Continuity of f(x) = 4 if rational and x^2 if irrational












0














Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis







share|cite|improve this question






















  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56
















0














Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis







share|cite|improve this question






















  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56














0












0








0







Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis







share|cite|improve this question













Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 '18 at 20:30









Aishwarya Deore

324




324












  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56


















  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56
















"I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
– Arthur
Nov 27 '18 at 20:46






"I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
– Arthur
Nov 27 '18 at 20:46














Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
– Aishwarya Deore
Nov 27 '18 at 20:56




Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
– Aishwarya Deore
Nov 27 '18 at 20:56










2 Answers
2






active

oldest

votes


















0














note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



hint



Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



Now ask yourself, what if $c$ is irrational?






share|cite|improve this answer





















  • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:58










  • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
    – Anurag A
    Nov 27 '18 at 21:02












  • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
    – Aishwarya Deore
    Nov 27 '18 at 21:07



















0














Let $a$ be a real.



$$lim_{xto a,xinBbb Q}f(x)=4$$



$$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



the function is continuous only at $x=pm 2$.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?






    share|cite|improve this answer





















    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07
















    0














    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?






    share|cite|improve this answer





















    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07














    0












    0








    0






    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?






    share|cite|improve this answer












    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 20:54









    Anurag A

    25.6k12249




    25.6k12249












    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07


















    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07
















    Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:58




    Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:58












    @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
    – Anurag A
    Nov 27 '18 at 21:02






    @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
    – Anurag A
    Nov 27 '18 at 21:02














    <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
    – Aishwarya Deore
    Nov 27 '18 at 21:07




    <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
    – Aishwarya Deore
    Nov 27 '18 at 21:07











    0














    Let $a$ be a real.



    $$lim_{xto a,xinBbb Q}f(x)=4$$



    $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



    the function is continuous only at $x=pm 2$.






    share|cite|improve this answer


























      0














      Let $a$ be a real.



      $$lim_{xto a,xinBbb Q}f(x)=4$$



      $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



      the function is continuous only at $x=pm 2$.






      share|cite|improve this answer
























        0












        0








        0






        Let $a$ be a real.



        $$lim_{xto a,xinBbb Q}f(x)=4$$



        $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



        the function is continuous only at $x=pm 2$.






        share|cite|improve this answer












        Let $a$ be a real.



        $$lim_{xto a,xinBbb Q}f(x)=4$$



        $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



        the function is continuous only at $x=pm 2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 21:06









        hamam_Abdallah

        37.9k21634




        37.9k21634






























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