Expressing friend group in set theory language and its equivalence relation












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$begingroup$


(A) Let X be the set of people in a social network, and for each x ∈ friends(x) denote the set of friends of x. For each W ⊆ X define



friends(W) = ∪w∈Wfriends(w)



For example if friends(Lea) = {Per} and friends(Kai) = {Lea,Ron}, friends({Lea,Kai}) = {Per,Lea,Ron}. Express the following in the language of set theory.



Does it seem like what I'm doing is correct? Also I'm not sure about the last two.



(i) Some of John's friends are also Kat's friends



                 friends(John) ∩ friends(Kat) ≠ ∅



(ii) Per is a mutual friend of Lea and Kai



                 Per ∈ friends(Lea) ∩ friends(Kai)



(iii) Ron is friends with everybody



                 Ron ∈ ∩w∈Wfriends(w)



(iv) Per's friends who are not friends with Kat are friends with Rob



                 friends(Per) - friends(Kat) ∈ friends(Rob)



(v) Kat is a friend of a friend of Lea.



(vi) Kat is friends with all of her friends' friends



(B) Let X be as before and friends be as in part (a). Define a relation on X by {x,y) | y ∈ friends(x)}. What properties must the function friends satisfy in order for this relation to be:



(i) reflexive



       ∀x ∈ friends(x). x~x



(ii) symmetric



       ∀x,y ∈ friends(x). if x~y then y~x



(iii)transitive



       ∀x,y,z ∈ friends(x). if x~y and y~z then x~z



I'm not sure if this is along the lines of what they are looking for.










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    0












    $begingroup$


    (A) Let X be the set of people in a social network, and for each x ∈ friends(x) denote the set of friends of x. For each W ⊆ X define



    friends(W) = ∪w∈Wfriends(w)



    For example if friends(Lea) = {Per} and friends(Kai) = {Lea,Ron}, friends({Lea,Kai}) = {Per,Lea,Ron}. Express the following in the language of set theory.



    Does it seem like what I'm doing is correct? Also I'm not sure about the last two.



    (i) Some of John's friends are also Kat's friends



                     friends(John) ∩ friends(Kat) ≠ ∅



    (ii) Per is a mutual friend of Lea and Kai



                     Per ∈ friends(Lea) ∩ friends(Kai)



    (iii) Ron is friends with everybody



                     Ron ∈ ∩w∈Wfriends(w)



    (iv) Per's friends who are not friends with Kat are friends with Rob



                     friends(Per) - friends(Kat) ∈ friends(Rob)



    (v) Kat is a friend of a friend of Lea.



    (vi) Kat is friends with all of her friends' friends



    (B) Let X be as before and friends be as in part (a). Define a relation on X by {x,y) | y ∈ friends(x)}. What properties must the function friends satisfy in order for this relation to be:



    (i) reflexive



           ∀x ∈ friends(x). x~x



    (ii) symmetric



           ∀x,y ∈ friends(x). if x~y then y~x



    (iii)transitive



           ∀x,y,z ∈ friends(x). if x~y and y~z then x~z



    I'm not sure if this is along the lines of what they are looking for.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      (A) Let X be the set of people in a social network, and for each x ∈ friends(x) denote the set of friends of x. For each W ⊆ X define



      friends(W) = ∪w∈Wfriends(w)



      For example if friends(Lea) = {Per} and friends(Kai) = {Lea,Ron}, friends({Lea,Kai}) = {Per,Lea,Ron}. Express the following in the language of set theory.



      Does it seem like what I'm doing is correct? Also I'm not sure about the last two.



      (i) Some of John's friends are also Kat's friends



                       friends(John) ∩ friends(Kat) ≠ ∅



      (ii) Per is a mutual friend of Lea and Kai



                       Per ∈ friends(Lea) ∩ friends(Kai)



      (iii) Ron is friends with everybody



                       Ron ∈ ∩w∈Wfriends(w)



      (iv) Per's friends who are not friends with Kat are friends with Rob



                       friends(Per) - friends(Kat) ∈ friends(Rob)



      (v) Kat is a friend of a friend of Lea.



