Prove piecewise function is only continuous at $x=0$ [closed]

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A function $f(x)$ is defined:



$$f(x)=casesx,& $x in mathbbQ $cr 0, &$~x notin mathbbQ$$$



where $mathbbQ$ denotes the set of rational numbers.



Prove that $f(x)$ is only continuous at $x=0$.






calculus limits analysis continuity







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closed as off-topic by José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants Sep 10 at 15:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    Done! What else?
    – José Carlos Santos
    Sep 10 at 9:11






  • 1




    What you have tried?
    – Chinnapparaj R
    Sep 10 at 9:11






  • 1




    Have you tried to find the proof when $x=1$ through the web?
    – mrs
    Sep 10 at 9:18










  • I have seen this question before in this site but I cannot find it.
    – Dog_69
    Sep 10 at 10:10














up vote
-2
down vote

favorite












A function $f(x)$ is defined:



$$f(x)=casesx,& $x in mathbbQ $cr 0, &$~x notin mathbbQ$$$



where $mathbbQ$ denotes the set of rational numbers.



Prove that $f(x)$ is only continuous at $x=0$.






calculus limits analysis continuity







share|cite|improve this question















closed as off-topic by José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants Sep 10 at 15:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    Done! What else?
    – José Carlos Santos
    Sep 10 at 9:11






  • 1




    What you have tried?
    – Chinnapparaj R
    Sep 10 at 9:11






  • 1




    Have you tried to find the proof when $x=1$ through the web?
    – mrs
    Sep 10 at 9:18










  • I have seen this question before in this site but I cannot find it.
    – Dog_69
    Sep 10 at 10:10












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











A function $f(x)$ is defined:



$$f(x)=casesx,& $x in mathbbQ $cr 0, &$~x notin mathbbQ$$$



where $mathbbQ$ denotes the set of rational numbers.



Prove that $f(x)$ is only continuous at $x=0$.






calculus limits analysis continuity







share|cite|improve this question















A function $f(x)$ is defined:



$$f(x)=casesx,& $x in mathbbQ $cr 0, &$~x notin mathbbQ$$$



where $mathbbQ$ denotes the set of rational numbers.



Prove that $f(x)$ is only continuous at $x=0$.






calculus limits analysis continuity




calculus limits analysis continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 10 at 9:11









mrs

58.6k750143




58.6k750143










asked Sep 10 at 9:09









Angle Qian

193




193




closed as off-topic by José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants Sep 10 at 15:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants Sep 10 at 15:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, mechanodroid, Adrian Keister, Jendrik Stelzner, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 3




    Done! What else?
    – José Carlos Santos
    Sep 10 at 9:11






  • 1




    What you have tried?
    – Chinnapparaj R
    Sep 10 at 9:11






  • 1




    Have you tried to find the proof when $x=1$ through the web?
    – mrs
    Sep 10 at 9:18










  • I have seen this question before in this site but I cannot find it.
    – Dog_69
    Sep 10 at 10:10












  • 3




    Done! What else?
    – José Carlos Santos
    Sep 10 at 9:11






  • 1




    What you have tried?
    – Chinnapparaj R
    Sep 10 at 9:11






  • 1




    Have you tried to find the proof when $x=1$ through the web?
    – mrs
    Sep 10 at 9:18










  • I have seen this question before in this site but I cannot find it.
    – Dog_69
    Sep 10 at 10:10







3




3




Done! What else?
– José Carlos Santos
Sep 10 at 9:11




Done! What else?
– José Carlos Santos
Sep 10 at 9:11




1




1




What you have tried?
– Chinnapparaj R
Sep 10 at 9:11




What you have tried?
– Chinnapparaj R
Sep 10 at 9:11




1




1




Have you tried to find the proof when $x=1$ through the web?
– mrs
Sep 10 at 9:18




Have you tried to find the proof when $x=1$ through the web?
– mrs
Sep 10 at 9:18












I have seen this question before in this site but I cannot find it.
– Dog_69
Sep 10 at 10:10




I have seen this question before in this site but I cannot find it.
– Dog_69
Sep 10 at 10:10










1 Answer
1






active

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up vote
3
down vote













HINT:



It follows from the density of $Bbb Q$ in $Bbb R$ namely:




Every open interval of real numbers contains rational numbers.




