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Estimates on number of topologies on a finite set

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Estimates on number of topologies on a finite set


Examples of topologies in which all open sets are regular?What can be said about the lattice of topologies on a given set?Are there non-Hausdorff examples of maximal compact topologies in the lattice of topologies on a set?Topologies on a finite setWhat is more probable for a number? being the smallest element in the set or being the median.Upper and/or lower Bound for Numbers of different topologies on the set $1,…n $How many topologies exist on a finite set?Topologies on a finite set. An open problem?Tychonoff's theorem for products of finite discrete topologies?Prove that a finite set has exactly one Hausdorff topology






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


I've seen on 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 $Xtimes X$? 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 $X$.










share|cite|improve this question











$endgroup$


















    2












    $begingroup$


    I've seen on 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 $Xtimes X$? 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 $X$.










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      I've seen on 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 $Xtimes X$? 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 $X$.










      share|cite|improve this question











      $endgroup$




      I've seen on 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 $Xtimes X$? 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 $X$.







      combinatorics general-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      Asaf Karagila

      313k34 gold badges449 silver badges784 bronze badges




      313k34 gold badges449 silver badges784 bronze badges










      asked 8 hours ago









      Keen-ameteurKeen-ameteur

      1,6236 silver badges16 bronze badges




      1,6236 silver badges16 bronze badges




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          See The number of finite topologies by Kleitman and Rothschild. If $T_n$ is the number of topologies on a set with $n$ elements, they show that
          $$log_2 T_nsim n^2/4$$
          This is quite an old paper and there may be a better asymptotic formula at this point. Asymptotically, the logarithm of the number of topologies is also equal to the logarithm of the number of partial orders on the set. The same authors have a sequel to the paper that gives a better formula for the number of partial orders, but without the logarithm I believe they cease to be asymptotically equal.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Are you referring to this article? researchgate.net/publication/…
            $endgroup$
            – Keen-ameteur
            4 hours ago











          • $begingroup$
            @Keen Yes. I guess I was remembering the wrong title.
            $endgroup$
            – Matt Samuel
            4 hours ago


















          1












          $begingroup$

          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
          $$tag1ale biff overlineasubseteq overlineb$$
          (or equivalently, $ain overlineb$).
          This need not be a partial order because we can have $ale b$ and $ble a$ with $ane b$. 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
          $$tag2 Atext closediffforall ain Acolon forall xin Scolon xle ato xin A.$$
          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
          $$ T_n=P_n+nchoose 2P_n-1+left(nchoose 3+frac12nchoose 2n-2choose 2right)P_n-2+ldots+2^n-1P_2+P_1.$$






          share|cite|improve this answer











          $endgroup$















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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            See The number of finite topologies by Kleitman and Rothschild. If $T_n$ is the number of topologies on a set with $n$ elements, they show that
            $$log_2 T_nsim n^2/4$$
            This is quite an old paper and there may be a better asymptotic formula at this point. Asymptotically, the logarithm of the number of topologies is also equal to the logarithm of the number of partial orders on the set. The same authors have a sequel to the paper that gives a better formula for the number of partial orders, but without the logarithm I believe they cease to be asymptotically equal.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Are you referring to this article? researchgate.net/publication/…
              $endgroup$
              – Keen-ameteur
              4 hours ago











            • $begingroup$
              @Keen Yes. I guess I was remembering the wrong title.
              $endgroup$
              – Matt Samuel
              4 hours ago















            4












            $begingroup$

            See The number of finite topologies by Kleitman and Rothschild. If $T_n$ is the number of topologies on a set with $n$ elements, they show that
            $$log_2 T_nsim n^2/4$$
            This is quite an old paper and there may be a better asymptotic formula at this point. Asymptotically, the logarithm of the number of topologies is also equal to the logarithm of the number of partial orders on the set. The same authors have a sequel to the paper that gives a better formula for the number of partial orders, but without the logarithm I believe they cease to be asymptotically equal.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Are you referring to this article? researchgate.net/publication/…
              $endgroup$
              – Keen-ameteur
              4 hours ago











            • $begingroup$
              @Keen Yes. I guess I was remembering the wrong title.
              $endgroup$
              – Matt Samuel
              4 hours ago













            4












            4








            4





            $begingroup$

            See The number of finite topologies by Kleitman and Rothschild. If $T_n$ is the number of topologies on a set with $n$ elements, they show that
            $$log_2 T_nsim n^2/4$$
            This is quite an old paper and there may be a better asymptotic formula at this point. Asymptotically, the logarithm of the number of topologies is also equal to the logarithm of the number of partial orders on the set. The same authors have a sequel to the paper that gives a better formula for the number of partial orders, but without the logarithm I believe they cease to be asymptotically equal.






