\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{lemma}{Lemma}[section] \begin{center} \vskip 1cm{\LARGE\bf Direct and Elementary Approach to \\ \vskip .1in Enumerate Topologies on a Finite Set} \vskip 1cm \large Messaoud Kolli\\ Faculty of Science\\ Department of Mathematics\\ King Khaled University\\ Abha\\ Saudi Arabia\\ \href{mailto:kmessaud@kku.edu.sa}{\tt kmessaud@kku.edu.sa} \\ \end{center} \vskip .1in \begin{abstract} Let $\mathbb{E}$ be a set with $n$ elements, and let $\tau (n,k)$ be the set of all labelled topologies on $\mathbb{E}$, having $k$ open sets, and $T(n,k)=\left\vert \tau (n,k)\right\vert $. In this paper, we use a direct approach to compute $T(n,k)$ for all $n\geq 4$ and $k\geq 6\cdot 2^{n-4}$. \end{abstract} \vskip .1in %\newtheorem{theorem}{Theorem}[section] %\newtheorem{acknowledgement}[theorem]{Acknowledgement} %\newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} %\newtheorem{lemma}[theorem]{Lemma} %\newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Let $\mathbb{E}$ be a set with $n$ elements. The problem of determining the total number of labelled topologies $T(n)$ one can define on $\mathbb{E}$ is still an open question. Sharp \cite{Sharp}, and Stephen \cite{Stephen} had shown that every topology which is not discrete contains $k\leq 3\cdot 2^{n-2}$ open sets, and that this bound is optimal. Stanley \cite{Stanley} computed all labelled topologies on $\mathbb{E},$ with $k\geq 7\cdot 2^{n-4}$ open sets. In the opposite sense, Benoumhani \cite{Benoumhani} computed, for all $n$, the total number of labelled topologies with $k \leq 12$ open sets. In the other hand, Ern\'e and Stege \cite{Erne-Stege} computed the total number of topologies, for $n \leq 14$. In this paper, we use a direct approach to compute all labelled topologies on $\mathbb{E}$ having $k\geq 6\cdot 2^{n-4}$ open sets. Furthermore, we confirm the results in \cite{Sharp, Stanley, Stephen}. This work is a continuation of the results of \cite{Benoumhani, Stanley}. Here is our approach. The set $\tau (n,k)$ is partitioned into two disjoint parts as follows:% \begin{equation*} \tau (n,k)=\tau _{1}(n,k)\cup \tau _{2}(n,k), \end{equation*} where \begin{eqnarray*} \tau _{1}(n,k) &=&\left\{ \tau =\left\{ \varnothing ,A_{1},\ldots,A_{k-2},~% \mathbb{E}\right\} \in \tau (n,k),\text{ \ such that \ }\bigcap% \limits_{i=1}^{k-2}A_{i}\neq \varnothing \right\} ,~\ \ \ \ \ \ \\ \ \tau _{2}(n,k) &=&\tau (n,k)-\tau _{1}(n,k).