How to list all possible topologies I am begginer and have no more idea about listing all the topologies on a given set. Precisely, given a finite set $X=\{a,b,c\}$ kindly guide me to list all its possible topologies.
 A: The power set (set of subsets) only has $8$ elements.   So the possibilities are very limited. 
Write out all possible collections of subsets which are closed under unions and intersections.
For instance,  $\{\{a,b,c\},\emptyset,\{a\}\}$ would be a topology. 
So would $\{\{a,b,c\},\emptyset,\{b\}\}$, and $\{\{a,b,c\},\emptyset,\{c\}\}$.
And $\{\{a,b,c\},\emptyset,\{a,b\}\}$.
And $\{\{a,b,c\},\emptyset,\{a,b\},\{b\}\} $.
See if you can locate the remaining few ($24$ to be exact, $6$ up to homeomorphism, see @Henno Brandsma's answer).
A: Any topology on $X = \{ a,b,c \}$ will contain $\emptyset$ and $X$, so I won't mention them again.


*

*If these are all the open sets, we have the indiscrete (sometimes called trivial) topology.

*We can have one more open set, then there are essentially two options: a singleton is open, so $\{a\}$ is the one extra open set, or $\{b\}$ or $\{c\}$. That are 3 more topologies (all homeomorphic to each other, if you already know that notion).

*One extra set that is a doubleton, so $\{a,b\}$, $\{a,c\}$ or $\{b,c\}$, again three extra topologies depending on which we have, all homeomorphic.

*That covers all topologies with three elements. We can also have $2$ extra sets besides the two compulsory ones (so $4$ in total): two singletons, a singleton and a doubleton, or two doubletons. The last case, however, has an extra open set in the topology: the singleton in which they must intersect (as we have only three elements). The two singleton topologies must also include the doubleton union of them as an extra set, so we get $3$ extra sets as well. So that leaves the topologies with extra sets $\{\{x\},\{x,y\}\}$ (singleton, doubleton, where the singleton must be a subset of the doubleton (as then their intersection, the singleton is also in the topology) and those with extra sets $\{\{x\},\{y,z\}\}$, disjoint (so the intersections are already in it as well). Of the former type there are $6$ instances, of the latter just $3$. So $9$ topologies with four sets. 

*The ones mentioned in the previous point: extra sets $\{x\}, \{x,y\},\{x,z\}$ (two doubletons and their intersection), where we pick $x,y,z$ distinct, which has 3 instances (depending on what point we choose as $x$) and those with extra sets $\{x\},\{y\},\{x,y\}$ (two singletons and their union), also three instances, depending on which point $z$ we omit from these sets. So $6$ topologies with $5$ sets.

*What happens if we have $4$ or more extra sets? As soon as we have all singletons, we get all subsets (the discrete topology) by taking unions, and as soon as we have 3 doubletons we get all singletons from their intersections and the same applies. So we get a topology with extra sets like $\{x\},\{y\},\{x,y\},\{y,z\}$, where the intersections are the singletons we already have or empty, and the unions $X$ or the doubleton we already have. This is basically the only type left and there are three instances (depending on what point is $z$, the rest is then determined). Or we can choose to make $x$ instead of $y$ the point in the last doubleton, so $\{x\},\{y\},\{x,y\},\{x,z\}$ as extra sets. So $3+3=6$ topologies with $6$ sets (all homeomorphic). 

*And finally the one topology with $8$ (all subsets) sets, the discrete one.


All in all $1 + 3 + 3 + 9 + 6 + 6+1 = 29$ different (as sets) topologies, but fewer ($9$) "essential types" (non-homeomorphic ones).
For more info, see the OEIS entry (which has the full enumeration towards the bottom for the case $n=3$ as well). No formula is known for the exact number as a function of the number of points in the set...
A: A "topology" for a given set, X, is a collection of subsets of X such that
1) the collection includes X itself and the empty set.
2) if the collection includes set A and B then it also includes their intersection.
3) if the collection includes set A and B then it also includes their union.
Two obvious such sets are the "discreet topology" that includes ALL subsets and the "indiscreet topology" that includes only X and the empty set.  Here, with X= {a, b, c} those are {{a, b, c}, {}, {a}, {b}, {c}, {ab}, {ac}, {bc}} and {{a, b, c}, {}}
Are there others?  Is there a topology that includes, say, {a} and {b} but not {c}?  Yes, {{a, b, c}, {}, {a}, {b}, {ab}} is a topology.  Similarly, {{a, b, c}, {}, {a}, {c}, {ac}} includes {a} and {c} but not b.  And, of course, {{a, b, c}, {}, {b}, {c}, {bc}} includs {b} and {c} but not {a}.
Is there a topology that includes {a} but neither {b} nor {c}? Yes, it is {{a, b, c}, {}, {a}}.  And, of course, {{a, b, c}, {}, {b}} and {{a, b, c}, {}, {c}} are also topologies. 
Do you see why those are the only possible topologies on {a, b, c}?
Notice that there are exactly $8= 2^3$ topologies  on {a, b, c}.  In general if set X contains n, a finite number, members, then there are $2^n$ topologies on X.