            share|cite|improve this answer











            $endgroup$



            See The number of finite topologies by Kleitman and Rothschild. If $T_n$ is the number of topologies on a set with $n$ elements, they show that
            $$log_2 T_nsim n^2/4$$
            This is quite an old paper and there may be a better asymptotic formula at this point. Asymptotically, the logarithm of the number of topologies is also equal to the logarithm of the number of partial orders on the set. The same authors have a sequel to the paper that gives a better formula for the number of partial orders, but without the logarithm I believe they cease to be asymptotically equal.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago

























            answered 8 hours ago









            Matt SamuelMatt Samuel

            40.4k6 gold badges38 silver badges71 bronze badges




            40.4k6 gold badges38 silver badges71 bronze badges











            • $begingroup$
              Are you referring to this article? researchgate.net/publication/…
              $endgroup$
              – Keen-ameteur
              4 hours ago











            • $begingroup$
              @Keen Yes. I guess I was remembering the wrong title.
              $endgroup$
              – Matt Samuel
              4 hours ago
















            • $begingroup$
              Are you referring to this article? researchgate.net/publication/…
              $endgroup$
              – Keen-ameteur
              4 hours ago











            • $begingroup$
              @Keen Yes. I guess I was remembering the wrong title.
              $endgroup$
              – Matt Samuel
              4 hours ago















            $begingroup$
            Are you referring to this article? researchgate.net/publication/…
            $endgroup$
            – Keen-ameteur
            4 hours ago





            $begingroup$
            Are you referring to this article? researchgate.net/publication/…
            $endgroup$
            – Keen-ameteur
            4 hours ago













            $begingroup$
            @Keen Yes. I guess I was remembering the wrong title.
            $endgroup$
            – Matt Samuel
            4 hours ago




            $begingroup$
            @Keen Yes. I guess I was remembering the wrong title.
            $endgroup$
            – Matt Samuel
            4 hours ago













            1












            $begingroup$

            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
            $$tag1ale biff overlineasubseteq overlineb$$
            (or equivalently, $ain overlineb$).
            This need not be a partial order because we can have $ale b$ and $ble a$ with $ane b$. 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
            $$tag2 Atext closediffforall ain Acolon forall xin Scolon xle ato xin A.$$
            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
            $$ T_n=P_n+nchoose 2P_n-1+left(nchoose 3+frac12nchoose 2n-2choose 2right)P_n-2+ldots+2^n-1P_2+P_1.$$






            share|cite|improve this answer











            $endgroup$

















              1












              $begingroup$

              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
              $$tag1ale biff overlineasubseteq overlineb$$
              (or equivalently, $ain overlineb$).
              This need not be a partial order because we can have $ale b$ and $ble a$ with $ane b$. 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
              $$tag2 Atext closediffforall ain Acolon forall xin Scolon xle ato xin A.$$
              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
              $$ T_n=P_n+nchoose 2P_n-1+left(nchoose 3+frac12nchoose 2n-2choose 2right)P_n-2+ldots+2^n-1P_2+P_1.$$






              share|cite|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                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
                $$tag1ale biff overlineasubseteq overlineb$$
                (or equivalently, $ain overlineb$).
                This need not be a partial order because we can have $ale b$ and $ble a$ with $ane b$. 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
                $$tag2 Atext closediffforall ain Acolon forall xin Scolon xle ato xin A.$$
                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
                $$ T_n=P_n+nchoose 2P_n-1+left(nchoose 3+frac12nchoose 2n-2choose 2right)P_n-2+ldots+2^n-1P_2+P_1.$$






                share|cite|improve this answer











                $endgroup$



                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
                $$tag1ale biff overlineasubseteq overlineb$$
                (or equivalently, $ain overlineb$).
                This need not be a partial order because we can have $ale b$ and $ble a$ with $ane b$. 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
                $$tag2 Atext closediffforall ain Acolon forall xin Scolon xle ato xin A.$$
                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
                $$ T_n=P_n+nchoose 2P_n-1+left(nchoose 3+frac12nchoose 2n-2choose 2right)P_n-2+ldots+2^n-1P_2+P_1.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 7 hours ago

























                answered 7 hours ago









                Hagen von EitzenHagen von Eitzen

                292k23 gold badges279 silver badges515 bronze badges




                292k23 gold badges279 silver badges515 bronze badges



























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