Characteristic function / Indicator function












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I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.










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    0












    $begingroup$


    I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.










      share|cite|improve this question











      $endgroup$




      I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.







      elementary-number-theory elementary-set-theory






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      edited Jan 6 at 13:39









      Andrés E. Caicedo

      65.4k8158249




      65.4k8158249










      asked Jan 6 at 13:38









      DraxlerDraxler

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      122






















          2 Answers
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          $begingroup$

          The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :




          $1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.




          But a function is a set of pairs, i.e. a subset of the cartesian product.



          Thus :




          $1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$




          and : $1_A subseteq X times { 0,1 }$.






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            $begingroup$

            For definition of characteristic function see the answer of Mauro.



            If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.



            This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$



            Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$



            Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.





            The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$



            For a proof of $(1)$ see this answer.



            For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$






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              2 Answers
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              2 Answers
              2






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              active

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              1












              $begingroup$

              The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :




              $1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.




              But a function is a set of pairs, i.e. a subset of the cartesian product.



              Thus :




              $1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$




              and : $1_A subseteq X times { 0,1 }$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :




                $1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.




                But a function is a set of pairs, i.e. a subset of the cartesian product.



                Thus :




                $1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$




                and : $1_A subseteq X times { 0,1 }$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :




                  $1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.




                  But a function is a set of pairs, i.e. a subset of the cartesian product.



                  Thus :




                  $1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$




                  and : $1_A subseteq X times { 0,1 }$.






                  share|cite|improve this answer









                  $endgroup$



                  The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :




                  $1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.




                  But a function is a set of pairs, i.e. a subset of the cartesian product.



                  Thus :




                  $1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$




                  and : $1_A subseteq X times { 0,1 }$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 13:56









                  Mauro ALLEGRANZAMauro ALLEGRANZA

                  65.9k449114




                  65.9k449114























                      0












                      $begingroup$

                      For definition of characteristic function see the answer of Mauro.



                      If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.



                      This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$



                      Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$



                      Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.





                      The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$



                      For a proof of $(1)$ see this answer.



                      For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        For definition of characteristic function see the answer of Mauro.



                        If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.



                        This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$



                        Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$



                        Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.





                        The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$



                        For a proof of $(1)$ see this answer.



                        For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          For definition of characteristic function see the answer of Mauro.



                          If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.



                          This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$



                          Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$



                          Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.





                          The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$



                          For a proof of $(1)$ see this answer.



                          For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$






                          share|cite|improve this answer











                          $endgroup$



                          For definition of characteristic function see the answer of Mauro.



                          If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.



                          This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$



                          Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$



                          Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.





                          The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$



                          For a proof of $(1)$ see this answer.



                          For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$







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                          share|cite|improve this answer








                          edited Jan 6 at 14:38

























                          answered Jan 6 at 14:23









                          drhabdrhab

                          101k545136




                          101k545136






























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