Give an example of a bounded function on $[0,1]$, which doesnot achieve its infimum on any $[a,b] subset [0,1], a < b$

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Give an example of a bounded function on $[0,1]$, which doesnot achieve its infimum on any $[a,b] subset [0,1], a < b$



MY attempts :



i take beginalign
f(x)=begincases
x &text if xneq 0,1\
1/2 &textotherwise
endcases
endalign



Is its corrects ??







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    $textYes.hspace0pt$
    – Theo Bendit
    Aug 18 at 5:21










  • thanks @TheoBendit
    – stupid
    Aug 18 at 5:34














up vote
0
down vote

favorite












Give an example of a bounded function on $[0,1]$, which doesnot achieve its infimum on any $[a,b] subset [0,1], a < b$



MY attempts :



i take beginalign
f(x)=begincases
x &text if xneq 0,1\
1/2 &textotherwise
endcases
endalign



Is its corrects ??







share|cite|improve this question


















  • 1




    $textYes.hspace0pt$
    – Theo Bendit
    Aug 18 at 5:21










  • thanks @TheoBendit
    – stupid
    Aug 18 at 5:34












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Give an example of a bounded function on $[0,1]$, which doesnot achieve its infimum on any $[a,b] subset [0,1], a < b$



MY attempts :



i take beginalign
f(x)=begincases
x &text if xneq 0,1\
1/2 &textotherwise
endcases
endalign



Is its corrects ??







share|cite|improve this question














Give an example of a bounded function on $[0,1]$, which doesnot achieve its infimum on any $[a,b] subset [0,1], a < b$



MY attempts :



i take beginalign
f(x)=begincases
x &text if xneq 0,1\
1/2 &textotherwise
endcases
endalign



Is its corrects ??









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 5:30









Arthur

100k793175




100k793175










asked Aug 18 at 5:16









stupid

627110




627110







  • 1




    $textYes.hspace0pt$
    – Theo Bendit
    Aug 18 at 5:21










  • thanks @TheoBendit
    – stupid
    Aug 18 at 5:34












  • 1




    $textYes.hspace0pt$
    – Theo Bendit
    Aug 18 at 5:21










  • thanks @TheoBendit
    – stupid
    Aug 18 at 5:34







1




1




$textYes.hspace0pt$
– Theo Bendit
Aug 18 at 5:21




$textYes.hspace0pt$
– Theo Bendit
Aug 18 at 5:21












thanks @TheoBendit
– stupid
Aug 18 at 5:34




thanks @TheoBendit
– stupid
Aug 18 at 5:34










1 Answer
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Yes. The infimum of $f$ is $0$ and the supremum is $1$, but no value of $x$ gives $0$ or $1$. So not only have you avoid the infimum, but also the supremum.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

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    up vote
    3
    down vote



    accepted










    Yes. The infimum of $f$ is $0$ and the supremum is $1$, but no value of $x$ gives $0$ or $1$. So not only have you avoid the infimum, but also the supremum.






    share|cite|improve this answer
























      up vote
      3
      down vote



      accepted










      Yes. The infimum of $f$ is $0$ and the supremum is $1$, but no value of $x$ gives $0$ or $1$. So not only have you avoid the infimum, but also the supremum.






      share|cite|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Yes. The infimum of $f$ is $0$ and the supremum is $1$, but no value of $x$ gives $0$ or $1$. So not only have you avoid the infimum, but also the supremum.






        share|cite|improve this answer












        Yes. The infimum of $f$ is $0$ and the supremum is $1$, but no value of $x$ gives $0$ or $1$. So not only have you avoid the infimum, but also the supremum.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 18 at 5:31









        Arthur

        100k793175




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