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 $X$


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



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

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

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

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

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

$$\tau_{D} $$

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?

  • $\begingroup$ Also, try not to use $a$ both for an element and a single-element set. For example: $\tau_{a}=\{\emptyset,\{a\},X\}$ $\endgroup$ – GEdgar Mar 11 '13 at 16:45
  • $\begingroup$ Hi GEdgar thanks. Are those sets the same? $\{\emptyset,a,X\}$ and $\{\emptyset,\{a\},X\}$ $\endgroup$ – Jorge Lavín Mar 11 '13 at 16:47
  • $\begingroup$ No, they aren’t, because $a\ne\{a\}$; that’s the point of @GEdgar’s comment. $\endgroup$ – Brian M. Scott Mar 11 '13 at 16:49
  • $\begingroup$ Hi Brian thanks. So in the definition of the topology things are like $\tau=\{\emptyset,\{a\},X\}$ rather than $\tau=\{\emptyset,a,X\}$ $\endgroup$ – Jorge Lavín Mar 11 '13 at 16:53
  • 1
    $\begingroup$ Yes, that’s right: the open sets have to be subsets of $\{a,b,c\}$, not elements. $\endgroup$ – Brian M. Scott Mar 11 '13 at 17:02

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 $\{a,c\}$.

  • $\begingroup$ Thanks again! The inclusion hint is a very good one, it will save some work for sure. $\endgroup$ – Jorge Lavín Mar 11 '13 at 16:59
  • $\begingroup$ @Nivalth: You’re welcome! $\endgroup$ – Brian M. Scott Mar 11 '13 at 17:01

Your definition of the $\tau_i$ leave something to be desired, as GEdgar points out, but I take your meaning.

We need closure under unions and (finite) intersections. For example, $\tau_{a,b}$ isn't a topology, for though $\{a\},\{b\}\in\tau_{a,b},$ we have $\{a,b\}=\{a\}\cup\{b\}\notin\tau_{a,b}$, so we don't have closure under unions. For an example of one that isn't closed under (finite) intersections, consider $\tau_{b,c,ab,ac}.$

It's also worth noting that you didn't cover all the $\tau_i$ in your list. There are $6$ proper non-empty subsets of $X$, and so $2^6=64$ different $\tau_i$. I count only $37$ in your list (and your list has some accidental duplicates, as well).

  • $\begingroup$ Hi thanks, I'm starting to view the difference between $a$ and $\{a\}$ which I'd say is (at least in a naive way of thinking about it) the difference between an element of a set and a set. So my definitions should be $\tau_{a}=\{\emptyset,\{a\},X\}$ -Edit- Thanks for pointing out the error in the $\tau_{i}$ list $\endgroup$ – Jorge Lavín Mar 11 '13 at 16:56
  • $\begingroup$ Absolutely right. $\endgroup$ – Cameron Buie Mar 11 '13 at 17:01

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