Set of ordered pairs [closed]












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Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






elementary-set-theory







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closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, amWhy Dec 25 '18 at 15:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






    elementary-set-theory







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, amWhy Dec 25 '18 at 15:06


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, amWhy

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


      Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






      elementary-set-theory







      share|cite|improve this question











      $endgroup$




      Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






      elementary-set-theory




      elementary-set-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 11:31









      Asaf Karagila♦

      304k32430763




      304k32430763










      asked Dec 25 '18 at 7:57









      t.ysnt.ysn

      1397




      1397




      closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, amWhy Dec 25 '18 at 15:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, amWhy

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, amWhy Dec 25 '18 at 15:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, amWhy

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          Just take $A={amidexists b; (a,b)in X}$.



          Any idea of your own now what to take for $B $?






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






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            $endgroup$




















              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Just take $A={amidexists b; (a,b)in X}$.



              Any idea of your own now what to take for $B $?






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Just take $A={amidexists b; (a,b)in X}$.



                Any idea of your own now what to take for $B $?






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Just take $A={amidexists b; (a,b)in X}$.



                  Any idea of your own now what to take for $B $?






                  share|cite|improve this answer









                  $endgroup$



                  Just take $A={amidexists b; (a,b)in X}$.



                  Any idea of your own now what to take for $B $?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 8:04









                  drhabdrhab

                  101k544130




                  101k544130























                      1












                      $begingroup$

                      Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






                          share|cite|improve this answer









                          $endgroup$



                          Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 25 '18 at 8:01









                          FrostFrost

                          1776




                          1776















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