Partially ordered sets cardinality












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What is the cardinality of the set of all partially ordered sets of natural numbers which have one least element and infinity number of maximal elements?



I only noticed that upperbound for this set is $P(mathbb{N}timesmathbb{N}) $ which cardinality is $mathfrak{C} $.










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    0












    $begingroup$


    What is the cardinality of the set of all partially ordered sets of natural numbers which have one least element and infinity number of maximal elements?



    I only noticed that upperbound for this set is $P(mathbb{N}timesmathbb{N}) $ which cardinality is $mathfrak{C} $.










    share|cite|improve this question











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      0








      0





      $begingroup$


      What is the cardinality of the set of all partially ordered sets of natural numbers which have one least element and infinity number of maximal elements?



      I only noticed that upperbound for this set is $P(mathbb{N}timesmathbb{N}) $ which cardinality is $mathfrak{C} $.










      share|cite|improve this question











      $endgroup$




      What is the cardinality of the set of all partially ordered sets of natural numbers which have one least element and infinity number of maximal elements?



      I only noticed that upperbound for this set is $P(mathbb{N}timesmathbb{N}) $ which cardinality is $mathfrak{C} $.







      elementary-set-theory order-theory cardinals






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      edited Jan 18 at 3:47









      Andrés E. Caicedo

      65.5k8159250




      65.5k8159250










      asked Jan 17 at 21:55









      avan1235avan1235

      3297




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          Given $Ssubset Bbb N$, we turn it into a poset by defining $apreceq biff amid b$. Now let $S$ contain $1$, all prime-squares, and an arbitrary set of primes. Then $S$ is of the kind we are interested in. As we can pick an arbitrary subset of the primes, there are $2^{aleph_0}$ such posets.






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            $begingroup$

            Given $Ssubset Bbb N$, we turn it into a poset by defining $apreceq biff amid b$. Now let $S$ contain $1$, all prime-squares, and an arbitrary set of primes. Then $S$ is of the kind we are interested in. As we can pick an arbitrary subset of the primes, there are $2^{aleph_0}$ such posets.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Given $Ssubset Bbb N$, we turn it into a poset by defining $apreceq biff amid b$. Now let $S$ contain $1$, all prime-squares, and an arbitrary set of primes. Then $S$ is of the kind we are interested in. As we can pick an arbitrary subset of the primes, there are $2^{aleph_0}$ such posets.






              share|cite|improve this answer









              $endgroup$
















                3












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                3





                $begingroup$

                Given $Ssubset Bbb N$, we turn it into a poset by defining $apreceq biff amid b$. Now let $S$ contain $1$, all prime-squares, and an arbitrary set of primes. Then $S$ is of the kind we are interested in. As we can pick an arbitrary subset of the primes, there are $2^{aleph_0}$ such posets.






                share|cite|improve this answer









                $endgroup$



                Given $Ssubset Bbb N$, we turn it into a poset by defining $apreceq biff amid b$. Now let $S$ contain $1$, all prime-squares, and an arbitrary set of primes. Then $S$ is of the kind we are interested in. As we can pick an arbitrary subset of the primes, there are $2^{aleph_0}$ such posets.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 22:07









                Hagen von EitzenHagen von Eitzen

                281k23272505




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