Construction of uncountably many non-isomorphic linear (total) orderings of natural numbers












0














I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).



I've already constructed two non-isomorphic total orderings. They are
$$
1 < 2 < 3 <ldots\
1 > 2 > 3 >ldots
$$
where "$<$" and "$>$" are strict total ordering relations.



Thanks in advance for your help.






elementary-set-theory order-theory natural-numbers







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    0














    I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).



    I've already constructed two non-isomorphic total orderings. They are
    $$
    1 < 2 < 3 <ldots\
    1 > 2 > 3 >ldots
    $$
    where "$<$" and "$>$" are strict total ordering relations.



    Thanks in advance for your help.






    elementary-set-theory order-theory natural-numbers







    share|cite|improve this question



























      0












      0








      0


      1





      I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).



      I've already constructed two non-isomorphic total orderings. They are
      $$
      1 < 2 < 3 <ldots\
      1 > 2 > 3 >ldots
      $$
      where "$<$" and "$>$" are strict total ordering relations.



      Thanks in advance for your help.






      elementary-set-theory order-theory natural-numbers







      share|cite|improve this question















      I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).



      I've already constructed two non-isomorphic total orderings. They are
      $$
      1 < 2 < 3 <ldots\
      1 > 2 > 3 >ldots
      $$
      where "$<$" and "$>$" are strict total ordering relations.



      Thanks in advance for your help.






      elementary-set-theory order-theory natural-numbers




      elementary-set-theory order-theory natural-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 29 '14 at 20:21









      Asaf Karagila♦

      302k32427757




      302k32427757










      asked Oct 29 '14 at 20:08









      EncorteEncorte

      112




      112






















          2 Answers
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          2














          HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.



          However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.






          share|cite|improve this answer





























            0














            You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.






            share|cite|improve this answer





















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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.



              However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.






              share|cite|improve this answer


























                2














                HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.



                However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.



                  However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.






                  share|cite|improve this answer












                  HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.



                  However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 29 '14 at 20:12









                  Asaf Karagila♦Asaf Karagila

                  302k32427757




                  302k32427757























                      0














                      You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.






                      share|cite|improve this answer


























                        0














                        You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.






                          share|cite|improve this answer












                          You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 24 '18 at 17:50









                          The Driven manThe Driven man

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