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











    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












    $begingroup$

    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


















            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.




            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

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            Image
            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
            Read more

            ComboBox Display Member on multiple fields

            Image
            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
            Read more

            Is it possible to collect Nectar points via Trainline?

            Image
            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
            Read more