need help understanding a continuity proof












3












$begingroup$


Can someone help with intuition behind this statement :



statement



Let $f : [0,1]rightarrow mathbb{R}$ be defined by:



$$f(x)=begin{cases}
0 & xnotin mathbb{Q} \
frac{1}{n} & text{if} , x=frac{m}{n}in mathbb{Q}
end{cases}
$$



If $xin [0,1]$ $mathbb{Q}$, then $f$ is continuous at $x$



Proof:



$forall varepsilon >0 , exists , N : frac{1}{N}<varepsilon$, since $xneq frac{m}{n} forall , m,n in mathbb{N},$ there is some $delta_k$ such that the interval $(x-delta_k,x+delta_k)$ has no points of the form $frac{m}{k+1}$. let $delta= min{delta_1,...,delta_{n-1}}$. Then the interval $(x-delta, x+delta)$ contains no points of the form $frac{m}{n}$ for $n=2,3,...,N$. Hence if $yin (x-delta,x+delta), |f(y)-f(x)|=|f(y)-0|leq frac{1}{n}$ for $ngeq N+1$ so $|f(y)-0|<varepsilon$ which proves that $f$ is continuous at $x.$



Question



Why does there exist a $delta_k$ such that the interval has no points of the form $frac{m}{k+1}$?






real-analysis continuity proof-explanation







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Can someone help with intuition behind this statement :



    statement



    Let $f : [0,1]rightarrow mathbb{R}$ be defined by:



    $$f(x)=begin{cases}
    0 & xnotin mathbb{Q} \
    frac{1}{n} & text{if} , x=frac{m}{n}in mathbb{Q}
    end{cases}
    $$



    If $xin [0,1]$ $mathbb{Q}$, then $f$ is continuous at $x$



    Proof:



    $forall varepsilon >0 , exists , N : frac{1}{N}<varepsilon$, since $xneq frac{m}{n} forall , m,n in mathbb{N},$ there is some $delta_k$ such that the interval $(x-delta_k,x+delta_k)$ has no points of the form $frac{m}{k+1}$. let $delta= min{delta_1,...,delta_{n-1}}$. Then the interval $(x-delta, x+delta)$ contains no points of the form $frac{m}{n}$ for $n=2,3,...,N$. Hence if $yin (x-delta,x+delta), |f(y)-f(x)|=|f(y)-0|leq frac{1}{n}$ for $ngeq N+1$ so $|f(y)-0|<varepsilon$ which proves that $f$ is continuous at $x.$



    Question



    Why does there exist a $delta_k$ such that the interval has no points of the form $frac{m}{k+1}$?






    real-analysis continuity proof-explanation







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Can someone help with intuition behind this statement :



      statement



      Let $f : [0,1]rightarrow mathbb{R}$ be defined by:



      $$f(x)=begin{cases}
      0 & xnotin mathbb{Q} \
      frac{1}{n} & text{if} , x=frac{m}{n}in mathbb{Q}
      end{cases}
      $$



      If $xin [0,1]$ $mathbb{Q}$, then $f$ is continuous at $x$



      Proof:



      $forall varepsilon >0 , exists , N : frac{1}{N}<varepsilon$, since $xneq frac{m}{n} forall , m,n in mathbb{N},$ there is some $delta_k$ such that the interval $(x-delta_k,x+delta_k)$ has no points of the form $frac{m}{k+1}$. let $delta= min{delta_1,...,delta_{n-1}}$. Then the interval $(x-delta, x+delta)$ contains no points of the form $frac{m}{n}$ for $n=2,3,...,N$. Hence if $yin (x-delta,x+delta), |f(y)-f(x)|=|f(y)-0|leq frac{1}{n}$ for $ngeq N+1$ so $|f(y)-0|<varepsilon$ which proves that $f$ is continuous at $x.$



      Question



      Why does there exist a $delta_k$ such that the interval has no points of the form $frac{m}{k+1}$?






      real-analysis continuity proof-explanation







      share|cite|improve this question











      $endgroup$




      Can someone help with intuition behind this statement :



      statement



      Let $f : [0,1]rightarrow mathbb{R}$ be defined by:



      $$f(x)=begin{cases}
      0 & xnotin mathbb{Q} \
      frac{1}{n} & text{if} , x=frac{m}{n}in mathbb{Q}
      end{cases}
      $$



