If $a notin A setminus B$ and $a in A$ , show that $a in B $.












0












$begingroup$


If $a notin A setminus B$ and $a in A$ , show that $a in B $.



hi. i am studying the basic set theory.



I want to prove that statement. but precisely i don't know the proof.






elementary-set-theory







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  • $begingroup$
    $Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
    $endgroup$
    – babemcnuggets
    Feb 3 at 0:05


















0












$begingroup$


If $a notin A setminus B$ and $a in A$ , show that $a in B $.



hi. i am studying the basic set theory.



I want to prove that statement. but precisely i don't know the proof.






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
    $endgroup$
    – babemcnuggets
    Feb 3 at 0:05
















0












0








0





$begingroup$


If $a notin A setminus B$ and $a in A$ , show that $a in B $.



hi. i am studying the basic set theory.



I want to prove that statement. but precisely i don't know the proof.






elementary-set-theory







share|cite|improve this question









$endgroup$




If $a notin A setminus B$ and $a in A$ , show that $a in B $.



hi. i am studying the basic set theory.



I want to prove that statement. but precisely i don't know the proof.






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 Feb 2 at 23:53









y2noy2no

204




204












  • $begingroup$
    $Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
    $endgroup$
    – babemcnuggets
    Feb 3 at 0:05




















  • $begingroup$
    $Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
    $endgroup$
    – babemcnuggets
    Feb 3 at 0:05


















$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05






$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05












3 Answers
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$begingroup$

Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
$$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$



So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$






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    0












    $begingroup$

    The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).



    But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $A - B = {a : a in A land anotin B}$.



      Let's say $a in A$. Then $a notin A - B$ implies directly:



      $a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..






      share|cite|improve this answer









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        3 Answers
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        3 Answers
        3






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        2












        $begingroup$

        Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
        $$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$



        So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
          $$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$



          So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
            $$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$



            So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$






            share|cite|improve this answer









            $endgroup$



            Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
            $$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$



            So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 3 at 0:10









            greeliousgreelious

            472112




            472112























                0












                $begingroup$

                The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).



                But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).



                  But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).



                    But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.






                    share|cite|improve this answer









                    $endgroup$



                    The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).



                    But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 3 at 0:07









                    davidlowryduda♦davidlowryduda

                    75.2k7121256




                    75.2k7121256























                        0












                        $begingroup$

                        $A - B = {a : a in A land anotin B}$.



                        Let's say $a in A$. Then $a notin A - B$ implies directly:



                        $a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $A - B = {a : a in A land anotin B}$.



                          Let's say $a in A$. Then $a notin A - B$ implies directly:



                          $a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $A - B = {a : a in A land anotin B}$.



                            Let's say $a in A$. Then $a notin A - B$ implies directly:



                            $a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..






                            share|cite|improve this answer









                            $endgroup$



                            $A - B = {a : a in A land anotin B}$.



                            Let's say $a in A$. Then $a notin A - B$ implies directly:



                            $a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 3 at 0:33









                            MariahMariah

                            2,1471718




                            2,1471718






























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