\ \ \ \ \end{eqnarray*}% In Theorem~\ref{theorem2.1}, we prove that the cardinal $T_{1}(n,k)=\left\vert \tau _{1}(n,k)\right\vert $ satisfies \begin{equation*} T_{1}(n,k)=\sum_{l=1}^{n-1}\binom{n}{l}T(l,k-1),~\ \ \ \ \ \ \ %T_{1}(n,k)=\sum_{l=1}^{n-1}\binom{n}{k}T(l,k-1),~\ \ \ \ \ \ \ \forall ~n\geq 1. \end{equation*}% This relation enables us to compute $T_{1}(n,k)$ for $k>5\cdot 2^{n-4}$. For the determination of the cardinal $T_{2}(n,k)=\left\vert \tau _{2}(n,k)\right\vert ,$ we introduce the notion of minimal open set (Definition~\ref{definition2.2}), and we designate by $\tau _{2}(n,k,\alpha )$\textit{\ }the labelled topologies in $\tau _{2}(n,k)$ having $\alpha \geq 2$\ minimal open sets. In Lemma~\ref{lem2.6}, it is proved that if $k>5\cdot 2^{n-4}$ such that $% k\neq 6\cdot 2^{n-4},~$and $k\neq 2^{n-1}$, then all the minimal open sets of $% \tau $ are necessarily singletons. So, we can compute the numbers $% T_{2}(n,k,\alpha )$ for all $~n\geq 4,~~~k\geq 6\cdot 2^{n-4},$ and $~\alpha \geq 2.$ \bigskip \section{Basic Results} \begin{theorem} \label{theorem2.1} For every integer $n>1,$ and $2\leq k\leq 2^{n},$ we have \begin{equation*} T_{1}(n,k)=\sum_{l=1}^{n-1}\binom{n}{l}T(l,k-1), \end{equation*} with the convention that $T(l,1)=0.$ \end{theorem} \begin{proof} Let $A\subset \mathbb{E}$, with $\left\vert A\right\vert =l\leq n-1,$ and let $\tau ^{\prime }$ be a topology on $A$, and having $k-1$ open sets. To this topology we associate the following one% \begin{equation*} \Phi _{A}(\tau ^{\prime })=\tau =\left\{ O\cup A^{c},~\ \ \ \ O\in \tau ^{\prime }\right\} \cup \{\varnothing \}\text{.} \end{equation*}% Obviously $\Phi _{A}$ is an injective mapping on $\tau (l,k-1)$ into $\tau _{1}(n,k).$ In the other hand, if $\left\vert A\right\vert =\left\vert B\right\vert =l\leq n-1$ and $A\neq B$, then \begin{equation*} R(\Phi _{A})\cap R(\Phi _{B})=\varnothing , \end{equation*}% where $R(\Phi _{A})$ is the image of $\Phi _{A}$. This shows that% \begin{equation*} T_{1}(n,k)\geq \sum_{l=1}^{n-1}\binom{n}{l}T(l,k-1). \end{equation*} Conversely, if $\tau =\left\{ \varnothing ,~A_{1},\ldots,\ A_{k-2},~\mathbb{E}% \right\} \in \tau _{1}(n,k)$, with $A_{1}=\bigcap\limits_{i=1}^{k-2}A_{i},$ then $\tau ^{\prime }=\left\{ O-A_{1},~\ \ O\in \tau \right\} $ is a topology on $A_{1}^{c},$ having $k-1$ open sets, and $\Phi _{A_{1}^{c}}(\tau ^{\prime })=\tau $. This shows the other inequality, and completes the proof. \end{proof} \bigskip The following definition will be needed in the sequel. \begin{definition} \label{definition2.2} Let $\tau =\left\{ \varnothing ,~A_{1},\ldots,\ A_{k-2},~\mathbb{E}\right\} \in \tau (n,k).