      If $xin [0,1]$ $mathbb{Q}$, then $f$ is continuous at $x$



      Proof:



      $forall varepsilon >0 , exists , N : frac{1}{N}<varepsilon$, since $xneq frac{m}{n} forall , m,n in mathbb{N},$ there is some $delta_k$ such that the interval $(x-delta_k,x+delta_k)$ has no points of the form $frac{m}{k+1}$. let $delta= min{delta_1,...,delta_{n-1}}$. Then the interval $(x-delta, x+delta)$ contains no points of the form $frac{m}{n}$ for $n=2,3,...,N$. Hence if $yin (x-delta,x+delta), |f(y)-f(x)|=|f(y)-0|leq frac{1}{n}$ for $ngeq N+1$ so $|f(y)-0|<varepsilon$ which proves that $f$ is continuous at $x.$



      Question



      Why does there exist a $delta_k$ such that the interval has no points of the form $frac{m}{k+1}$?






      real-analysis continuity proof-explanation




      real-analysis continuity proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 3:28









      Gaby Alfonso

      972317




      972317










      asked Dec 16 '18 at 2:04









      FrankFrank

      16210




      16210






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          $x$ is irrational and also once we fix $k$, there are only finitely many $m$ that you can pick.



          Hence if a neighborhood of $x$ is small enough, we can exclude those numbers of the form of $frac{m}{k+1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:49










          • $begingroup$
            First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
            $endgroup$
            – Siong Thye Goh
            Dec 17 '18 at 1:19



















          0












          $begingroup$

          We know that $x$ is between two numbers divided by $k+1$, meaning $frac {m_1}{k+1}leq xleq frac {m_2}{k+1}$ for some $m_1,m_2leq k+1$, then if we take $$delta_kleq min (|x-frac {m_1}{k+1}|,|x-frac {m_2}{k+1}|)$$ we get the required the in a $delta_k$ neighbourhood of $x$, there are no points of the mentioned form.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:50






          • 1




            $begingroup$
            I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
            $endgroup$
            – NL1992
            Dec 19 '18 at 16:15













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          $x$ is irrational and also once we fix $k$, there are only finitely many $m$ that you can pick.



          Hence if a neighborhood of $x$ is small enough, we can exclude those numbers of the form of $frac{m}{k+1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:49










          • $begingroup$
            First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
            $endgroup$
            – Siong Thye Goh
            Dec 17 '18 at 1:19
















          0












          $begingroup$

          $x$ is irrational and also once we fix $k$, there are only finitely many $m$ that you can pick.



          Hence if a neighborhood of $x$ is small enough, we can exclude those numbers of the form of $frac{m}{k+1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:49










          • $begingroup$
            First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
            $endgroup$
            – Siong Thye Goh
            Dec 17 '18 at 1:19














          0












          0








          0





          $begingroup$

          $x$ is irrational and also once we fix $k$, there are only finitely many $m$ that you can pick.



          Hence if a neighborhood of $x$ is small enough, we can exclude those numbers of the form of $frac{m}{k+1}$.






          share|cite|improve this answer









          $endgroup$



          $x$ is irrational and also once we fix $k$, there are only finitely many $m$ that you can pick.



          Hence if a neighborhood of $x$ is small enough, we can exclude those numbers of the form of $frac{m}{k+1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 2:09









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118












          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:49










          • $begingroup$
            First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
            $endgroup$
            – Siong Thye Goh
            Dec 17 '18 at 1:19


















          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:49










          • $begingroup$
            First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
            $endgroup$
            – Siong Thye Goh
            Dec 17 '18 at 1:19
















          $begingroup$
          Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
          $endgroup$
          – Frank
          Dec 16 '18 at 18:49




          $begingroup$
          Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
          $endgroup$
          – Frank
          Dec 16 '18 at 18:49












          $begingroup$
          First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
          $endgroup$
          – Siong Thye Goh
          Dec 17 '18 at 1:19




