Show that $Z(I_{A})=A$












0












$begingroup$


Problem:



Consider $X$ a compact Hausdorff space, with $C(X)$ the set of continuous functions on $X$. If $A$ is closed subset of $X$, define $I_{A}={f in C(X)| f|_{A} = 0 }$,
and $Z(I):={x in X|f(x)=0 $ for all $f in I }$.



$1$ )Show that $Z(I_{A})=A$.



First if $x in A$, then for every $f in I_{A}, f(x)=0$, so $x in Z(I_{A})$. The other inclusion is confusing me (it could be false). Is it not possible for an ideal of functions on $C(X)$ to be zero on $A$ but also on other sets? Hint appreciated. I know that $I_{A}$ is closed with respect to the norm topology (the sup norm).






general-topology ring-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Urysohn's lemma is your friend.
    $endgroup$
    – Eric Wofsey
    Nov 24 '18 at 3:39
















0












$begingroup$


Problem:



Consider $X$ a compact Hausdorff space, with $C(X)$ the set of continuous functions on $X$. If $A$ is closed subset of $X$, define $I_{A}={f in C(X)| f|_{A} = 0 }$,
and $Z(I):={x in X|f(x)=0 $ for all $f in I }$.



$1$ )Show that $Z(I_{A})=A$.



First if $x in A$, then for every $f in I_{A}, f(x)=0$, so $x in Z(I_{A})$. The other inclusion is confusing me (it could be false). Is it not possible for an ideal of functions on $C(X)$ to be zero on $A$ but also on other sets? Hint appreciated. I know that $I_{A}$ is closed with respect to the norm topology (the sup norm).






general-topology ring-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Urysohn's lemma is your friend.
    $endgroup$
    – Eric Wofsey
    Nov 24 '18 at 3:39














0












0








0





$begingroup$


Problem:



Consider $X$ a compact Hausdorff space, with $C(X)$ the set of continuous functions on $X$. If $A$ is closed subset of $X$, define $I_{A}={f in C(X)| f|_{A} = 0 }$,
and $Z(I):={x in X|f(x)=0 $ for all $f in I }$.



$1$ )Show that $Z(I_{A})=A$.



First if $x in A$, then for every $f in I_{A}, f(x)=0$, so $x in Z(I_{A})$. The other inclusion is confusing me (it could be false). Is it not possible for an ideal of functions on $C(X)$ to be zero on $A$ but also on other sets? Hint appreciated. I know that $I_{A}$ is closed with respect to the norm topology (the sup norm).






general-topology ring-theory







share|cite|improve this question











$endgroup$




Problem:



Consider $X$ a compact Hausdorff space, with $C(X)$ the set of continuous functions on $X$. If $A$ is closed subset of $X$, define $I_{A}={f in C(X)| f|_{A} = 0 }$,
and $Z(I):={x in X|f(x)=0 $ for all $f in I }$.



$1$ )Show that $Z(I_{A})=A$.



First if $x in A$, then for every $f in I_{A}, f(x)=0$, so $x in Z(I_{A})$. The other inclusion is confusing me (it could be false). Is it not possible for an ideal of functions on $C(X)$ to be zero on $A$ but also on other sets? Hint appreciated. I know that $I_{A}$ is closed with respect to the norm topology (the sup norm).






general-topology ring-theory




general-topology ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 3:52







IntegrateThis

















asked Nov 24 '18 at 3:12









IntegrateThisIntegrateThis

1,7131717




1,7131717








  • 3




    $begingroup$
    Urysohn's lemma is your friend.
    $endgroup$
    – Eric Wofsey
    Nov 24 '18 at 3:39














  • 3




    $begingroup$
    Urysohn's lemma is your friend.
    $endgroup$
    – Eric Wofsey
    Nov 24 '18 at 3:39








3




3




$begingroup$
Urysohn's lemma is your friend.
$endgroup$
– Eric Wofsey
Nov 24 '18 at 3:39




$begingroup$
Urysohn's lemma is your friend.
$endgroup$
– Eric Wofsey
Nov 24 '18 at 3:39










2 Answers
2






active

oldest

votes


















2












$begingroup$

For any point $xnotin A$, the sets ${x}$ and $A$ are closed and can be separated by a continuous function






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
    $endgroup$
    – IntegrateThis
    Nov 24 '18 at 3:44












  • $begingroup$
    You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
    $endgroup$
    – user25959
    Nov 24 '18 at 3:47










  • $begingroup$
    So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
    $endgroup$
    – IntegrateThis
    Nov 24 '18 at 4:11






  • 2




    $begingroup$
    More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
    $endgroup$
    – user25959
    Nov 24 '18 at 4:14



















0












$begingroup$

If $x notin A$, then there is by Urysohn's lemma a continuous function $f:X to [0,1]$ such that $f(x) = 1$ and $f[A] = {0}$. Then $f in I_A$ (as $f|A equiv 0$) and $x notin Z(I_A)$ is witnessed by this $f$.



This shows that $Z(I_A) subseteq A$ by contraposition.