$ The element $A_{i}$ is called a minimal open set, if it satisfies: \begin{equation*} A_{i}\cap A_{j}=A_{i}~\ ~\ or ~\ ~\ \varnothing, \qquad \forall ~j=1,\ldots,k-2. \end{equation*} \end{definition} \begin{remark}\rm{ i) A topology on $\mathbb{E}$ is a bounded lattice with $(1=\mathbb{E},$ $0=\varnothing ).$ A minimal open set is in fact an atom. Recall that an atom in a partially ordered set is an element which covers $0$. So, every topology has at least one minimal open set, and $% \tau _{1}(n,k)$ is the subset of topologies having exactly one minimal open set. \smallskip ii) If $\tau \in \tau _{2}(n,k)$, then $\tau $ has at least two minimal open sets. \smallskip iii) The space $\mathbb{E}$ is a union of $\alpha $ minimal open sets for the topology $\tau \in \tau (n,k)$ if and only if $% k=2^{\alpha }.$ \smallskip iv) If $\tau $ has $\alpha $ minimal open sets, then $% k\geq 2^{\alpha }.$ } \end{remark} \begin{definition} For $\alpha \geq 2$, we define \begin{equation*} \tau _{2}(n,k,\alpha )=\left\{ \tau \in \tau _{2}(n,k),~~\ \tau \hbox{ has } \alpha \hbox{ minimal open sets } \right\} . \end{equation*} \end{definition} \noindent Note that if $\alpha _{1}\neq \alpha _{2}$, then $\tau _{2}(n,k,\alpha _{1})\cap \tau _{2}(n,k,\alpha _{2})=\varnothing $. So% \begin{equation*} T_{2}(n,k)=\sum_{\alpha \geq 2,~\ 2^{\alpha }\leq ~k}T_{2}(n,k,\alpha ). \end{equation*}% The computation of $T_{2}(n,k)$ is then equivalent to the computation of $% T_{2}(n,k,\alpha ),$ for $\alpha \geq 2,$ under the condition $2^{\alpha }\leq ~k$ $.$ If $k=2^{\alpha},$ then% \begin{equation*} T_{2}(n,2^{\alpha },\alpha )=S(n,\alpha ), \end{equation*}% where $S(n,\alpha )$ is the Stirling number of the second kind. \bigskip \begin{lemma} Let $n\geq 1$, $\alpha \geq 2$. Then $\tau _{2}(n,k,\alpha )$ is empty, for $k>2^{n-1}+2^{\alpha -1}.$ In addition, this bound is optimal: \begin{equation*} \tau _{2}(n,2^{n-1}+2^{\alpha -1},\alpha )\neq \varnothing . \end{equation*} \end{lemma} \begin{proof} We argue by contradiction. Suppose that $\tau \in \tau _{2}(n,k,\alpha )$, and write it as \begin{equation*} \tau =\left\{ \varnothing ,~A_{1},\ldots,A_{\alpha },\ldots,\mathbb{E}\right\} , \end{equation*}% where $A_{1},...,A_{\alpha }$ are the $\alpha $ minimal open sets of $\tau $% . Put $A=\bigcup\limits_{i=1}^{\alpha }A_{i}$, the topology $\tau ^{\prime }=\left\{ O-A,~\ \ O\in \tau \right\} $ on $A^{c}$ has at least $\left\lceil k~2^{1-\alpha }-1\right\rceil $ open sets. In the other hand, $\left\vert A^{c}\right\vert \leq n-\alpha $, and since $\tau ^{\prime }$ is at most the discrete topology, we obtain \begin{equation*} k~2^{1-\alpha }-1\leq \left\vert \tau ^{\prime }\right\vert \leq 2^{n-\alpha }. \end{equation*}% This contradiction proves that $\tau _{2}(n,k,\alpha )$ is empty. The second assertion will be proved in the next section. \end{proof} \bigskip \begin{lemma} \label{lem2.6} Let $\tau \in \tau _{2}(n,k,\alpha),$ with $k>5\cdot 2^{n-4}$, $k\neq 6\cdot 2^{n-4}$, and $k\neq 2^{n-1}$. Then, all the minimal open sets of $\tau $ are singletons. \end{lemma} \bigskip \begin{proof} Let $\tau =\left\{ \varnothing ,\ A_{1},\ldots,A_{\alpha },\ldots,\mathbb{E}\right\} \in \tau _{2}(n,k,\alpha ),$ where $A_{1},\ldots,A_{\alpha }$ are its minimal open sets, and suppose that $% A=\bigcup\limits_{i=1}^{\alpha }A_{i}$ has more than $\alpha +1$ elements. The same argument used in the previous Lemma gives $5\cdot 2^{n-4}n$, then $(n)_l=0$. We start by the number of topologies $\tau \in \tau _{2}(n,k,\alpha )$% , such that $\tau $ has at least one minimal open set, which is not a singleton. For this, the previous Lemma gives $k=2^{n-1}$ or $k=6\cdot 2^{n-4}$. If $k=2^{n-1}$, then $\alpha =n-1$ and the number of these topologies is $$ S(n,n-1)=\dfrac{(n)_{2}}{2}. $$ If $k=6\cdot 2^{n-4}$, we have $\alpha =n-2,$ and the number of these topologies is $$ 2(n-2)\binom{n}{n-2}\binom{n-2}{1}=(n-2)~(n)_{3}. $$ \noindent The remaining topologies of $\tau _{2}(n,k,\alpha )$ have the property that all their minimal open sets are singletons. For this, let $\tau \in $ $\tau _{2}(n,k,\alpha )$ \begin{equation*} \tau =\left\{ \varnothing ,\ A_{1},\ldots,A_{\alpha },\ldots,\mathbb{E}\right\} . \end{equation*}% Put $\alpha =n-i,~~0\leq i\leq n-2$, and $A=\cup _{i=1}^{\alpha }A_{i}$ . The topology $\tau ^{\prime }=\left\{ O-A,~\ \ O\in \tau \right\} $ (on $% A^{c}$), can be written as follows:% \begin{equation*} \tau ^{\prime }=\left\{ \varnothing ,~C_{1},\ldots,C_{m}\right\} ,~\ \ \ m\in \left\{ 0,1,2,\ldots,5\cdot 2^{i-3}-1,~3\cdot 2^{i-2}-1,~2^{i}-1\right\} . \end{equation*} To reconstruct $\tau $ from $\tau ^{\prime }$, we remark that every $C_{j}$, if it exists, generates $2^{i_{j}}$ open sets in $\tau $, with $i_{j}\leq n-i-1.$ So, the number $k$ has necessarily the form:% \begin{equation*} k=2^{n-i}+2^{i_{1}}+2^{i_{2}}+...+2^{i_{m}}, \end{equation*}% where the integers $i_{j}$ , $\ 1\leq j\leq m$ can be dependant. Our approach is that for all $\alpha ,~2\leq \alpha \leq n$, we determine all possibilities of the number $k$, and next the number of all these topologies. For \underline{$\alpha =n.$} $A^{c}=\varnothing $; so $m=0$, $k=2^{n}$ and $T_{2}(n,2^{n},n)=1$. This case corresponds to the discrete topology. \bigskip For \underline{$\alpha =n-1.