Showing that $M := {f in C([-1,1]) : f(0) > 0}$ is open












3












$begingroup$


Given $f in C([-1,1],mathbb{R})$ equipped with sup norm metric. I am trying to find out whether the subset $$M := {f in C([-1,1]) : f(0) > 0} $$ is open/closed in $C([-1,1])$



My work:



$M$ is open. Since any function in $M$ doesn't cross the negative y-axis, we can find an $epsilon$-ball around the axis, such that $B_{epsilon}(f)$ is completely contained within $M$. I am having trouble formalizing this, i.e finding such $epsilon$



$M$ is not closed. Consider the sequence of functions $f_{n}(x) := frac{x+1}{n}$



$f_{n}(0) > 0$ for all $n in mathbb{N}$ but $f_{n}$ converges to the zero function, which is not an element of $M$. So $M$ doesn't include its limit points.






real-analysis metric-spaces







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$endgroup$








  • 1




    $begingroup$
    For a simpler counterexample: the constant functions $f_n(x) = 1/n$ also do the trick! They converge to the $0$ function, which is not in $M$.
    $endgroup$
    – dafinguzman
    Dec 11 '18 at 20:24
















3












$begingroup$


Given $f in C([-1,1],mathbb{R})$ equipped with sup norm metric. I am trying to find out whether the subset $$M := {f in C([-1,1]) : f(0) > 0} $$ is open/closed in $C([-1,1])$



My work:



$M$ is open. Since any function in $M$ doesn't cross the negative y-axis, we can find an $epsilon$-ball around the axis, such that $B_{epsilon}(f)$ is completely contained within $M$. I am having trouble formalizing this, i.e finding such $epsilon$



$M$ is not closed. Consider the sequence of functions $f_{n}(x) := frac{x+1}{n}$



$f_{n}(0) > 0$ for all $n in mathbb{N}$ but $f_{n}$ converges to the zero function, which is not an element of $M$. So $M$ doesn't include its limit points.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    For a simpler counterexample: the constant functions $f_n(x) = 1/n$ also do the trick! They converge to the $0$ function, which is not in $M$.
    $endgroup$
    – dafinguzman
    Dec 11 '18 at 20:24














3












3








3





$begingroup$


Given $f in C([-1,1],mathbb{R})$ equipped with sup norm metric. I am trying to find out whether the subset $$M := {f in C([-1,1]) : f(0) > 0} $$ is open/closed in $C([-1,1])$



My work:



$M$ is open. Since any function in $M$ doesn't cross the negative y-axis, we can find an $epsilon$-ball around the axis, such that $B_{epsilon}(f)$ is completely contained within $M$. I am having trouble formalizing this, i.e finding such $epsilon$



$M$ is not closed. Consider the sequence of functions $f_{n}(x) := frac{x+1}{n}$



$f_{n}(0) > 0$ for all $n in mathbb{N}$ but $f_{n}$ converges to the zero function, which is not an element of $M$. So $M$ doesn't include its limit points.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$




Given $f in C([-1,1],mathbb{R})$ equipped with sup norm metric. I am trying to find out whether the subset $$M := {f in C([-1,1]) : f(0) > 0} $$ is open/closed in $C([-1,1])$



My work:



$M$ is open. Since any function in $M$ doesn't cross the negative y-axis, we can find an $epsilon$-ball around the axis, such that $B_{epsilon}(f)$ is completely contained within $M$. I am having trouble formalizing this, i.e finding such $epsilon$



$M$ is not closed. Consider the sequence of functions $f_{n}(x) := frac{x+1}{n}$



$f_{n}(0) > 0$ for all $n in mathbb{N}$ but $f_{n}$ converges to the zero function, which is not an element of $M$. So $M$ doesn't include its limit points.






real-analysis metric-spaces




real-analysis metric-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 20:10









yarafoudahyarafoudah

1137




1137








  • 1




    $begingroup$
    For a simpler counterexample: the constant functions $f_n(x) = 1/n$ also do the trick! They converge to the $0$ function, which is not in $M$.
    $endgroup$
    – dafinguzman
    Dec 11 '18 at 20:24














  • 1




    $begingroup$
    For a simpler counterexample: the constant functions $f_n(x) = 1/n$ also do the trick! They converge to the $0$ function, which is not in $M$.
    $endgroup$
    – dafinguzman
    Dec 11 '18 at 20:24








1




1




$begingroup$
For a simpler counterexample: the constant functions $f_n(x) = 1/n$ also do the trick! They converge to the $0$ function, which is not in $M$.
$endgroup$
– dafinguzman
Dec 11 '18 at 20:24




$begingroup$
For a simpler counterexample: the constant functions $f_n(x) = 1/n$ also do the trick! They converge to the $0$ function, which is not in $M$.
$endgroup$
– dafinguzman
Dec 11 '18 at 20:24










2 Answers
2






active

oldest

votes


















6












$begingroup$

The set is indeed open. If $f in M$, then $f(0) = epsilon > 0$. But then if $|g -f |_infty < frac{epsilon}{2}$ , we will find that $g(0) ge frac{epsilon}{2} > 0$. This shows that $M$ contains a ball around of $f$ of radius $epsilon/2$, and so $M$ is open.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! that was helpful :)
    $endgroup$
    – yarafoudah
    Dec 11 '18 at 20:32










