How to prove that for any nonempty set $A$, there exists a maximal partial order $lambda$ with the axiom of...












1












$begingroup$


The ``maximal partial order'' means for any partial order $alphainmathscr{B}(A)$, $lambdasubseteqalpha$ implies $lambda=alpha$.



It is obvious if we apply the well-ordering theorem or Zorn's lemma. Since the equivalence among well-ordering theorem, Zorn's lemma and the axiom of choice, we can give a proof directly from the axiom of choice, I mean, not a pretended proof( to prove Zorn's lemma first and then use the lemma).



Could any one help me?






order-theory







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$endgroup$

















    1












    $begingroup$


    The ``maximal partial order'' means for any partial order $alphainmathscr{B}(A)$, $lambdasubseteqalpha$ implies $lambda=alpha$.



    It is obvious if we apply the well-ordering theorem or Zorn's lemma. Since the equivalence among well-ordering theorem, Zorn's lemma and the axiom of choice, we can give a proof directly from the axiom of choice, I mean, not a pretended proof( to prove Zorn's lemma first and then use the lemma).



    Could any one help me?






    order-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      The ``maximal partial order'' means for any partial order $alphainmathscr{B}(A)$, $lambdasubseteqalpha$ implies $lambda=alpha$.



      It is obvious if we apply the well-ordering theorem or Zorn's lemma. Since the equivalence among well-ordering theorem, Zorn's lemma and the axiom of choice, we can give a proof directly from the axiom of choice, I mean, not a pretended proof( to prove Zorn's lemma first and then use the lemma).



      Could any one help me?






      order-theory







      share|cite|improve this question









      $endgroup$




      The ``maximal partial order'' means for any partial order $alphainmathscr{B}(A)$, $lambdasubseteqalpha$ implies $lambda=alpha$.



      It is obvious if we apply the well-ordering theorem or Zorn's lemma. Since the equivalence among well-ordering theorem, Zorn's lemma and the axiom of choice, we can give a proof directly from the axiom of choice, I mean, not a pretended proof( to prove Zorn's lemma first and then use the lemma).



      Could any one help me?






      order-theory




      order-theory






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      share|cite|improve this question











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      asked Jan 21 at 6:36









      闫嘉琦闫嘉琦

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          $begingroup$

          Apply Zorn's lemma to the collection $mathcal{C}$ of partial orders in $A$. Here $mathcal{C}$ is ordered by $subseteq$.






          share|cite|improve this answer









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            $begingroup$

            Apply Zorn's lemma to the collection $mathcal{C}$ of partial orders in $A$. Here $mathcal{C}$ is ordered by $subseteq$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Apply Zorn's lemma to the collection $mathcal{C}$ of partial orders in $A$. Here $mathcal{C}$ is ordered by $subseteq$.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                Apply Zorn's lemma to the collection $mathcal{C}$ of partial orders in $A$. Here $mathcal{C}$ is ordered by $subseteq$.






                share|cite|improve this answer









                $endgroup$



                Apply Zorn's lemma to the collection $mathcal{C}$ of partial orders in $A$. Here $mathcal{C}$ is ordered by $subseteq$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 19:52









                Alberto TakaseAlberto Takase

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