Can we find a well ordering on an infinite set with no largest element?












5












$begingroup$


According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.

My question is:




For a given infinite set, can we find a well ordering on it such that there is no largest element?







elementary-set-theory well-orders







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Set of natural numbers.
    $endgroup$
    – Wuestenfux
    Jul 2 '17 at 18:54






  • 2




    $begingroup$
    Induce the well-order by a bijection with a cardinal number.
    $endgroup$
    – Daniel Fischer
    Jul 2 '17 at 18:55










  • $begingroup$
    Use Choice + transfinite induction to define it one at a time?
    $endgroup$
    – Chappers
    Jul 2 '17 at 19:02










  • $begingroup$
    math.stackexchange.com/questions/6501/…
    $endgroup$
    – Count Iblis
    Jul 2 '17 at 19:16










  • $begingroup$
    @CountIblis That question is not related to this one at all.
    $endgroup$
    – Noah Schweber
    Jul 2 '17 at 20:02
















5












$begingroup$


According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.

My question is:




For a given infinite set, can we find a well ordering on it such that there is no largest element?







elementary-set-theory well-orders







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Set of natural numbers.
    $endgroup$
    – Wuestenfux
    Jul 2 '17 at 18:54






  • 2




    $begingroup$
    Induce the well-order by a bijection with a cardinal number.
    $endgroup$
    – Daniel Fischer
    Jul 2 '17 at 18:55










  • $begingroup$
    Use Choice + transfinite induction to define it one at a time?
    $endgroup$
    – Chappers
    Jul 2 '17 at 19:02










  • $begingroup$
    math.stackexchange.com/questions/6501/…
    $endgroup$
    – Count Iblis
    Jul 2 '17 at 19:16










  • $begingroup$
    @CountIblis That question is not related to this one at all.
    $endgroup$
    – Noah Schweber
    Jul 2 '17 at 20:02














5












5








5


2



$begingroup$


According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.

My question is:




For a given infinite set, can we find a well ordering on it such that there is no largest element?







elementary-set-theory well-orders







share|cite|improve this question











$endgroup$




According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.

My question is:




For a given infinite set, can we find a well ordering on it such that there is no largest element?







elementary-set-theory well-orders




elementary-set-theory well-orders






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 2 '17 at 22:52









RQM

1213




1213










asked Jul 2 '17 at 18:53









BumblebeeBumblebee

9,74912551




9,74912551








  • 2




    $begingroup$
    Set of natural numbers.
    $endgroup$
    – Wuestenfux
    Jul 2 '17 at 18:54






  • 2




    $begingroup$
    Induce the well-order by a bijection with a cardinal number.
    $endgroup$
    – Daniel Fischer
    Jul 2 '17 at 18:55










  • $begingroup$
    Use Choice + transfinite induction to define it one at a time?
    $endgroup$
    – Chappers
    Jul 2 '17 at 19:02










  • $begingroup$
    math.stackexchange.com/questions/6501/…
    $endgroup$
    – Count Iblis
    Jul 2 '17 at 19:16










  • $begingroup$
    @CountIblis That question is not related to this one at all.
    $endgroup$
    – Noah Schweber
    Jul 2 '17 at 20:02














  • 2




    $begingroup$
    Set of natural numbers.
    $endgroup$
    – Wuestenfux
    Jul 2 '17 at 18:54






  • 2




    $begingroup$
    Induce the well-order by a bijection with a cardinal number.
    $endgroup$
    – Daniel Fischer
    Jul 2 '17 at 18:55










  • $begingroup$
    Use Choice + transfinite induction to define it one at a time?
    $endgroup$
    – Chappers
    Jul 2 '17 at 19:02










  • $begingroup$
    math.stackexchange.com/questions/6501/…
    $endgroup$
    – Count Iblis
    Jul 2 '17 at 19:16










  • $begingroup$
    @CountIblis That question is not related to this one at all.
    $endgroup$
    – Noah Schweber
    Jul 2 '17 at 20:02








2




2




$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54




$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54




2




2




$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer
Jul 2 '17 at 18:55




$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer
Jul 2 '17 at 18:55












$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02




$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02












$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16




$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16












$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02




$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02










3 Answers
3






active

oldest

votes


















9












$begingroup$

Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.



Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.



It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.