  • $begingroup$
    To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
    $endgroup$
    – yarafoudah
    Dec 11 '18 at 20:32










  • $begingroup$
    Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
    $endgroup$
    – User8128
    Dec 11 '18 at 20:34





















4












$begingroup$

The map $phi: C([-1,1]) to mathbb R$ given by $phi(f) = f(0)$ is continuous, since $|f(0)-g(0)| le |f - g|$. Since $(0,infty)$ is open in $mathbb R$,
$M = phi^{-1}((0,infty))$ is open in $C([-1,1])$.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    6












    $begingroup$

    The set is indeed open. If $f in M$, then $f(0) = epsilon > 0$. But then if $|g -f |_infty < frac{epsilon}{2}$ , we will find that $g(0) ge frac{epsilon}{2} > 0$. This shows that $M$ contains a ball around of $f$ of radius $epsilon/2$, and so $M$ is open.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! that was helpful :)
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
      $endgroup$
      – User8128
      Dec 11 '18 at 20:34


















    6












    $begingroup$

    The set is indeed open. If $f in M$, then $f(0) = epsilon > 0$. But then if $|g -f |_infty < frac{epsilon}{2}$ , we will find that $g(0) ge frac{epsilon}{2} > 0$. This shows that $M$ contains a ball around of $f$ of radius $epsilon/2$, and so $M$ is open.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! that was helpful :)
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
      $endgroup$
      – User8128
      Dec 11 '18 at 20:34
















    6












    6








    6





    $begingroup$

    The set is indeed open. If $f in M$, then $f(0) = epsilon > 0$. But then if $|g -f |_infty < frac{epsilon}{2}$ , we will find that $g(0) ge frac{epsilon}{2} > 0$. This shows that $M$ contains a ball around of $f$ of radius $epsilon/2$, and so $M$ is open.






    share|cite|improve this answer









    $endgroup$



    The set is indeed open. If $f in M$, then $f(0) = epsilon > 0$. But then if $|g -f |_infty < frac{epsilon}{2}$ , we will find that $g(0) ge frac{epsilon}{2} > 0$. This shows that $M$ contains a ball around of $f$ of radius $epsilon/2$, and so $M$ is open.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 11 '18 at 20:13









    User8128User8128

    10.8k1622




    10.8k1622












    • $begingroup$
      Thank you! that was helpful :)
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
      $endgroup$
      – User8128
      Dec 11 '18 at 20:34




















    • $begingroup$
      Thank you! that was helpful :)
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
      $endgroup$
      – yarafoudah
      Dec 11 '18 at 20:32










    • $begingroup$
      Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
      $endgroup$
      – User8128
      Dec 11 '18 at 20:34


















    $begingroup$
    Thank you! that was helpful :)
    $endgroup$
    – yarafoudah
    Dec 11 '18 at 20:32




    $begingroup$
    Thank you! that was helpful :)
    $endgroup$
    – yarafoudah
    Dec 11 '18 at 20:32












    $begingroup$
    To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
    $endgroup$
    – yarafoudah
    Dec 11 '18 at 20:32




    $begingroup$
    To make sure I filled in details correctly: Since $|g -f |_infty < frac{epsilon}{2}$ it follows $|g(0)-f(0)| < frac{epsilon}{2}$ then $frac{epsilon}{2} < g(0) < frac{3epsilon}{2}$ which means $0 < frac{f(0)}{2} < g(0)$
    $endgroup$
    – yarafoudah
    Dec 11 '18 at 20:32












    $begingroup$
    Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
    $endgroup$
    – User8128
    Dec 11 '18 at 20:34






    $begingroup$
    Exactly! Or another way to look at at: $$g(0) = f(0) + (g(0) - f(0)) ge f(0) - | g - f|_infty ge epsilon - epsilon /2 = epsilon / 2.$$
    $endgroup$
    – User8128
    Dec 11 '18 at 20:34













    4












    $begingroup$

    The map $phi: C([-1,1]) to mathbb R$ given by $phi(f) = f(0)$ is continuous, since $|f(0)-g(0)| le |f - g|$. Since $(0,infty)$ is open in $mathbb R$,
    $M = phi^{-1}((0,infty))$ is open in $C([-1,1])$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      The map $phi: C([-1,1]) to mathbb R$ given by $phi(f) = f(0)$ is continuous, since $|f(0)-g(0)| le |f - g|$. Since $(0,infty)$ is open in $mathbb R$,
      $M = phi^{-1}((0,infty))$ is open in $C([-1,1])$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        The map $phi: C([-1,1]) to mathbb R$ given by $phi(f) = f(0)$ is continuous, since $|f(0)-g(0)| le |f - g|$. Since $(0,infty)$ is open in $mathbb R$,
        $M = phi^{-1}((0,infty))$ is open in $C([-1,1])$.






        share|cite|improve this answer









        $endgroup$



        The map $phi: C([-1,1]) to mathbb R$ given by $phi(f) = f(0)$ is continuous, since $|f(0)-g(0)| le |f - g|$. Since $(0,infty)$ is open in $mathbb R$,
        $M = phi^{-1}((0,infty))$ is open in $C([-1,1])$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 20:59









        Robert IsraelRobert Israel

        330k23218473




        330k23218473






























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