Sequences of sequences: question about Cauchy's construction of the real numbers












1














As is well known, one way of constructing the real numbers is to consider Cauchy sequences and call two of them equivalent if they have the same limit. I got to thinking about the Cauchy sequences themselves (ignoring the equivalence relation).



If we have a Cauchy sequence, we can make it shorter by removing the first element, or adding the first two elements. In fact we can do this arbitrarily often, creating a sequence of sequences.



Let the Cauchy sequence be $(a_n)_n$ and define $b_1 = (a_n)_n$ and $b_{n+1}$ by adding the first two elements of $b_n$ and attaching the remaining elements. Thus $b_2=(a_1+a_2,a_3,a_4,dots), b_3=(a_1+a_2+a_3,a_4,dots)$ etc.



I am having a hard time determining what the limit of $(b_n)_n$ would be as $n$ tends to infinity. Am I correct in surmising that the limit of $(b_n)_n$ generally does not exist, and even if it exists, the limit is not a sequence? Is it therefore correct to state that the set of Cauchy sequences is not complete, even though the set of their equivalence classes (= the real numbers) obviously is complete?






real-analysis cauchy-sequences







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    1














    As is well known, one way of constructing the real numbers is to consider Cauchy sequences and call two of them equivalent if they have the same limit. I got to thinking about the Cauchy sequences themselves (ignoring the equivalence relation).



    If we have a Cauchy sequence, we can make it shorter by removing the first element, or adding the first two elements. In fact we can do this arbitrarily often, creating a sequence of sequences.



    Let the Cauchy sequence be $(a_n)_n$ and define $b_1 = (a_n)_n$ and $b_{n+1}$ by adding the first two elements of $b_n$ and attaching the remaining elements. Thus $b_2=(a_1+a_2,a_3,a_4,dots), b_3=(a_1+a_2+a_3,a_4,dots)$ etc.



    I am having a hard time determining what the limit of $(b_n)_n$ would be as $n$ tends to infinity. Am I correct in surmising that the limit of $(b_n)_n$ generally does not exist, and even if it exists, the limit is not a sequence? Is it therefore correct to state that the set of Cauchy sequences is not complete, even though the set of their equivalence classes (= the real numbers) obviously is complete?






    real-analysis cauchy-sequences







    share|cite|improve this question

























      1












      1








      1







      As is well known, one way of constructing the real numbers is to consider Cauchy sequences and call two of them equivalent if they have the same limit. I got to thinking about the Cauchy sequences themselves (ignoring the equivalence relation).



      If we have a Cauchy sequence, we can make it shorter by removing the first element, or adding the first two elements. In fact we can do this arbitrarily often, creating a sequence of sequences.



      Let the Cauchy sequence be $(a_n)_n$ and define $b_1 = (a_n)_n$ and $b_{n+1}$ by adding the first two elements of $b_n$ and attaching the remaining elements. Thus $b_2=(a_1+a_2,a_3,a_4,dots), b_3=(a_1+a_2+a_3,a_4,dots)$ etc.



      I am having a hard time determining what the limit of $(b_n)_n$ would be as $n$ tends to infinity. Am I correct in surmising that the limit of $(b_n)_n$ generally does not exist, and even if it exists, the limit is not a sequence? Is it therefore correct to state that the set of Cauchy sequences is not complete, even though the set of their equivalence classes (= the real numbers) obviously is complete?






      real-analysis cauchy-sequences







      share|cite|improve this question













      As is well known, one way of constructing the real numbers is to consider Cauchy sequences and call two of them equivalent if they have the same limit. I got to thinking about the Cauchy sequences themselves (ignoring the equivalence relation).



      If we have a Cauchy sequence, we can make it shorter by removing the first element, or adding the first two elements. In fact we can do this arbitrarily often, creating a sequence of sequences.



      Let the Cauchy sequence be $(a_n)_n$ and define $b_1 = (a_n)_n$ and $b_{n+1}$ by adding the first two elements of $b_n$ and attaching the remaining elements. Thus $b_2=(a_1+a_2,a_3,a_4,dots), b_3=(a_1+a_2+a_3,a_4,dots)$ etc.



