$ A = (Acap E^c) cup (Acap E)$? The 2019 Stack Overflow Developer Survey Results Are InIs $(X cup Y)cap Z subset X cup(Y cap Z)$?Does $(Acup B)cap(Ccup D)=(Acap C)cup(Bcap D)$?Proving $A cup (B cap C) = (A cup B) cap (A cup C)$.$A=(A cap B) cup(A cap B^mathsfc) $Prove that $ (A cup B) cap C subseteq A cup (B cap C)$Simple set theory: $Acap (Bcup C) subset (A cap B) cup C$Is $Acup B=Acup Bcap A^c$?Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Proof verification: prove $Acup (Bcap C) = (Acup B) cap (A cup C)$.How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$

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$ A = (Acap E^c) cup (Acap E)$?



The 2019 Stack Overflow Developer Survey Results Are InIs $(X cup Y)cap Z subset X cup(Y cap Z)$?Does $(Acup B)cap(Ccup D)=(Acap C)cup(Bcap D)$?Proving $A cup (B cap C) = (A cup B) cap (A cup C)$.$A=(A cap B) cup(A cap B^mathsfc) $Prove that $ (A cup B) cap C subseteq A cup (B cap C)$Simple set theory: $Acap (Bcup C) subset (A cap B) cup C$Is $Acup B=Acup Bcap A^c$?Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Proof verification: prove $Acup (Bcap C) = (Acup B) cap (A cup C)$.How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$










1












$begingroup$


$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?



Here is my proof:



Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.






elementary-set-theory







share|cite|improve this question









$endgroup$











  • $begingroup$
    $A subset A cap E$ <- this is false
    $endgroup$
    – Benitok
    Mar 23 at 17:34















1












$begingroup$


$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?



Here is my proof:



Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.






elementary-set-theory







share|cite|improve this question









$endgroup$











  • $begingroup$
    $A subset A cap E$ <- this is false
    $endgroup$
    – Benitok
    Mar 23 at 17:34













1












1








1





$begingroup$


$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?



Here is my proof:



Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.






elementary-set-theory







share|cite|improve this question









$endgroup$




$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?



Here is my proof:



Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.






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 Mar 23 at 17:21









José MarínJosé Marín

253211




253211











  • $begingroup$
    $A subset A cap E$ <- this is false
    $endgroup$
    – Benitok
    Mar 23 at 17:34
















  • $begingroup$
    $A subset A cap E$ <- this is false
    $endgroup$
    – Benitok
    Mar 23 at 17:34















$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34




$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34










2 Answers
2






active

oldest

votes


















2












$begingroup$

There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.



I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.



Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.



To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.



If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    That's true. Thank you.
    $endgroup$
    – José Marín
    Mar 23 at 18:03


















2












$begingroup$

That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.



$$
beginalign
A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
&= A cap ( E cup E^c) hskip1cm text (Complement law)\
&= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
endalign
$$






share|cite|improve this answer









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.



    I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.



    Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.



    To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.



    If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's true. Thank you.
      $endgroup$
      – José Marín
      Mar 23 at 18:03















    2












    $begingroup$

    There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.



    I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.



    Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.



    To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.



    If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's true. Thank you.
      $endgroup$
      – José Marín
      Mar 23 at 18:03













    2












    2








    2





    $begingroup$

    There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.



    I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.



    Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.



    To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.



    If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.






    share|cite|improve this answer









    $endgroup$



    There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.



    I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.



    Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.



    To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.



    If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 23 at 17:36









    AspiringMathematicianAspiringMathematician

    1,878621




    1,878621











    • $begingroup$
      That's true. Thank you.
      $endgroup$
      – José Marín
      Mar 23 at 18:03
















    • $begingroup$
      That's true. Thank you.
      $endgroup$
      – José Marín
      Mar 23 at 18:03















    $begingroup$
    That's true. Thank you.
    $endgroup$
    – José Marín
    Mar 23 at 18:03




    $begingroup$
    That's true. Thank you.
    $endgroup$
    – José Marín
    Mar 23 at 18:03











    2












    $begingroup$

    That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.



    $$
    beginalign
    A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
    &= A cap ( E cup E^c) hskip1cm text (Complement law)\
    &= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
    endalign
    $$






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.



      $$
      beginalign
      A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
      &= A cap ( E cup E^c) hskip1cm text (Complement law)\
      &= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
      endalign
      $$






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.



        $$
        beginalign
        A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
        &= A cap ( E cup E^c) hskip1cm text (Complement law)\
        &= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
        endalign
        $$






        share|cite|improve this answer









        $endgroup$



        That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.



        $$
        beginalign
        A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
        &= A cap ( E cup E^c) hskip1cm text (Complement law)\
        &= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
        endalign
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 23 at 17:36









        leonbloyleonbloy

        42.2k647108




        42.2k647108



























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