          $begingroup$
          First, I think there is a typo, I think $delta = min{ delta+1, delta_2, ldots, delta_{N-1} }$. $delta_i$ is defined to be small enough such that there is no number of the form of $frac{m}{i+1}$ in the corresponding neighborhood. By taking the minimum of them, we make sure that any rational number inside the neighborhood, has a denominator that is bigger than $N+1$. that is $f(y)$ is either $0$ or being map to $frac1m$ where $m ge N+1$.
          $endgroup$
          – Siong Thye Goh
          Dec 17 '18 at 1:19











          0












          $begingroup$

          We know that $x$ is between two numbers divided by $k+1$, meaning $frac {m_1}{k+1}leq xleq frac {m_2}{k+1}$ for some $m_1,m_2leq k+1$, then if we take $$delta_kleq min (|x-frac {m_1}{k+1}|,|x-frac {m_2}{k+1}|)$$ we get the required the in a $delta_k$ neighbourhood of $x$, there are no points of the mentioned form.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:50






          • 1




            $begingroup$
            I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
            $endgroup$
            – NL1992
            Dec 19 '18 at 16:15


















          0












          $begingroup$

          We know that $x$ is between two numbers divided by $k+1$, meaning $frac {m_1}{k+1}leq xleq frac {m_2}{k+1}$ for some $m_1,m_2leq k+1$, then if we take $$delta_kleq min (|x-frac {m_1}{k+1}|,|x-frac {m_2}{k+1}|)$$ we get the required the in a $delta_k$ neighbourhood of $x$, there are no points of the mentioned form.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:50






          • 1




            $begingroup$
            I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
            $endgroup$
            – NL1992
            Dec 19 '18 at 16:15
















          0












          0








          0





          $begingroup$

          We know that $x$ is between two numbers divided by $k+1$, meaning $frac {m_1}{k+1}leq xleq frac {m_2}{k+1}$ for some $m_1,m_2leq k+1$, then if we take $$delta_kleq min (|x-frac {m_1}{k+1}|,|x-frac {m_2}{k+1}|)$$ we get the required the in a $delta_k$ neighbourhood of $x$, there are no points of the mentioned form.






          share|cite|improve this answer









          $endgroup$



          We know that $x$ is between two numbers divided by $k+1$, meaning $frac {m_1}{k+1}leq xleq frac {m_2}{k+1}$ for some $m_1,m_2leq k+1$, then if we take $$delta_kleq min (|x-frac {m_1}{k+1}|,|x-frac {m_2}{k+1}|)$$ we get the required the in a $delta_k$ neighbourhood of $x$, there are no points of the mentioned form.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 2:12









          NL1992NL1992

          7311




          7311












          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:50






          • 1




            $begingroup$
            I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
            $endgroup$
            – NL1992
            Dec 19 '18 at 16:15




















          • $begingroup$
            Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
            $endgroup$
            – Frank
            Dec 16 '18 at 18:50






          • 1




            $begingroup$
            I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
            $endgroup$
            – NL1992
            Dec 19 '18 at 16:15


















          $begingroup$
          Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
          $endgroup$
          – Frank
          Dec 16 '18 at 18:50




          $begingroup$
          Why do we choose $delta=min{delta_1,delta_2,...delta_{n-1}}$ How do we know, for example, there does not exist a $m_1$ such that $frac{m_1}{10}<frac{m_2}{1000} forall m_2$? If such an $m_1$ exists, then when we choose $y in {x-delta_k,x+delta_k} rightarrow frac{1}{n}>epsilon$, so the function would not converge.
          $endgroup$
          – Frank
          Dec 16 '18 at 18:50




          1




          1




          $begingroup$
          I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
          $endgroup$
          – NL1992
          Dec 19 '18 at 16:15






          $begingroup$
          I'm afraid I haven't completely understood your question but I'll answer something and if that doesn't answer it, please explain what you meant. For any number $x$, we have the two closest numbers to it with denominator $k+1$, and if we take the distance between $x$ and the closer one of these two, we are guaranteed not to meet any number with denominator $k+1$ in our set region. If we take it to be the minimal distance of all such denominators (up to $k+1$), then it must be that any of the denominators up to $k+1$ will not be met in the set region.
          $endgroup$
          – NL1992
          Dec 19 '18 at 16:15




















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