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

            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