      (vi) Kat is friends with all of her friends' friends



      (B) Let X be as before and friends be as in part (a). Define a relation on X by {x,y) | y ∈ friends(x)}. What properties must the function friends satisfy in order for this relation to be:



      (i) reflexive



             ∀x ∈ friends(x). x~x



      (ii) symmetric



             ∀x,y ∈ friends(x). if x~y then y~x



      (iii)transitive



             ∀x,y,z ∈ friends(x). if x~y and y~z then x~z



      I'm not sure if this is along the lines of what they are looking for.










      share|cite|improve this question









      $endgroup$




      (A) Let X be the set of people in a social network, and for each x ∈ friends(x) denote the set of friends of x. For each W ⊆ X define



      friends(W) = ∪w∈Wfriends(w)



      For example if friends(Lea) = {Per} and friends(Kai) = {Lea,Ron}, friends({Lea,Kai}) = {Per,Lea,Ron}. Express the following in the language of set theory.



      Does it seem like what I'm doing is correct? Also I'm not sure about the last two.



      (i) Some of John's friends are also Kat's friends



                       friends(John) ∩ friends(Kat) ≠ ∅



      (ii) Per is a mutual friend of Lea and Kai



                       Per ∈ friends(Lea) ∩ friends(Kai)



      (iii) Ron is friends with everybody



                       Ron ∈ ∩w∈Wfriends(w)



      (iv) Per's friends who are not friends with Kat are friends with Rob



                       friends(Per) - friends(Kat) ∈ friends(Rob)



      (v) Kat is a friend of a friend of Lea.



      (vi) Kat is friends with all of her friends' friends



      (B) Let X be as before and friends be as in part (a). Define a relation on X by {x,y) | y ∈ friends(x)}. What properties must the function friends satisfy in order for this relation to be:



      (i) reflexive



             ∀x ∈ friends(x). x~x



      (ii) symmetric



             ∀x,y ∈ friends(x). if x~y then y~x



      (iii)transitive



             ∀x,y,z ∈ friends(x). if x~y and y~z then x~z



      I'm not sure if this is along the lines of what they are looking for.







      elementary-set-theory






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      asked Jan 7 at 13:17









      Jamie WhitakerJamie Whitaker

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          $begingroup$

          Most of what you do is correct, here are some adjustments:



          (iii) Ron is a friend of everybody thus $forall x in X: Ron in friends(x)$ or equivalent
          $$Ron in bigcap_{xin X} friends(x) $$
          (iv) Here you should use the subset symbol $"subseteq"$ instead of $"in"$.



          (v)
          $Kat in friends(friends(Lea))$



          (vi) This one is tricky and in the formulation it is actually unclear which way the friendsships go. The friends of Kat's friends is the set friends(friends(Kat)).
          If Kat is a friend of all of her friends' friends then
          $$ forall x in friends(friends(Kat)): Kat in friends(x) $$
          thus
          $$ Kat in bigcap_{xin friends(friends(Kat))} friends(x) $$



          The other interpretation is that that every friend of a friend of Kat is also a friend of Kat, which can be expressed as
          $$ friends(friends(Kat)) subseteq friends(Kat) $$



          In part B.



          Your relation is defined by $R={(x,y) | yin friends(x)} $, in other words $xsim y$ iff y is a friend of x.



          Thus for R to be reflexive we must have
          $$forall xin X : (x,x) in R $$
          Note that $(x,x)in R$ if and only if $x in friends(x)$, in other words, everyone must be friends with themselves.



          For symmetry. We note that $(x,y)in R$ iff $yin friends(x)$ and that $(y,x)in R$ iff $xin friends(y)$. Thus symmetry can be formulated as:
          $$ forall x,y in X: (y in friends(x) Leftrightarrow x in friends(y)) $$
          In other words friendship goes both ways.



