Question about a function being continuousA problem on summing real numbers all taken to the same exponentProving a function is continuous on all irrational numbersHow can I prove this function is not continuous for every point other than 0?At what points is this piecewise function continuous?Explain about proofcontinuous but not absolutely continuous functionProb. 17, Chap. 4 in Baby Rudin: A real function can have at most countably many simple discontinuities on a given segmentGiven two real numbers $a$ and $b$ such that $a<b$, what about the convergence of these two sequences?Understanding a functionAlmost everywhere differentiability of a function

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Question about a function being continuous


A problem on summing real numbers all taken to the same exponentProving a function is continuous on all irrational numbersHow can I prove this function is not continuous for every point other than 0?At what points is this piecewise function continuous?Explain about proofcontinuous but not absolutely continuous functionProb. 17, Chap. 4 in Baby Rudin: A real function can have at most countably many simple discontinuities on a given segmentGiven two real numbers $a$ and $b$ such that $a<b$, what about the convergence of these two sequences?Understanding a functionAlmost everywhere differentiability of a function













0












$begingroup$


Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and



$g_n(x)=begincasesfrac1sqrtx-r_n & xneq r_n; \ 3 & x=r_n.endcases$



This is continuous by definition ?
for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and



    $g_n(x)=begincasesfrac1sqrtx-r_n & xneq r_n; \ 3 & x=r_n.endcases$



    This is continuous by definition ?
    for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and



      $g_n(x)=begincasesfrac1sqrtx-r_n & xneq r_n; \ 3 & x=r_n.endcases$



      This is continuous by definition ?
      for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.










      share|cite|improve this question











      $endgroup$




      Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and



      $g_n(x)=begincasesfrac1sqrtx-r_n & xneq r_n; \ 3 & x=r_n.endcases$



      This is continuous by definition ?
      for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.







      analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 29 at 10:49









      J. W. Tanner

      4,4691320




      4,4691320










      asked Mar 29 at 10:29









      kim wenasakim wenasa

      194




      194




















          2 Answers
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          0












          $begingroup$

          $g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrtx-r_n$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              0












              $begingroup$

              $g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                $g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  $g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$






                  share|cite|improve this answer









                  $endgroup$



                  $g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 29 at 10:39









                  FredFred

                  48.6k11849




                  48.6k11849





















                      1












                      $begingroup$

                      It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrtx-r_n$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrtx-r_n$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrtx-r_n$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.






                          share|cite|improve this answer









                          $endgroup$



                          It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrtx-r_n$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 29 at 10:39









                          Maja BlumensteinMaja Blumenstein

                          1096




                          1096



























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