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









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    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






























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