Inverse of set operations












1












$begingroup$


$A cup B equiv C $ then what is $A$ in terms of $B,C$?



I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.



From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.



Complement is easy as it is it's own inverse $(A^C)^C=A$



not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.



when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?










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$endgroup$












  • $begingroup$
    What does $equiv$ mean in this context? Why aren't you saying they are equal?
    $endgroup$
    – fleablood
    5 hours ago
















1












$begingroup$


$A cup B equiv C $ then what is $A$ in terms of $B,C$?



I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.



From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.



Complement is easy as it is it's own inverse $(A^C)^C=A$



not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.



when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?










share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $equiv$ mean in this context? Why aren't you saying they are equal?
    $endgroup$
    – fleablood
    5 hours ago














1












1








1





$begingroup$


$A cup B equiv C $ then what is $A$ in terms of $B,C$?



I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.



From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.



Complement is easy as it is it's own inverse $(A^C)^C=A$



not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.



when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?










share|cite|improve this question











$endgroup$




$A cup B equiv C $ then what is $A$ in terms of $B,C$?



I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.



From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.



Complement is easy as it is it's own inverse $(A^C)^C=A$



not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.



when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?







elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







Arjang

















asked 5 hours ago









ArjangArjang

5,59262363




5,59262363












  • $begingroup$
    What does $equiv$ mean in this context? Why aren't you saying they are equal?
    $endgroup$
    – fleablood
    5 hours ago


















  • $begingroup$
    What does $equiv$ mean in this context? Why aren't you saying they are equal?
    $endgroup$
    – fleablood
    5 hours ago
















$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
5 hours ago




$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
5 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Inverse operators for union, intersection, or set difference are impossible in general.



Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".





On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.



If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.



This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$



What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.



(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice counter example!
    $endgroup$
    – fleablood
    4 hours ago










  • $begingroup$
    So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
    $endgroup$
    – fleablood
    3 hours ago



















2












$begingroup$

It's not possible.



Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.



1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)



2) ${x not in C, x in B} = emptyset$ (as $B subset C$)



3) ${x in C, x not in B} = Csetminus B$.



4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).



Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.



Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.



====



FWIW



We can list all possible sets possible to describe they are



The four above:



1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.



[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.



[1 and 4] and [1 and 2 and 4] = $C^c cup B$



[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$



[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.






share|cite|improve this answer











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    2 Answers
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    2 Answers
    2






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    active

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    2












    $begingroup$

    Inverse operators for union, intersection, or set difference are impossible in general.



    Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
    $$ ? cup {1,2} = {1,2,3} $$
    Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".





    On the other hand symmetric difference
    $$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
    has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.



    If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.



    This gives an opportunity to use more of the usual algebraic rules on sets.
    And you can express the remaining set operations in this vocabulary too:
    $$ A^complement = 1 mathoptriangle A qquadqquad
    Acup B = Amathoptriangle B mathoptriangle(Acap B) $$



    What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.



    (The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice counter example!
      $endgroup$
      – fleablood
      4 hours ago










    • $begingroup$
      So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
      $endgroup$
      – fleablood
      3 hours ago
















    2












    $begingroup$

    Inverse operators for union, intersection, or set difference are impossible in general.



    Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
    $$ ? cup {1,2} = {1,2,3} $$
    Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".





    On the other hand symmetric difference
    $$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
    has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.



    If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.



    This gives an opportunity to use more of the usual algebraic rules on sets.
    And you can express the remaining set operations in this vocabulary too:
    $$ A^complement = 1 mathoptriangle A qquadqquad
    Acup B = Amathoptriangle B mathoptriangle(Acap B) $$



    What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.



    (The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice counter example!
      $endgroup$
      – fleablood
      4 hours ago










    • $begingroup$
      So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
      $endgroup$
      – fleablood
      3 hours ago














    2












    2








    2





    $begingroup$

    Inverse operators for union, intersection, or set difference are impossible in general.



    Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
    $$ ? cup {1,2} = {1,2,3} $$
    Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".





    On the other hand symmetric difference
    $$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
    has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.



    If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.



    This gives an opportunity to use more of the usual algebraic rules on sets.
    And you can express the remaining set operations in this vocabulary too:
    $$ A^complement = 1 mathoptriangle A qquadqquad
    Acup B = Amathoptriangle B mathoptriangle(Acap B) $$



    What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.



    (The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).






    share|cite|improve this answer











    $endgroup$



    Inverse operators for union, intersection, or set difference are impossible in general.



    Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
    $$ ? cup {1,2} = {1,2,3} $$
    Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".





    On the other hand symmetric difference
    $$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
    has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.



    If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.



    This gives an opportunity to use more of the usual algebraic rules on sets.
    And you can express the remaining set operations in this vocabulary too:
    $$ A^complement = 1 mathoptriangle A qquadqquad
    Acup B = Amathoptriangle B mathoptriangle(Acap B) $$



    What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.



    (The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 4 hours ago

























    answered 4 hours ago









    Henning MakholmHenning Makholm

    239k17304541




    239k17304541












    • $begingroup$
      Nice counter example!
      $endgroup$
      – fleablood
      4 hours ago










    • $begingroup$
      So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
      $endgroup$
      – fleablood
      3 hours ago


















    • $begingroup$
      Nice counter example!
      $endgroup$
      – fleablood
      4 hours ago










    • $begingroup$
      So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
      $endgroup$
      – fleablood
      3 hours ago
















    $begingroup$
    Nice counter example!
    $endgroup$
    – fleablood
    4 hours ago




    $begingroup$
    Nice counter example!
    $endgroup$
    – fleablood
    4 hours ago












    $begingroup$
    So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
    $endgroup$
    – fleablood
    3 hours ago




    $begingroup$
    So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
    $endgroup$
    – fleablood
    3 hours ago











    2












    $begingroup$

    It's not possible.



    Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.



    1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)



    2) ${x not in C, x in B} = emptyset$ (as $B subset C$)



    3) ${x in C, x not in B} = Csetminus B$.



    4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).



    Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.



    Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.



    ====



    FWIW



    We can list all possible sets possible to describe they are



    The four above:



    1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.



    [1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.



    [1 and 4] and [1 and 2 and 4] = $C^c cup B$



    [3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$



    [1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      It's not possible.



      Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.



      1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)



      2) ${x not in C, x in B} = emptyset$ (as $B subset C$)



      3) ${x in C, x not in B} = Csetminus B$.



      4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).



      Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.



      Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.



      ====



      FWIW



      We can list all possible sets possible to describe they are



      The four above:



      1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.



      [1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.



      [1 and 4] and [1 and 2 and 4] = $C^c cup B$



      [3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$



      [1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        It's not possible.



        Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.



        1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)



        2) ${x not in C, x in B} = emptyset$ (as $B subset C$)



        3) ${x in C, x not in B} = Csetminus B$.



        4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).



        Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.



        Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.



        ====



        FWIW



        We can list all possible sets possible to describe they are



        The four above:



        1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.



        [1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.



        [1 and 4] and [1 and 2 and 4] = $C^c cup B$



        [3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$



        [1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.






        share|cite|improve this answer











        $endgroup$



        It's not possible.



        Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.



        1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)



        2) ${x not in C, x in B} = emptyset$ (as $B subset C$)



        3) ${x in C, x not in B} = Csetminus B$.



        4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).



        Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.



        Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.



        ====



        FWIW



        We can list all possible sets possible to describe they are



        The four above:



        1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.



        [1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.



        [1 and 4] and [1 and 2 and 4] = $C^c cup B$



        [3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$



        [1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 3 hours ago

























        answered 4 hours ago









        fleabloodfleablood

        69.1k22685




        69.1k22685






























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