If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff




  • $a, bin I_X$ and $a<b$ in the original sense of $X$; or


  • $a, bin F_X$ and $a<b$ in the original sense of $X$; or


  • $ain F_X, bin I_X$.







share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
      $endgroup$
      – Noah Schweber
      Jul 2 '17 at 21:48






    • 1




      $begingroup$
      @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
      $endgroup$
      – Orest Bucicovschi
      Jul 2 '17 at 23:44



















    2












    $begingroup$

    In the usual development of ZFC, a cardinal is defined to the the
    least ordinal with a given cardinality. Infinite cardinals are "limit
    ordinals" --- as ordered sets they have no largest element.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9












      $begingroup$

      Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.



      Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.



      It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.



      If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff




      • $a, bin I_X$ and $a<b$ in the original sense of $X$; or


      • $a, bin F_X$ and $a<b$ in the original sense of $X$; or


      • $ain F_X, bin I_X$.







      share|cite|improve this answer











      $endgroup$


















        9












        $begingroup$

        Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.



        Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.



        It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.



        If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff




        • $a, bin I_X$ and $a<b$ in the original sense of $X$; or


        • $a, bin F_X$ and $a<b$ in the original sense of $X$; or


        • $ain F_X, bin I_X$.







        share|cite|improve this answer











        $endgroup$
















          9












          9








          9





          $begingroup$

          Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.



          Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.



          It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.



          If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff




          • $a, bin I_X$ and $a<b$ in the original sense of $X$; or


          • $a, bin F_X$ and $a<b$ in the original sense of $X$; or


          • $ain F_X, bin I_X$.







          share|cite|improve this answer











          $endgroup$



          Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.



          Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.



          It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.



          If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff




          • $a, bin I_X$ and $a<b$ in the original sense of $X$; or


          • $a, bin F_X$ and $a<b$ in the original sense of $X$; or


          • $ain F_X, bin I_X$.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 2 '17 at 21:47

























          answered Jul 2 '17 at 20:05









          Noah SchweberNoah Schweber

          128k10152294




          128k10152294























              3












              $begingroup$

              If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
                $endgroup$
                – Noah Schweber
                Jul 2 '17 at 21:48






              • 1




                $begingroup$
                @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
                $endgroup$
                – Orest Bucicovschi
                Jul 2 '17 at 23:44
















              3












              $begingroup$

              If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
                $endgroup$
                – Noah Schweber
                Jul 2 '17 at 21:48






              • 1




                $begingroup$
                @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
                $endgroup$
                – Orest Bucicovschi
                Jul 2 '17 at 23:44














              3












              3








              3





              $begingroup$

              If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.






              share|cite|improve this answer









              $endgroup$



              If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jul 2 '17 at 19:58









              Orest BucicovschiOrest Bucicovschi

              28.6k31748




              28.6k31748








              • 2




                $begingroup$
                As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
                $endgroup$
                – Noah Schweber
                Jul 2 '17 at 21:48






              • 1




                $begingroup$
                @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
                $endgroup$
                – Orest Bucicovschi
                Jul 2 '17 at 23:44














              • 2




                $begingroup$
                As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
                $endgroup$
                – Noah Schweber
                Jul 2 '17 at 21:48






              • 1




                $begingroup$
                @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
                $endgroup$
                – Orest Bucicovschi
                Jul 2 '17 at 23:44








              2




              2




              $begingroup$
              As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
              $endgroup$
              – Noah Schweber
              Jul 2 '17 at 21:48




              $begingroup$
              As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
              $endgroup$
              – Noah Schweber
              Jul 2 '17 at 21:48




              1




              1




              $begingroup$
              @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
              $endgroup$
              – Orest Bucicovschi
              Jul 2 '17 at 23:44




              $begingroup$
              @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
              $endgroup$
              – Orest Bucicovschi
              Jul 2 '17 at 23:44











              2












              $begingroup$

              In the usual development of ZFC, a cardinal is defined to the the
              least ordinal with a given cardinality. Infinite cardinals are "limit
              ordinals" --- as ordered sets they have no largest element.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                In the usual development of ZFC, a cardinal is defined to the the
                least ordinal with a given cardinality. Infinite cardinals are "limit
                ordinals" --- as ordered sets they have no largest element.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  In the usual development of ZFC, a cardinal is defined to the the
                  least ordinal with a given cardinality. Infinite cardinals are "limit
                  ordinals" --- as ordered sets they have no largest element.






                  share|cite|improve this answer









                  $endgroup$



                  In the usual development of ZFC, a cardinal is defined to the the
                  least ordinal with a given cardinality. Infinite cardinals are "limit
                  ordinals" --- as ordered sets they have no largest element.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 2 '17 at 18:56









                  Lord Shark the UnknownLord Shark the Unknown

                  108k1162135




                  108k1162135






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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