$} $A^{c}$ $=\left\{ x\right\} $; so $m=1$, and $% \tau ^{\prime }=\left\{ \varnothing ,C_{1}=\left\{ x\right\} \right\} $. All the possibilities of $k$ are given by% \begin{equation*} k=2^{n-1}+2^{n-1-j},~\ \ 1\leq j\leq n-1. \end{equation*}% The number of these topologies is \begin{equation*} T_{2}(n,2^{n-1}+2^{n-1-j},n-1)=n\binom{n-1}{j}=\frac{(n)_{j+1}}{j!},~\ \ 1\leq j\leq n-1. \end{equation*} For \underline{$\alpha =n-2.$} $A^{c}$ $=\left\{ x,y\right\},$ $\tau ^{\prime }=\left\{ \varnothing ,C_{1},\ldots,C_{m}\right\} ,~$with$\ \ \ m=1,2$ or $3.$ If $m=1,$ $\tau ^{\prime }=\left\{ \varnothing ,~C_{1}=\left\{ x,y\right\} \right\} $. Since we are supposing $k\geq 6\cdot 2^{n-4}$, the unique possibility is that $C_{1}$ generates $2^{n-3}$ open sets. So, $% k=2^{n-2}+2^{n-3}=6\cdot 2^{n-4},$ and the number of these topologies is $$ \binom{n}{n-2}\binom{n-2}{1}=\dfrac{(n)_{3}}{2}. $$ \noindent If $m=2,$ $\tau ^{\prime }$ $=\left\{ \varnothing ,C_{1}=\left\{ x\right\} ,C_{2}=\left\{ x,y\right\} \right\} $ \ \ \ \ or \ \ \ \ $\tau ^{\prime }=\left\{ \varnothing ,C_{1}=\left\{ y\right\} ,C_{2}=\left\{ x,y\right\} \right\}.$ Here we have two categories of solutions: \smallskip a) $C_{1}$ generates $2^{n-3}$ open sets, and $C_{2}$ generates $% 2^{n-3-j},0\leq j\leq n-3$, \ open sets. Hence% \begin{equation*} k=2^{n-2}+2^{n-3}+2^{n-3-j}=6\cdot 2^{n-4}+2^{n-3-j},~\ \ 0\leq j\leq n-3. \end{equation*}% The number of such topologies is $$ 2(j+1)\binom{n}{n-2}\binom{n-2}{j+1}=\dfrac{(n)_{j+3}}{j!}. $$ b)$~$\ $C_{1}$ generates $2^{n-4}$ open sets and also $C_{2}$ generates $% 2^{n-4}$. So, $\ k=2^{n-2}+2^{n-4}+2^{n-4}=6\cdot 2^{n-4},$ and the number in this case is $$ 2~\binom{n}{n-2}\binom{n-2}{2}=\dfrac{(n)_{3}}{2}. $$ \noindent If $m=3,$ $~\ \tau ^{\prime }=\left\{ \varnothing ,C_{1}=\left\{ x\right\} ,C_{2}=\left\{ y\right\} ,C_{3}=\left\{ x,y\right\} \right\} $. There are $8$ categories of solutions: \bigskip a) Each $C_{j}$, $j=1,2,3$ generates $2^{n-3}$ open sets. So, $k=2^{n-2}+2^{n-3}+2^{n-3}+2^{n-3}=10\cdot 2^{n-4},$ and the wanted number is $$ \binom{n}{n-2}\binom{n-2}{1}=\dfrac{(n)_{3}}{2}. $$ b) $C_{1}$ generates $2^{n-3}$ open sets, $C_{2}$ and $C_{3}$ each one generates $% 2^{n-3-j}\ $\ open sets, with $1\leq j\leq n-3$. So, $% k=2^{n-2}+2^{n-3}+2^{n-3-j}+2^{n-3-j}=6\cdot 2^{n-4}+2^{n-2-j},~\ \ \ \ 1\leq j\leq n-3$, and the number of these topologies is $$ 2(j+1)\binom{n}{n-2} \binom{n-2}{j+1}=\dfrac{(n)_{j+3}}{j!