$A subseteq Z(I_A)$ is true basically by definition: $x in A$ and for all $fin I_A$ we have $f(x)=0$ (by definition of $I_A$), so that $x in Z(I_A)$, by that set's definition. This you already saw.






share|cite|improve this answer









$endgroup$













    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%2f3011129%2fshow-that-zi-a-a%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









    2












    $begingroup$

    For any point $xnotin A$, the sets ${x}$ and $A$ are closed and can be separated by a continuous function






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 3:44












    • $begingroup$
      You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
      $endgroup$
      – user25959
      Nov 24 '18 at 3:47










    • $begingroup$
      So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 4:11






    • 2




      $begingroup$
      More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
      $endgroup$
      – user25959
      Nov 24 '18 at 4:14
















    2












    $begingroup$

    For any point $xnotin A$, the sets ${x}$ and $A$ are closed and can be separated by a continuous function






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 3:44












    • $begingroup$
      You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
      $endgroup$
      – user25959
      Nov 24 '18 at 3:47










    • $begingroup$
      So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 4:11






    • 2




      $begingroup$
      More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
      $endgroup$
      – user25959
      Nov 24 '18 at 4:14














    2












    2








    2





    $begingroup$

    For any point $xnotin A$, the sets ${x}$ and $A$ are closed and can be separated by a continuous function






    share|cite|improve this answer











    $endgroup$



    For any point $xnotin A$, the sets ${x}$ and $A$ are closed and can be separated by a continuous function







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 24 '18 at 3:46

























    answered Nov 24 '18 at 3:40









    user25959user25959

    1,573816




    1,573816












    • $begingroup$
      Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 3:44












    • $begingroup$
      You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
      $endgroup$
      – user25959
      Nov 24 '18 at 3:47










    • $begingroup$
      So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 4:11






    • 2




      $begingroup$
      More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
      $endgroup$
      – user25959
      Nov 24 '18 at 4:14


















    • $begingroup$
      Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 3:44












    • $begingroup$
      You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
      $endgroup$
      – user25959
      Nov 24 '18 at 3:47










    • $begingroup$
      So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
      $endgroup$
      – IntegrateThis
      Nov 24 '18 at 4:11






    • 2




      $begingroup$
      More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
      $endgroup$
      – user25959
      Nov 24 '18 at 4:14
















    $begingroup$
    Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
    $endgroup$
    – IntegrateThis
    Nov 24 '18 at 3:44






    $begingroup$
    Thanks for the answer. I know Urysohn's lemma but I'm not sure how this relates to my problem other than that $A$ is closed and so there would be a function that would be $0$ on $A$ and $1$ otherwise for some other closed set, in this case ${x}$
    $endgroup$
    – IntegrateThis
    Nov 24 '18 at 3:44














    $begingroup$
    You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
    $endgroup$
    – user25959
    Nov 24 '18 at 3:47




    $begingroup$
    You would use $A$ and ${x}$ as the disjoint closed sets in the statement of Urysohn's lemma
    $endgroup$
    – user25959
    Nov 24 '18 at 3:47












    $begingroup$
    So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
    $endgroup$
    – IntegrateThis
    Nov 24 '18 at 4:11




    $begingroup$
    So there would be a function in the ideal which would be zero on A but 1 on {x}, and then I guess the product of that function with some other function would lead to a contradiction?
    $endgroup$
    – IntegrateThis
    Nov 24 '18 at 4:11




    2




    2




    $begingroup$
    More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
    $endgroup$
    – user25959
    Nov 24 '18 at 4:14




    $begingroup$
    More simply - that function shows that $x$ doesn't belong to $Z(I_A)$
    $endgroup$
    – user25959
    Nov 24 '18 at 4:14











    0












    $begingroup$

    If $x notin A$, then there is by Urysohn's lemma a continuous function $f:X to [0,1]$ such that $f(x) = 1$ and $f[A] = {0}$. Then $f in I_A$ (as $f|A equiv 0$) and $x notin Z(I_A)$ is witnessed by this $f$.



    This shows that $Z(I_A) subseteq A$ by contraposition.



    $A subseteq Z(I_A)$ is true basically by definition: $x in A$ and for all $fin I_A$ we have $f(x)=0$ (by definition of $I_A$), so that $x in Z(I_A)$, by that set's definition. This you already saw.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $x notin A$, then there is by Urysohn's lemma a continuous function $f:X to [0,1]$ such that $f(x) = 1$ and $f[A] = {0}$. Then $f in I_A$ (as $f|A equiv 0$) and $x notin Z(I_A)$ is witnessed by this $f$.



      This shows that $Z(I_A) subseteq A$ by contraposition.



      $A subseteq Z(I_A)$ is true basically by definition: $x in A$ and for all $fin I_A$ we have $f(x)=0$ (by definition of $I_A$), so that $x in Z(I_A)$, by that set's definition. This you already saw.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $x notin A$, then there is by Urysohn's lemma a continuous function $f:X to [0,1]$ such that $f(x) = 1$ and $f[A] = {0}$. Then $f in I_A$ (as $f|A equiv 0$) and $x notin Z(I_A)$ is witnessed by this $f$.



        This shows that $Z(I_A) subseteq A$ by contraposition.



        $A subseteq Z(I_A)$ is true basically by definition: $x in A$ and for all $fin I_A$ we have $f(x)=0$ (by definition of $I_A$), so that $x in Z(I_A)$, by that set's definition. This you already saw.






        share|cite|improve this answer









        $endgroup$



        If $x notin A$, then there is by Urysohn's lemma a continuous function $f:X to [0,1]$ such that $f(x) = 1$ and $f[A] = {0}$. Then $f in I_A$ (as $f|A equiv 0$) and $x notin Z(I_A)$ is witnessed by this $f$.



        This shows that $Z(I_A) subseteq A$ by contraposition.



        $A subseteq Z(I_A)$ is true basically by definition: $x in A$ and for all $fin I_A$ we have $f(x)=0$ (by definition of $I_A$), so that $x in Z(I_A)$, by that set's definition. This you already saw.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 10:57









        Henno BrandsmaHenno Brandsma

        106k347114




        106k347114






























            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%2f3011129%2fshow-that-zi-a-a%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more