          For R to be transtive we must have
          $$ xsim y : and : ysim z Rightarrow xsim z $$
          , which is equivalent to saying
          $$ yin friends(x) : and : zin friends(y) Rightarrow zin friends(x) $$
          In other words if Alice and Bob are friends and Bob and Casper are friends, then Alice and Casper must be friends. (if the relation is transitive)






          share|cite|improve this answer










          New contributor




          Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













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            1 Answer
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            $begingroup$

            Most of what you do is correct, here are some adjustments:



            (iii) Ron is a friend of everybody thus $forall x in X: Ron in friends(x)$ or equivalent
            $$Ron in bigcap_{xin X} friends(x) $$
            (iv) Here you should use the subset symbol $"subseteq"$ instead of $"in"$.



            (v)
            $Kat in friends(friends(Lea))$



            (vi) This one is tricky and in the formulation it is actually unclear which way the friendsships go. The friends of Kat's friends is the set friends(friends(Kat)).
            If Kat is a friend of all of her friends' friends then
            $$ forall x in friends(friends(Kat)): Kat in friends(x) $$
            thus
            $$ Kat in bigcap_{xin friends(friends(Kat))} friends(x) $$



            The other interpretation is that that every friend of a friend of Kat is also a friend of Kat, which can be expressed as
            $$ friends(friends(Kat)) subseteq friends(Kat) $$



            In part B.



            Your relation is defined by $R={(x,y) | yin friends(x)} $, in other words $xsim y$ iff y is a friend of x.



            Thus for R to be reflexive we must have
            $$forall xin X : (x,x) in R $$
            Note that $(x,x)in R$ if and only if $x in friends(x)$, in other words, everyone must be friends with themselves.



            For symmetry. We note that $(x,y)in R$ iff $yin friends(x)$ and that $(y,x)in R$ iff $xin friends(y)$. Thus symmetry can be formulated as:
            $$ forall x,y in X: (y in friends(x) Leftrightarrow x in friends(y)) $$
            In other words friendship goes both ways.



            For R to be transtive we must have
            $$ xsim y : and : ysim z Rightarrow xsim z $$
            , which is equivalent to saying
            $$ yin friends(x) : and : zin friends(y) Rightarrow zin friends(x) $$
            In other words if Alice and Bob are friends and Bob and Casper are friends, then Alice and Casper must be friends. (if the relation is transitive)






            share|cite|improve this answer










            New contributor




            Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$


















              0












              $begingroup$

              Most of what you do is correct, here are some adjustments:



              (iii) Ron is a friend of everybody thus $forall x in X: Ron in friends(x)$ or equivalent
              $$Ron in bigcap_{xin X} friends(x) $$
              (iv) Here you should use the subset symbol $"subseteq"$ instead of $"in"$.



              (v)
              $Kat in friends(friends(Lea))$



              (vi) This one is tricky and in the formulation it is actually unclear which way the friendsships go. The friends of Kat's friends is the set friends(friends(Kat)).
              If Kat is a friend of all of her friends' friends then
              $$ forall x in friends(friends(Kat)): Kat in friends(x) $$
              thus
              $$ Kat in bigcap_{xin friends(friends(Kat))} friends(x) $$



              The other interpretation is that that every friend of a friend of Kat is also a friend of Kat, which can be expressed as
              $$ friends(friends(Kat)) subseteq friends(Kat) $$



              In part B.



              Your relation is defined by $R={(x,y) | yin friends(x)} $, in other words $xsim y$ iff y is a friend of x.



              Thus for R to be reflexive we must have
              $$forall xin X : (x,x) in R $$
              Note that $(x,x)in R$ if and only if $x in friends(x)$, in other words, everyone must be friends with themselves.



              For symmetry. We note that $(x,y)in R$ iff $yin friends(x)$ and that $(y,x)in R$ iff $xin friends(y)$. Thus symmetry can be formulated as:
              $$ forall x,y in X: (y in friends(x) Leftrightarrow x in friends(y)) $$
              In other words friendship goes both ways.



              For R to be transtive we must have
              $$ xsim y : and : ysim z Rightarrow xsim z $$
              , which is equivalent to saying
              $$ yin friends(x) : and : zin friends(y) Rightarrow zin friends(x) $$
              In other words if Alice and Bob are friends and Bob and Casper are friends, then Alice and Casper must be friends. (if the relation is transitive)






              share|cite|improve this answer










              New contributor




              Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$
















                0












                0








                0





                $begingroup$

                Most of what you do is correct, here are some adjustments:



                (iii) Ron is a friend of everybody thus $forall x in X: Ron in friends(x)$ or equivalent
                $$Ron in bigcap_{xin X} friends(x) $$
                (iv) Here you should use the subset symbol $"subseteq"$ instead of $"in"$.



