TITLE: Estimates on number of topologies on a finite set
QUESTION [7 upvotes]: I've seen that there is currently no known formula for the amount of distinct topologies on a finite set. However I was wondering there are some rough estimates, aside from the trivial ones on the amount.
Given a finite set $X$ of cardinality $n$ we have the trivial upper bound of $2^{2^n}$. I was wondering whether there are some stronger estimates. 
In particular, I was wondering whether one can show there are less topologies on $X$ then there are relations on $X\times X$? i.e:
Can we say that it is larger than $2^{n^2}$ or smaller than  $2^{n^2}$?
I would also welcome any help in estimates for an infinite set $X$.

REPLY [5 votes]: On a finite set $S$, the topology is uniquely determined by the closures of singletons: A subset of $S$ is closed iff it contains the closure of all its elements.
Given a topology, we can define
$$\tag1a\le b\iff \overline{\{a\}}\subseteq  \overline{\{b\}}$$
(or equivalently, $a\in \overline{\{b\}}$).
This need not be a partial order because we can have $a\le b$ and $b\le a$ with $a\ne b$. But it is a reflexive and transitive relation on $S$.
On the other hand, from a a reflexive and transitive relation $\le$ on $S$, we can define a topology by declaring 
$$\tag2 A\text{ closed}\iff\forall a\in A\colon \forall x\in S\colon x\le a\to x\in A.$$ 
One verifies that the union or intersection of two closed sets is closed and trivially $\emptyset$ and $S$ are closed. 
Moreover, the associations topology $\leftrightarrow$ reflexive transitive relation in $(1)$ and $(2)$ are inverse of each other, hence 

On a finite set, there are as many topologies as there are reflexive transitive binary relations.

The relation to partial orders is that a reflexive transitive binary relation on $S$ is the same as a partial order on a partition of $S$. In particular, if the counts of topologies and partial orders are $T_n$ and $P_n$, respectively, we have 
$$ T_n=P_n+{n\choose 2}P_{n-1}+\left({n\choose 3}+\frac12{n\choose 2}{n-2\choose 2}\right)P_{n-2}+\ldots+2^{n-1}P_2+P_1.$$