Upper and/or lower Bound for Numbers of different topologies on the set $\{1,...n \}$ As the title says I am looking for upper and lower bound for the cardinality of different topologies on a set $\{1,....n\}$ for natural n!
Are there some known bounds? My teacher says that there no formula which gives the exactly number.
Thanks in advance!
Feel free to add more tags I dont know which tags I should add

Let $X=\{1,....n,\}$and $m(n)$ is the number of topologies on $X$. We want to show that $m(n)\leq 2^{n(n-1)}$
We denote $\mathcal{M}$ for the set of all topologies on $X$ and we define for every $x \in X$ a function as follows:
$f_x:\mathcal{M}\rightarrow \mathcal{P}(X)$, $f(\tau )= \bigcap_{U\in\tau,x\in U}U$
So the map creates a topology on the "smallest" neighborhood of the point $x$
One can verify that the set $\{f_x(\tau):x\in X\}$ is a basis for $\tau$. If we pick up two topologies (e.g. $\tau_1$,$\tau_2\in \mathcal{M})$ on $X$ then they are equal if and only if the corresponding basis are equal. So its sufficient to count the numbers of the maps $f_x$ to get an upper bound.
For an arbitrary $x\in X$ there are $2^{n-1}$ sets in $\mathcal{P}(X)$ cointaining $x$. So for $f_x$ there are $2^{n-1}$ different $f_x$. We now can conclude:
$$m(n)\leq (2^{n-1})^n=2^{n(n-1)}$$  Because there are $n$ possibilities for $x$
