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












0














$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
























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










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












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
















0














$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
























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










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












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














0












0








0







$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















$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 at 15:26

























asked Nov 27 at 15:09









Student

334




334












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










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












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


















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










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












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
















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




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












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






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














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




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










2 Answers
2






active

oldest

votes


















1














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





























    0














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






    share|cite|improve this answer





















    • Wow very good answer
      – Akash Roy
      Nov 27 at 15:37











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015881%2fcheck-if-fa-cap-f-1b-is-a-subset-of-fa-cap-b%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    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


























      1














      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
























        1












        1








        1






        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












        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 at 15:24









        drhab

        97.8k544129




        97.8k544129























            0














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






            share|cite|improve this answer





















            • Wow very good answer
              – Akash Roy
              Nov 27 at 15:37
















            0














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






            share|cite|improve this answer





















            • Wow very good answer
              – Akash Roy
              Nov 27 at 15:37














            0












            0








            0






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






            share|cite|improve this answer












            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 at 15:14









            Robert Israel

            318k23208458




            318k23208458












            • Wow very good answer
              – Akash Roy
              Nov 27 at 15:37


















            • Wow very good answer
              – Akash Roy
              Nov 27 at 15:37
















            Wow very good answer
            – Akash Roy
            Nov 27 at 15:37




            Wow very good answer
            – Akash Roy
            Nov 27 at 15:37


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015881%2fcheck-if-fa-cap-f-1b-is-a-subset-of-fa-cap-b%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Index of /

            This page is only for reference, If you need detailed information, please check here
            Read more

            Tribalistas

            Image
              Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
            Read more

            Listed building

            Image
            Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
            Read more