$(A times B)^c = (A^c times Y) cup (X times B^c)$












1












$begingroup$


Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.



Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.



I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?



Any help is appreciated...






elementary-set-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.



    Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.



    I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?



    Any help is appreciated...






    elementary-set-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.



      Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.



      I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?



      Any help is appreciated...






      elementary-set-theory







      share|cite|improve this question









      $endgroup$




      Let $X$ and $Y$ be universal sets and $A subseteq X$ and $B subseteq Y$.



      Then, I want to prove that $(A times B)^c=(A^c times Y) cup (X times B^c)$.



      I think that i have to suppose that $(x,y) in (A times B)^c$, then $(x,y) in (X times Y) setminus (Atimes B) Rightarrow (x,y) in (A^c times Y) cup (X times B^c)$, but is this correct? And how should I prove that?



      Any help is appreciated...






      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










      asked Jan 2 at 10:46









      t.ysnt.ysn

      1397




      1397






















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          $begingroup$

          Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.



          For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.



          To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot. :)
            $endgroup$
            – t.ysn
            Jan 2 at 11:07











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          $begingroup$

          Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.



          For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.



          To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot. :)
            $endgroup$
            – t.ysn
            Jan 2 at 11:07
















          1












          $begingroup$

          Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.



          For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.



          To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot. :)
            $endgroup$
            – t.ysn
            Jan 2 at 11:07














          1












          1








          1





          $begingroup$

          Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.



          For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.



          To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.






          share|cite|improve this answer









          $endgroup$



          Let $(x,y)in(Atimes B)^c$. Then $(x,y)notin Atimes B$ which implies that $xnotin A$ or $ynotin B$. Hence, $(x,y)in A^ctimes Y$ or $(x,y)in Xtimes B^c$.



          For the other direction. Let $(x,y)in (A^c times Y) cup (X times B^c)$. This implies that $(x,y)in A^c times Y$ or $(x,y)in X times B^c$, namely $xnotin A$ or $ynotin B$. Thus, $(x,y)notin Atimes B$.



          To see this result and get some intuition check for example: $A=B=[0,1]$ and $X=Y=mathbb{R}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 10:53









          Test123Test123

          2,782828




          2,782828












          • $begingroup$
            Thanks a lot. :)
            $endgroup$
            – t.ysn
            Jan 2 at 11:07


















          • $begingroup$
            Thanks a lot. :)
            $endgroup$
            – t.ysn
            Jan 2 at 11:07
















          $begingroup$
          Thanks a lot. :)
          $endgroup$
          – t.ysn
          Jan 2 at 11:07




          $begingroup$
          Thanks a lot. :)
          $endgroup$
          – t.ysn
          Jan 2 at 11:07


















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