}. $$ c) $\ C_{1}$ and $C_{2}$ each one generates $2^{n-3}$ open sets, but $C_{3}$ generates $2^{n-4}$ open sets. So, $% k=2^{n-2}+2^{n-3}+2^{n-3}+2^{n-4}=9\cdot 2^{n-4}$, and the number of these topologies is $$ 2\binom{n}{n-2}\binom{n-2}{2}=\dfrac{(n)_{4}}{2}. $$ d) $C_{1}$ generates $2^{n-3}$ open sets, $C_{2}$ generates $2^{n-2-j},$ $2\leq j\leq n-3$ open sets. So, $C_{3}$ generates $2^{n-3-j}$ open sets, and $k=2^{n-2}+2^{n-3}+2^{n-2-j}+2^{n-3-j}=6\cdot 2^{n-4}+3\cdot 2^{n-3-j},~\ \ \ \ 2\leq j\leq n-3$. The number of these topologies is $$ 2(j+1)\binom{n}{n-2} \binom{n-2}{j+1}=\dfrac{(n)_{j+3}}{j!}. $$ e) $C_{1},~C_{2}$ and $C_{3}$ each one generates $2^{n-4}$ open sets. So, $% k=2^{n-2}+2^{n-4}+2^{n-4}+2^{n-4}=7\cdot 2^{n-4}$, and the number of these topologies is $$ \binom{n}{n-2}\binom{n-2}{2}=\dfrac{(n)_{4}}{4}. $$ f) $C_{1}$ and $C_{2}$, each one generates $2^{n-4}$ open sets, but $C_{3}$ generates $2^{n-5}$ open sets. In this case $% k=2^{n-2}+2^{n-4}+2^{n-4}+2^{n-5}=13\cdot 2^{n-5}$, and the number of these topologies is $$ 6 \binom{n}{n-2}\binom{n-2}{3}=\dfrac{(n)_{5}}{2}. $$ g) $C_{1}$ generates $2^{n-4}$ open sets, and each one of $C_{2}$, $C_{3}$ generates $2^{n-5}$. So, $\ k=2^{n-2}+2^{n-4}+2^{n-5}+2^{n-5}=6\cdot 2^{n-4}$, and the number of these topologies is $$ 6\binom{n}{n-2}\binom{n-2}{3}= \dfrac{(n)_{5}}{2}. $$ h) Each one of $C_{1},~C_{2}$ generates $2^{n-4}$ open sets, but $C_{3}$ generates $2^{n-6}$ . So, $k=2^{n-2}+2^{n-4}+2^{n-4}+2^{n-6}=25\cdot 2^{n-6}$, and the number of these topologies is $$ 6 \binom{n}{n-2}\binom{n-2}{4}= \dfrac{(n)_{6}}{8}. $$ All the other cases give $k<6\cdot 2^{n-4}$. We resume all these results in the next statement. \begin{theorem} Let $n\geq 4$, and $\alpha =n-2$. Then we have \begin{center} \begin{equation*} \begin{tabular}{|c|c|} \hline $k$ & $T_{2}(n,k,n-2)$ \\ \hline $6\cdot 2^{n-4}$ & $(n-1)(n)_{3}+\dfrac{1}{2}(n)_{5}$ \\ \hline $6\cdot 2^{n-4}+1$ & $(n)_{3}$ \\ \hline $6\cdot 2^{n-4}+2^{n-3-j},~\ \ \ \ 4\leq j\leq n-4$ & $\dfrac{(n-2)~(n)_{j+3}}{% (j+1)!}$ \\ \hline $6\cdot 2^{n-4}+3\cdot 2^{n-3-j},~\ \ 5\leq j\leq n-3$ & $\dfrac{(n)_{j+3}}{j!}$ \\ \hline $25\cdot 2^{n-6}$ & $\dfrac{7}{24}(n)_{6}+\dfrac{1}{24}(n)_{7}$ \\ \hline $51\cdot 2^{n-7}$ & $\dfrac{1}{24}(n)_{7}$ \\ \hline $13\cdot 2^{n-5}$ & $(n)_{5}+\dfrac{1}{6}(n)_{6}$ \\ \hline $27\cdot 2^{n-6}$ & $\dfrac{1}{6}(n)_{6}$ \\ \hline $7\cdot 2^{n-4}$ & $\dfrac{5~}{4}(n)_{4}+\dfrac{1}{2}(n)_{5}$ \\ \hline $15\cdot 2^{n-5}$ & $\dfrac{1}{2}(n)_{5}$ \\ \hline $2^{n-1}$ & $(n)_{3}+(n)_{4}$ \\ \hline $9\cdot 2^{n-4}$ & $\dfrac{1}{2}(n)_{4}$ \\ \hline $10\cdot 2^{n-4}$ & $\dfrac{1}{2}(n)_{3}$ \\ \hline \end{tabular}% \end{equation*} \end{center} All other topologies in $\tau _{2}(n,k,n-2)$ have $k<6\cdot 2^{n-4}$ open