Continuity at $x=0$ is clear:



Given $epsilon>0$ take $delta=epsilon$ and you'll have that
$$
|f(y)-f(x)|=|f(y)|<epsilon
$$
whenever $|y|<delta$ because $f(y)$ is either $0$ or $y$.



If $xneq0$ you use the density of $Bbb Q$ to show that you can find arbitrarily close $y$ near $x$ such that $|f(y)-f(x)|$ is larger than a suitably chosen $epsilon$.



To work this out carefully, you have to distinguish between the two cases $xinBbb Q$ and $xnotinBbb Q$.






share|cite|improve this answer





























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    HINT:



    It follows from the density of $Bbb Q$ in $Bbb R$ namely:




    Every open interval of real numbers contains rational numbers.




    Continuity at $x=0$ is clear:



    Given $epsilon>0$ take $delta=epsilon$ and you'll have that
    $$
    |f(y)-f(x)|=|f(y)|<epsilon
    $$
    whenever $|y|<delta$ because $f(y)$ is either $0$ or $y$.



    If $xneq0$ you use the density of $Bbb Q$ to show that you can find arbitrarily close $y$ near $x$ such that $|f(y)-f(x)|$ is larger than a suitably chosen $epsilon$.



    To work this out carefully, you have to distinguish between the two cases $xinBbb Q$ and $xnotinBbb Q$.






    share|cite|improve this answer


























      up vote
      3
      down vote













      HINT:



      It follows from the density of $Bbb Q$ in $Bbb R$ namely:




      Every open interval of real numbers contains rational numbers.




      Continuity at $x=0$ is clear:



      Given $epsilon>0$ take $delta=epsilon$ and you'll have that
      $$
      |f(y)-f(x)|=|f(y)|<epsilon
      $$
      whenever $|y|<delta$ because $f(y)$ is either $0$ or $y$.



      If $xneq0$ you use the density of $Bbb Q$ to show that you can find arbitrarily close $y$ near $x$ such that $|f(y)-f(x)|$ is larger than a suitably chosen $epsilon$.



      To work this out carefully, you have to distinguish between the two cases $xinBbb Q$ and $xnotinBbb Q$.






      share|cite|improve this answer
























        up vote
        3
        down vote










        up vote
        3
        down vote









        HINT:



        It follows from the density of $Bbb Q$ in $Bbb R$ namely:




        Every open interval of real numbers contains rational numbers.




        Continuity at $x=0$ is clear:



        Given $epsilon>0$ take $delta=epsilon$ and you'll have that
        $$
        |f(y)-f(x)|=|f(y)|<epsilon
        $$
        whenever $|y|<delta$ because $f(y)$ is either $0$ or $y$.



        If $xneq0$ you use the density of $Bbb Q$ to show that you can find arbitrarily close $y$ near $x$ such that $|f(y)-f(x)|$ is larger than a suitably chosen $epsilon$.



        To work this out carefully, you have to distinguish between the two cases $xinBbb Q$ and $xnotinBbb Q$.






        share|cite|improve this answer














        HINT:



        It follows from the density of $Bbb Q$ in $Bbb R$ namely:




        Every open interval of real numbers contains rational numbers.




        Continuity at $x=0$ is clear:



        Given $epsilon>0$ take $delta=epsilon$ and you'll have that
        $$
        |f(y)-f(x)|=|f(y)|<epsilon
        $$
        whenever $|y|<delta$ because $f(y)$ is either $0$ or $y$.



        If $xneq0$ you use the density of $Bbb Q$ to show that you can find arbitrarily close $y$ near $x$ such that $|f(y)-f(x)|$ is larger than a suitably chosen $epsilon$.



        To work this out carefully, you have to distinguish between the two cases $xinBbb Q$ and $xnotinBbb Q$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 10 at 9:39









        mrs

        58.6k750143




        58.6k750143










        answered Sep 10 at 9:29









        Andrea Mori

        19k13464




        19k13464












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