      I am having a hard time determining what the limit of $(b_n)_n$ would be as $n$ tends to infinity. Am I correct in surmising that the limit of $(b_n)_n$ generally does not exist, and even if it exists, the limit is not a sequence? Is it therefore correct to state that the set of Cauchy sequences is not complete, even though the set of their equivalence classes (= the real numbers) obviously is complete?






      real-analysis cauchy-sequences




      real-analysis cauchy-sequences






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      asked Jan 4 at 18:16









      StefanieStefanie

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      506






















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          You are mixing up two systems of enumeration different things.



          Your $a_n$ are numbers, given for each $ngeq1$, and the $n$ enumerates these numbers in the list $(a_n)_{ngeq1}$. Assume that $lim_{ntoinfty} a_n=alphain{mathbb R}$. On the other hand, your $b_n$ are not numbers, but sequences, and the $n$ enumerates these sequences. I'd write it in the following way:
          $$eqalign{
          {bf b}_1&=(a_1,a_2,a_3,a_4,ldots),cr
          {bf b}_2&=(a_1+a_2,a_3,a_4,a_5,ldots),cr
          {bf b}_3&=(a_1+a_2+a_3,a_4,a_5,ldots),cr
          &vdotscr
          {bf b}_j&=(a_1+a_2+ldots+a_j,a_{j+1},a_{j+2},a_{j+3},ldots)qquad(jgeq1).cr}$$
          Let ${bf b}_{j.k}$ be the $k^{rm th}$ element of the sequence ${bf b}_j$. Then it is easy to see that for every $jgeq1$ one has $lim_{ktoinfty}{bf b}_{j.k}=alpha$.



          That's all one can say about the sequences ${bf b}_j$, $>jgeq1$. In particular; this has nothing to do with the completeness of ${mathbb R}$.






          share|cite|improve this answer





















          • Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
            – Stefanie
            Jan 4 at 21:19











          Your Answer





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          1 Answer
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          active

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          active

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          0














          You are mixing up two systems of enumeration different things.



          Your $a_n$ are numbers, given for each $ngeq1$, and the $n$ enumerates these numbers in the list $(a_n)_{ngeq1}$. Assume that $lim_{ntoinfty} a_n=alphain{mathbb R}$. On the other hand, your $b_n$ are not numbers, but sequences, and the $n$ enumerates these sequences. I'd write it in the following way:
          $$eqalign{
          {bf b}_1&=(a_1,a_2,a_3,a_4,ldots),cr
          {bf b}_2&=(a_1+a_2,a_3,a_4,a_5,ldots),cr
          {bf b}_3&=(a_1+a_2+a_3,a_4,a_5,ldots),cr
          &vdotscr
          {bf b}_j&=(a_1+a_2+ldots+a_j,a_{j+1},a_{j+2},a_{j+3},ldots)qquad(jgeq1).cr}$$
          Let ${bf b}_{j.k}$ be the $k^{rm th}$ element of the sequence ${bf b}_j$. Then it is easy to see that for every $jgeq1$ one has $lim_{ktoinfty}{bf b}_{j.k}=alpha$.



          That's all one can say about the sequences ${bf b}_j$, $>jgeq1$. In particular; this has nothing to do with the completeness of ${mathbb R}$.






          share|cite|improve this answer





















          • Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
            – Stefanie
            Jan 4 at 21:19
















          0














          You are mixing up two systems of enumeration different things.



          Your $a_n$ are numbers, given for each $ngeq1$, and the $n$ enumerates these numbers in the list $(a_n)_{ngeq1}$. Assume that $lim_{ntoinfty} a_n=alphain{mathbb R}$. On the other hand, your $b_n$ are not numbers, but sequences, and the $n$ enumerates these sequences. I'd write it in the following way:
          $$eqalign{
          {bf b}_1&=(a_1,a_2,a_3,a_4,ldots),cr
          {bf b}_2&=(a_1+a_2,a_3,a_4,a_5,ldots),cr
          {bf b}_3&=(a_1+a_2+a_3,a_4,a_5,ldots),cr
          &vdotscr
          {bf b}_j&=(a_1+a_2+ldots+a_j,a_{j+1},a_{j+2},a_{j+3},ldots)qquad(jgeq1).cr}$$
          Let ${bf b}_{j.k}$ be the $k^{rm th}$ element of the sequence ${bf b}_j$. Then it is easy to see that for every $jgeq1$ one has $lim_{ktoinfty}{bf b}_{j.k}=alpha$.



