If $x in A$ why does it not follow that $x in A-B$?












1














I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?










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  • 2




    Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
    – amWhy
    Nov 28 at 23:55








  • 1




    Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
    – JonathanZ
    Nov 29 at 0:02












  • So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
    – cap
    Nov 29 at 0:11


















1














I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?










share|cite|improve this question


















  • 2




    Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
    – amWhy
    Nov 28 at 23:55








  • 1




    Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
    – JonathanZ
    Nov 29 at 0:02












  • So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
    – cap
    Nov 29 at 0:11
















1












1








1







I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?










share|cite|improve this question













I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?







elementary-set-theory






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asked Nov 28 at 23:52









cap

103




103








  • 2




    Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
    – amWhy
    Nov 28 at 23:55








  • 1




    Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
    – JonathanZ
    Nov 29 at 0:02












  • So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
    – cap
    Nov 29 at 0:11
















  • 2




    Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
    – amWhy
    Nov 28 at 23:55








  • 1




    Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
    – JonathanZ
    Nov 29 at 0:02












  • So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
    – cap
    Nov 29 at 0:11










2




2




Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55






Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55






1




1




Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02






Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02














So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11






So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11












1 Answer
1






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2














Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
$$
A = (Asetminus B )cup (Acap B)
$$

and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.



For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.






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    1 Answer
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    Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
    $$
    A = (Asetminus B )cup (Acap B)
    $$

    and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.



    For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.






    share|cite|improve this answer


























      2














      Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
      $$
      A = (Asetminus B )cup (Acap B)
      $$

      and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.



      For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.






      share|cite|improve this answer
























        2












        2








        2






        Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
        $$
        A = (Asetminus B )cup (Acap B)
        $$

        and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.



        For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.






        share|cite|improve this answer












        Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
        $$
        A = (Asetminus B )cup (Acap B)
        $$

        and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.



        For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 23:56









        Clement C.

        49.4k33785




        49.4k33785






























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