                (v)
                $Kat in friends(friends(Lea))$



                (vi) This one is tricky and in the formulation it is actually unclear which way the friendsships go. The friends of Kat's friends is the set friends(friends(Kat)).
                If Kat is a friend of all of her friends' friends then
                $$ forall x in friends(friends(Kat)): Kat in friends(x) $$
                thus
                $$ Kat in bigcap_{xin friends(friends(Kat))} friends(x) $$



                The other interpretation is that that every friend of a friend of Kat is also a friend of Kat, which can be expressed as
                $$ friends(friends(Kat)) subseteq friends(Kat) $$



                In part B.



                Your relation is defined by $R={(x,y) | yin friends(x)} $, in other words $xsim y$ iff y is a friend of x.



                Thus for R to be reflexive we must have
                $$forall xin X : (x,x) in R $$
                Note that $(x,x)in R$ if and only if $x in friends(x)$, in other words, everyone must be friends with themselves.



                For symmetry. We note that $(x,y)in R$ iff $yin friends(x)$ and that $(y,x)in R$ iff $xin friends(y)$. Thus symmetry can be formulated as:
                $$ forall x,y in X: (y in friends(x) Leftrightarrow x in friends(y)) $$
                In other words friendship goes both ways.



                For R to be transtive we must have
                $$ xsim y : and : ysim z Rightarrow xsim z $$
                , which is equivalent to saying
                $$ yin friends(x) : and : zin friends(y) Rightarrow zin friends(x) $$
                In other words if Alice and Bob are friends and Bob and Casper are friends, then Alice and Casper must be friends. (if the relation is transitive)






                share|cite|improve this answer










                New contributor




                Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$



                Most of what you do is correct, here are some adjustments:



                (iii) Ron is a friend of everybody thus $forall x in X: Ron in friends(x)$ or equivalent
                $$Ron in bigcap_{xin X} friends(x) $$
                (iv) Here you should use the subset symbol $"subseteq"$ instead of $"in"$.



                (v)
                $Kat in friends(friends(Lea))$



                (vi) This one is tricky and in the formulation it is actually unclear which way the friendsships go. The friends of Kat's friends is the set friends(friends(Kat)).
                If Kat is a friend of all of her friends' friends then
                $$ forall x in friends(friends(Kat)): Kat in friends(x) $$
                thus
                $$ Kat in bigcap_{xin friends(friends(Kat))} friends(x) $$



                The other interpretation is that that every friend of a friend of Kat is also a friend of Kat, which can be expressed as
                $$ friends(friends(Kat)) subseteq friends(Kat) $$



                In part B.



                Your relation is defined by $R={(x,y) | yin friends(x)} $, in other words $xsim y$ iff y is a friend of x.



                Thus for R to be reflexive we must have
                $$forall xin X : (x,x) in R $$
                Note that $(x,x)in R$ if and only if $x in friends(x)$, in other words, everyone must be friends with themselves.



                For symmetry. We note that $(x,y)in R$ iff $yin friends(x)$ and that $(y,x)in R$ iff $xin friends(y)$. Thus symmetry can be formulated as:
                $$ forall x,y in X: (y in friends(x) Leftrightarrow x in friends(y)) $$
                In other words friendship goes both ways.



                For R to be transtive we must have
                $$ xsim y : and : ysim z Rightarrow xsim z $$
                , which is equivalent to saying
                $$ yin friends(x) : and : zin friends(y) Rightarrow zin friends(x) $$
                In other words if Alice and Bob are friends and Bob and Casper are friends, then Alice and Casper must be friends. (if the relation is transitive)







                share|cite|improve this answer










                New contributor




                Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 15:26





















                New contributor




                Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                answered Jan 7 at 15:05









                Leander Tilsted KristensenLeander Tilsted Kristensen

                113




                113




                New contributor




                Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                New contributor





                Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






























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