[Math] Topologies on a finite set

general-topology

I'm having trouble while learning the topology basics. I need to find all the topologies that can be defined on a two and a three element set.

First case: two element set

For a two elements set, say $X=\{a,b\}$ there can be defined 4 different topologies

$$\tau_{T}=\{ \emptyset, X\} $$
$$ \tau_{a}=\{\emptyset,a,X\} $$
$$\tau_{b}=\{\emptyset,b,X \} $$
$$ \tau_{D}=\{\emptyset,a,b,X\}$$

Second case: three element set

Now we have to work with a three element set $X=\{a,b,c\}$ . The two extreme possibilities are the trivial topology $\tau_{T}$ and the discrete topology $\tau_{D}$ having $2$ and $8$ element respectively. I started working with the union of $\tau_{T}$ and $\{a\},\{b\},\{c\}$ which I denote as $\tau_{i}$ for $i=a,b,c$. All of them are topologies.

Now here comes where I doubt. Consider the set

$$\tau_{ab}=\{\emptyset, \{a,b\},X\}$$

Question 1 is $\tau_{ab}$ the same as $\tau_{a,ab}\equiv\{\emptyset,\{a\},\{a,b\},X\}$ ? I'd say it is not. I found that the number of topologies that can be defined on a three element set is 29.

Question 2 Extending the previous notation, I can construct the following collections of subsets of $X$

$$\tau_{T}$$

$$\tau_{a},\tau_{b},\tau_{c},\tau_{ab},\tau_{bc},\tau_{ac} $$

$$\tau_{a,b},\tau_{a,c},\tau_{a,ab},\tau_{a,bc},\tau_{a,ac}$$

$$\tau_{b,c},\tau_{b,ab},\tau_{b,bc},\tau_{b,ac}$$

$$\tau_{c,ab},\tau_{c,bc},\tau_{a,ac} $$

$$\tau_{a,b,c},\tau_{a,b,ab},\tau_{a,b,bc},\tau_{a,b,ac}, \tau_{b,c,ab},\tau_{b,c,bc},\tau_{b,c,ac} $$

$$\tau_{a,b,c,ab},\tau_{a,b,c,bc},\tau_{a,b,c,ac}, \tau_{b,c,ab,bc}, \tau_{b,c,ab,ac}, \tau_{c,ab,bc,ac} $$

$$\tau_{a,b,c,ab,bc} , \tau_{a,b,c,ab,ac}, \tau_{b,c,ab,bc,ac}$$

$$ \tau_{a,b,c,ab,bc,ac}$$

$$\tau_{D} $$

Of course there are more than 29 collections of subsets, hence some of them can't be topologies for $X$. How can I identify them?

Best Answer

Remember that a topology is by definition closed under finite intersections and arbitrary unions. Since these topologies are all finite anyway, it suffices to check that for all $U,V\in\tau$, $U\cup V\in\tau$ and $U\cap V\in\tau$. For instance, if $\{a,b\}\in\tau$ and $\{a,c\}\in\tau$, then their intersection, $\{a\}$, must belong to $\tau$. Similarly, if $\{a\}\in\tau$ and $\{b\}\in\tau$, then their union, $\{a,b\}$, must belong to $\tau$. These are the only restrictions apart from the requirement that $\varnothing,\{a,b,c\}\in\tau$.

Note that if $U,V\in\tau$ with $U\subseteq V$, then $U\cap V=U\in\tau$ and $U\cup V=V\in\tau$ automatically, so you need only check pairs such that $U\setminus V\ne\varnothing\ne V\setminus U$. Those will be pairs like $\{a\}$ and $\{b\}$, pairs like $\{a\}$ and $\{b,c\}$ (which cause no trouble anyway), and pairs like $\{a,b\}$ and $\{a,c\}$.