sets. \end{theorem} \bigskip \noindent We use the same reasoning as above, to show the following theorem. \begin{theorem} Let $n\geq 5,$ and $\alpha =n-i,~~3\leq i\leq ~n-2$. Then, the following results hold. For $\alpha =n-3$, if $n=5$, we have \begin{equation*} \begin{tabular}{|c|c|c|c|c|c|} \hline $k$ & $12$ & $13$ & $14$ & $15$ & $18$ \\ \hline $T_{2}(5,k,2)$ & $360$ & $60$ & $180$ & $60$ & $20$ \\ \hline \end{tabular}% \end{equation*}% If $n\geq 6$, we have \begin{equation*} \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline $k$ & $6\cdot 2^{n-4}$ & $25\cdot 2^{n-6}$ & $13\cdot 2^{n-5}$ & $27\cdot 2^{n-6}$ & $% 7\cdot 2^{n-4}$ & $15\cdot 2^{n-5}$ & $9\cdot 2^{n-4}$ \\ \hline $T_{2}(n,k,n-3)$ & $(n)_{4}+\frac{5}{2}(n)_{5}$ & $\frac{1}{4}(n)_{6}$ & $% \frac{1}{2}(n)_{5}$ & $\frac{1}{6}(n)_{6}$ & $(n)_{4}+\frac{1}{2}(n)_{5}$ & $% \frac{1}{2}(n)_{5}$ & $\frac{1}{6}(n)_{4}$ \\ & $+\frac{5}{4}(n)_{6}$ & & & & & & \\ \hline \end{tabular} \end{equation*} For $\alpha =n-4,$ and $n\geq 6$ \begin{equation*} \begin{tabular}{|c|c|c|c|c|} \hline $k$ & $25\cdot 2^{n-6}$ & $13\cdot 2^{n-5}$ & $27\cdot 2^{n-6}$ & $17\cdot 2^{n-5}$ \\ \hline $T_{2}(n,k,n-4)$ & $\frac{1}{8}(n)_{6}$ & $\frac{1}{2}(n)_{5}+\frac{1}{6}% (n)_{6}$ & $\frac{1}{6}(n)_{6}$ & $\frac{1}{24}(n)_{5}$ \\ \hline \end{tabular} \end{equation*} For $\alpha =n-i,~~5\leq i\leq n-2,$ and $n\geq 7$ \begin{equation*} \begin{tabular}{|c|c|c|c|} \hline $k$ & $6\cdot 2^{n-4}+2^{n-i-1}$ & $6\cdot 2^{n-4}+3\cdot 2^{n-i-2}$ & $% 2^{n-1}+2^{n-i-1} $ \\ \hline $T_{2}(n,k,n-i)$ & $\dfrac{\left( n-2\right) }{(i-1)!}~(n)_{i+1}$ & $\dfrac{1% }{(i-1)!}~(n)_{i+2}$ & $\dfrac{(n)_{i+1}}{i!}$ \\ \hline \end{tabular}% \end{equation*}% All other topologies in $\tau _{2}(n,k,n-i),~~3\leq i\leq n-2,$ have $k<6\cdot 2^{n-4}$ open sets. \end{theorem} Now, we compute $T_{1}(n,k)$, for $k>5\cdot ~2^{n-4}$. \begin{theorem} For all $n\geq 5,$ and $k>5\cdot ~2^{n-4}$, we have: \begin{eqnarray*} T_{1}(n,2^{n-1}+1)&=&n, \\ T_{1}(n,3\cdot 2^{n-3}+1)&=&(n)_{3}, \\ T_{1}(n,5\cdot 2^{n-4}+1)&=&(n)_{4}, \\ T_{1}(n,k)&=&0, otherwise. \end{eqnarray*} \end{theorem} \begin{proof} Obviously, we have $T_{1}(n,2^{n-1}+1)=nT(n-1,2^{n-1})=n$, $T_{1}(n,3\cdot 2^{n-3}+1)=nT(n-1,3\cdot 2^{n-3})=n\left( n-1\right) _{2}=(n)_{3},$ and $T_{1}(n,5\cdot 2^{n-4}+1)=nT(n-1,5\cdot 2^{n-4})=n(n-1)_{3}=(n)_{4}$. If $k>2^{n-1}+1$, we have $T(l,k-1)=0,$ for $1\leq l\leq n-1$, so $T_{1}(n,k)=0.$ If $5\cdot 2^{n-4}+1< k <2^{n-1}+1$, and $k$ $\neq 3\cdot 2^{n-3}+1$, the Theorem~\ref{theorem2.1} yields $T_{1}(n,k)=nT(n-1,k-1)$. But we know that $T(n-1,k-1)=0$, for $5\cdot 2^{n-4}