Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.












0












$begingroup$


$f : X rightarrow Y , A subset X, Bsubset Y $



Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.



$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!



$f^{-1}(B) = {x: f(x) = B }$



$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $



I don't know how to proceed further.






functions elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
    $endgroup$
    – Robert Israel
    Nov 27 '18 at 15:15










  • $begingroup$
    @Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
    $endgroup$
    – Yadati Kiran
    Nov 27 '18 at 15:21












  • $begingroup$
    Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
    $endgroup$
    – Student
    Nov 27 '18 at 15:25
















0












$begingroup$


$f : X rightarrow Y , A subset X, Bsubset Y $



Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.



$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!



$f^{-1}(B) = {x: f(x) = B }$



$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $



I don't know how to proceed further.






functions elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
    $endgroup$
    – Robert Israel
    Nov 27 '18 at 15:15










  • $begingroup$
    @Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
    $endgroup$
    – Yadati Kiran
    Nov 27 '18 at 15:21












  • $begingroup$
    Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
    $endgroup$
    – Student
    Nov 27 '18 at 15:25














0












0








0





$begingroup$


$f : X rightarrow Y , A subset X, Bsubset Y $



Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.



$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!



$f^{-1}(B) = {x: f(x) = B }$



$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $



I don't know how to proceed further.






functions elementary-set-theory







share|cite|improve this question











$endgroup$




$f : X rightarrow Y , A subset X, Bsubset Y $



Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.



$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!



$f^{-1}(B) = {x: f(x) = B }$



$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $



I don't know how to proceed further.






functions elementary-set-theory




functions elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 15:26







Student

















asked Nov 27 '18 at 15:09









StudentStudent

334




334












  • $begingroup$
    Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
    $endgroup$
    – Robert Israel
    Nov 27 '18 at 15:15










  • $begingroup$
    @Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
    $endgroup$
    – Yadati Kiran
    Nov 27 '18 at 15:21












  • $begingroup$
    Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
    $endgroup$
    – Student
    Nov 27 '18 at 15:25


















  • $begingroup$
    Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
    $endgroup$
    – Robert Israel
    Nov 27 '18 at 15:15










  • $begingroup$
    @Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
    $endgroup$
    – Yadati Kiran
    Nov 27 '18 at 15:21












  • $begingroup$
    Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
    $endgroup$
    – Student
    Nov 27 '18 at 15:25
















$begingroup$
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
$endgroup$
– Robert Israel
Nov 27 '18 at 15:15




$begingroup$
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
$endgroup$
– Robert Israel
Nov 27 '18 at 15:15












$begingroup$
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
$endgroup$
– Yadati Kiran
Nov 27 '18 at 15:21






$begingroup$
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
$endgroup$
– Yadati Kiran
Nov 27 '18 at 15:21














$begingroup$
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
$endgroup$
– Student
Nov 27 '18 at 15:25




$begingroup$
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
$endgroup$
– Student
Nov 27 '18 at 15:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.



The last definition states that $xin f^{-1}(B)iff f(x)in B$.



We can conclude directly that $f(f^{-1}(B))subseteq B$.



Now observe that:




  • $Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$


  • $Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$



Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Wow very good answer
      $endgroup$
      – Akash Roy
      Nov 27 '18 at 15:37











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.



    The last definition states that $xin f^{-1}(B)iff f(x)in B$.



    We can conclude directly that $f(f^{-1}(B))subseteq B$.



    Now observe that:




    • $Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$


    • $Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$



    Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.



      The last definition states that $xin f^{-1}(B)iff f(x)in B$.



      We can conclude directly that $f(f^{-1}(B))subseteq B$.



      Now observe that:




      • $Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$


      • $Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$



      Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.



        The last definition states that $xin f^{-1}(B)iff f(x)in B$.



        We can conclude directly that $f(f^{-1}(B))subseteq B$.



        Now observe that:




        • $Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$


        • $Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$



        Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$






        share|cite|improve this answer









        $endgroup$



        By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.



        The last definition states that $xin f^{-1}(B)iff f(x)in B$.



        We can conclude directly that $f(f^{-1}(B))subseteq B$.



        Now observe that:




        • $Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$


        • $Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$



        Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 15:24









        drhabdrhab

        100k544130




        100k544130























            0












            $begingroup$

            It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Wow very good answer
              $endgroup$
              – Akash Roy
              Nov 27 '18 at 15:37
















            0












            $begingroup$

            It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Wow very good answer
              $endgroup$
              – Akash Roy
              Nov 27 '18 at 15:37














            0












            0








            0





            $begingroup$

            It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...






            share|cite|improve this answer









            $endgroup$



            It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 '18 at 15:14









            Robert IsraelRobert Israel

            321k23210462




            321k23210462












            • $begingroup$
              Wow very good answer
              $endgroup$
              – Akash Roy
              Nov 27 '18 at 15:37


















            • $begingroup$
              Wow very good answer
              $endgroup$
              – Akash Roy
              Nov 27 '18 at 15:37
















            $begingroup$
            Wow very good answer
            $endgroup$
            – Akash Roy
            Nov 27 '18 at 15:37




            $begingroup$
            Wow very good answer
            $endgroup$
            – Akash Roy
            Nov 27 '18 at 15:37


















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