerset S}$ do not contain $S$ itself, and so are also not topologies on $S$.

We are left with $64$ subsets of $\powerset {\powerset S}$ to investigate.

In the following, they will be numbered arbitrarily.

First we have the discrete topology and indiscrete topology on $S$:

From here, we investigate topologies according to the number of its singletons.

Before doing that, note that a topology on $S$ with either $3$ singletons or $3$ doubletons is the discrete topology.

Hence in the below we need only investigate topologies with $0$, $1$ and $2$ singletons and doubletons.

Topologies with no Singletons
Let $\TT_0$ be a topology on $S$ which does not contain a singleton.

Unless $\TT_0$ is the indiscrete topology, $\TT_0$ must therefore contain at least one doubleton.

there exist $2$ distinct doubletons $S_1$ and $S_2$ of $\powerset S$ such that $S_1, S_2 \in \TT_0$.

Their intersection $S_1 \cap S_2$ cannot be $\O$ as then there would be at least $4$ distinct elements in $S$.

On the other hand, $S_1 \cap S_2$ cannot have $2$ distinct elements otherwise $S_1 = S_2$.

So $S_1 \cap S_2$ is a singleton.

But because $\TT_0$ be a topology, $S_1 \cap S_2 \in \TT_0$.

But this contradicts our assertion that $\TT_0$ does not contain a singleton.

So for $\TT_0$ to be a topology, $\TT_0$ must contain exactly $1$ doubleton.

Hence we have the included set topologies on $S$:

Topologies with $1$ Singleton
Let $\TT_1$ be a topology on $S$ which contains exactly $1$ singleton $U$.

First we note that by adding just $1$ singleton to the indiscrete topology on $S$ yields the excluded set topologies on $S$:

Now consider topologies $\TT_1$ on $S$ which contain exactly one doubleton.

Let us add one doubleton $D$ to each of these topologies which contains one singleton $U$.

First suppose that $D \cap U = \O$.

Thus we have the partition topologies on $S$:

Now suppose that $D \cap U \ne \O$.

Thus we find we have the order topologies on $S$:

Now consider topologies $\TT_1$ on $S$ which contain exactly $2$ doubletons $D_1$ and $D_2$ in addition to its one singleton $U$.

First consider the situation where $D_1 \cap D_2 = U$.

Thus we have the particular point topologies on $S$:

Now consider the situation where $D_1 \cap D_2 \ne U$.

Then $D_1 \cap D_2 \in \TT_1$ by definition of topology.

This would mean that $\TT_1$ contains $2$ singletons.

Hence it is not possible for a topology on $S$ with exactly one singleton $U$ to have $2$ doubletons $D_1$ and $D_2$ such that $D_1 \cap D_2 \ne U$.

Thus we have exhausted our list of topologies on $S$ with one singleton.

Topologies with $2$ Singletons
Let $\TT_2$ be a topology on $S$ which contains exactly $2$ singletons: $U_1$ and $U_2$.

We note that $U_1 \cup U_2$ is also in $\TT_2$.

Thus $\TT_2$ contains at least one doubleton.

This fact yields us the excluded point topologies on $S$:

Now consider topologies $\TT_2$ on $S$ which contain exactly $2$ doubletons $D_1$ and $D_2$ in addition to its $2$ singletons $U_1$ and $U_2$.

Suppose $D_1 \cap D_2 \notin \set {U_1, U_2}$.

Then $\TT_2$ would contain $3$ singletons.

So we note that $D_1 \cap D_2 = U_1$ or $D_1 \cap D_2 = U_2$.

This yields the following topologies, which do not appear to be instances of a topology of a particular variety:

That exhausts our possible topologies $\TT_2$ on $S$ which contain exactly $2$ singletons.

Thus we have enumerated all the different topologies on $S$, and the total comes to $29$.