          That's all one can say about the sequences ${bf b}_j$, $>jgeq1$. In particular; this has nothing to do with the completeness of ${mathbb R}$.






          share|cite|improve this answer





















          • Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
            – Stefanie
            Jan 4 at 21:19














          0












          0








          0






          You are mixing up two systems of enumeration different things.



          Your $a_n$ are numbers, given for each $ngeq1$, and the $n$ enumerates these numbers in the list $(a_n)_{ngeq1}$. Assume that $lim_{ntoinfty} a_n=alphain{mathbb R}$. On the other hand, your $b_n$ are not numbers, but sequences, and the $n$ enumerates these sequences. I'd write it in the following way:
          $$eqalign{
          {bf b}_1&=(a_1,a_2,a_3,a_4,ldots),cr
          {bf b}_2&=(a_1+a_2,a_3,a_4,a_5,ldots),cr
          {bf b}_3&=(a_1+a_2+a_3,a_4,a_5,ldots),cr
          &vdotscr
          {bf b}_j&=(a_1+a_2+ldots+a_j,a_{j+1},a_{j+2},a_{j+3},ldots)qquad(jgeq1).cr}$$
          Let ${bf b}_{j.k}$ be the $k^{rm th}$ element of the sequence ${bf b}_j$. Then it is easy to see that for every $jgeq1$ one has $lim_{ktoinfty}{bf b}_{j.k}=alpha$.



          That's all one can say about the sequences ${bf b}_j$, $>jgeq1$. In particular; this has nothing to do with the completeness of ${mathbb R}$.






          share|cite|improve this answer












          You are mixing up two systems of enumeration different things.



          Your $a_n$ are numbers, given for each $ngeq1$, and the $n$ enumerates these numbers in the list $(a_n)_{ngeq1}$. Assume that $lim_{ntoinfty} a_n=alphain{mathbb R}$. On the other hand, your $b_n$ are not numbers, but sequences, and the $n$ enumerates these sequences. I'd write it in the following way:
          $$eqalign{
          {bf b}_1&=(a_1,a_2,a_3,a_4,ldots),cr
          {bf b}_2&=(a_1+a_2,a_3,a_4,a_5,ldots),cr
          {bf b}_3&=(a_1+a_2+a_3,a_4,a_5,ldots),cr
          &vdotscr
          {bf b}_j&=(a_1+a_2+ldots+a_j,a_{j+1},a_{j+2},a_{j+3},ldots)qquad(jgeq1).cr}$$
          Let ${bf b}_{j.k}$ be the $k^{rm th}$ element of the sequence ${bf b}_j$. Then it is easy to see that for every $jgeq1$ one has $lim_{ktoinfty}{bf b}_{j.k}=alpha$.



          That's all one can say about the sequences ${bf b}_j$, $>jgeq1$. In particular; this has nothing to do with the completeness of ${mathbb R}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 19:14









          Christian BlatterChristian Blatter

          172k7113326




          172k7113326












          • Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
            – Stefanie
            Jan 4 at 21:19


















          • Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
            – Stefanie
            Jan 4 at 21:19
















          Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
          – Stefanie
          Jan 4 at 21:19




          Thank you! This helps a lot. Sorry for using $n$ twice. Still - suppose we have the sequence consisting of all ones. Then the first element of $b_j$ grows without bound as $jtoinfty$. Is the limit then even a sequence? Or is it not even possible to talk about the limit of a sequence of sequences? I find this topic really confusing.
          – Stefanie
          Jan 4 at 21:19


















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