What are the number of Hausdorff topologies on the set $X=\{x_1,x_2,x_3,x_4,x_5\}$?

Question

What are the number of Hausdorff topologies on the set $X=\{x_1,x_2,x_3,x_4,x_5\}$?

I had once computed the the topologies on the set $X=\{x_1,x_2,x_3\}$ which comes out to be $29$.But,that was very tedious job to do.

I also tried computing the number of topoologies on $X=\{x_1,x_2,x_3,x_4,x_5\}$,but this seems to be more painstaking job even from before.

Now,restricting condition of Hausdorffness on counting boggled my mind even more. I've tried a lot but every time I failed miserably.

Please suggest any elegant way so that I can classify which of the topology is Hausdorff.

Thank you

Answer 1

A finite set $X$ with a $T_1$ topology is discrete: if all $\{x\}$ are closed, all subsets of $X$ (which are finite too) are closed too (finite unions of closed sets are closed) and so all subsets are open (every complement is also closed).

There is only one discrete topology on $X$, namely $\mathscr{P}(X)$. No need for counting arguments: Hausdorff implies $T_1$, of course.

Answer 2

If any topology on $X$ is to be Hausdorff, for any two points $x_i \neq x_j$ there needs to be two disjoint sets, one containing $x_i$ and another containing $x_j$.

Since this has to be true for all pairs of points, each set of the form $\{x_i\}$ must be open, so by the axioms defining a topology, the power set on $X$ (the discrete topology) is the only Hausdorff topology on $X$.