$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational. Does a limit exist?












0














Does the following function have a limit as x approaches a?



$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.



Answers given in terms of delta-epsilon please!



My thoughts so far:



(1.) Its graph seems to show this function acting as though it were two constant functions



(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.



I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.






real-analysis







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  • Similar (math.stackexchange.com/questions/403338/…)
    – Yadati Kiran
    Nov 24 at 18:34










  • Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
    – Winther
    Nov 24 at 19:50


















0














Does the following function have a limit as x approaches a?



$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.



Answers given in terms of delta-epsilon please!



My thoughts so far:



(1.) Its graph seems to show this function acting as though it were two constant functions



(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.



I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.






real-analysis







share|cite|improve this question
























  • Similar (math.stackexchange.com/questions/403338/…)
    – Yadati Kiran
    Nov 24 at 18:34










  • Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
    – Winther
    Nov 24 at 19:50
















0












0








0







Does the following function have a limit as x approaches a?



$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.



Answers given in terms of delta-epsilon please!



My thoughts so far:



(1.) Its graph seems to show this function acting as though it were two constant functions



(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.



I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.






real-analysis







share|cite|improve this question















Does the following function have a limit as x approaches a?



$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.



Answers given in terms of delta-epsilon please!



My thoughts so far:



(1.) Its graph seems to show this function acting as though it were two constant functions



(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.



I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













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edited Nov 24 at 19:43

























asked Nov 24 at 18:28









Miguel Cumming-Romo

265




265












  • Similar (math.stackexchange.com/questions/403338/…)
    – Yadati Kiran
    Nov 24 at 18:34










  • Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
    – Winther
    Nov 24 at 19:50




















  • Similar (math.stackexchange.com/questions/403338/…)
    – Yadati Kiran
    Nov 24 at 18:34










  • Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
    – Winther
    Nov 24 at 19:50


















Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34




Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34












Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50






Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50












2 Answers
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You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.



Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.






share|cite|improve this answer





























    1














    Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.



    Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.



    With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.



      Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.






      share|cite|improve this answer


























        2














        You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.



        Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.






        share|cite|improve this answer
























          2












          2








          2






          You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.



          Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.






          share|cite|improve this answer












          You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.



          Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 18:36









          user3482749

          2,421414




          2,421414























              1














              Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.



              Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.



              With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).






              share|cite|improve this answer


























                1














                Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.



                Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.



                With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).






                share|cite|improve this answer
























                  1












                  1








                  1






                  Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.



                  Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.



                  With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).






                  share|cite|improve this answer












                  Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.



                  Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.



                  With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 18:56









                  Ovi

                  12.4k1038111




                  12.4k1038111






























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