Why does the support must be closed?












4












$begingroup$


Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?



The exercise:



Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,



$(eta cdot g): X rightarrow mathbb{R}$,



$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and



$(eta cdot g)(x) = 0$ if $x notin U$



is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.



I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.



Can anyone help me?






general-topology







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
    $endgroup$
    – Chessanator
    9 hours ago












  • $begingroup$
    I meant that $(eta cdot g)$ is continuous yes!
    $endgroup$
    – HK4
    8 hours ago
















4












$begingroup$


Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?



The exercise:



Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,



$(eta cdot g): X rightarrow mathbb{R}$,



$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and



$(eta cdot g)(x) = 0$ if $x notin U$



is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.



I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.



Can anyone help me?






general-topology







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
    $endgroup$
    – Chessanator
    9 hours ago












  • $begingroup$
    I meant that $(eta cdot g)$ is continuous yes!
    $endgroup$
    – HK4
    8 hours ago














4












4








4


1



$begingroup$


Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?



The exercise:



Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,



$(eta cdot g): X rightarrow mathbb{R}$,



$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and



$(eta cdot g)(x) = 0$ if $x notin U$



is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.



I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.



Can anyone help me?






general-topology







share|cite|improve this question









$endgroup$




Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?



The exercise:



Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,



$(eta cdot g): X rightarrow mathbb{R}$,



$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and



$(eta cdot g)(x) = 0$ if $x notin U$



is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.



I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.



Can anyone help me?






general-topology




general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 9 hours ago









HK4HK4

354




354








  • 1




    $begingroup$
    You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
    $endgroup$
    – Chessanator
    9 hours ago












  • $begingroup$
    I meant that $(eta cdot g)$ is continuous yes!
    $endgroup$
    – HK4
    8 hours ago














  • 1




    $begingroup$
    You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
    $endgroup$
    – Chessanator
    9 hours ago












  • $begingroup$
    I meant that $(eta cdot g)$ is continuous yes!
    $endgroup$
    – HK4
    8 hours ago








1




1




$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
9 hours ago






$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
9 hours ago














$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
8 hours ago




$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
8 hours ago










1 Answer
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$begingroup$

Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
$$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$



Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.



To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
$$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
0 & |x| geq 1. end{cases}$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.






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    5












    $begingroup$

    Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
    $$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$



    Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.



    To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
    $$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
    0 & |x| geq 1. end{cases}$$
    Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
      $$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$



      Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.



      To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
      $$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
      0 & |x| geq 1. end{cases}$$
      Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
        $$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$



        Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.



        To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
        $$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
        0 & |x| geq 1. end{cases}$$
        Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.






        share|cite|improve this answer









        $endgroup$



        Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
        $$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$



        Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.



        To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
        $$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
        0 & |x| geq 1. end{cases}$$
        Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 9 hours ago









        levaplevap

        47k23273




        47k23273






























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