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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?

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You are mixing up two systems of enumeration different things.

Your $a_n$ are numbers, given for each $n\geq1$, and the $n$ enumerates these numbers in the list $(a_n)_{n\geq1}$. Assume that $\lim_{n\to\infty} a_n=\alpha\in{\mathbb R}$. 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: $$\eqalign{ {\bf b}_1&=(a_1,a_2,a_3,a_4,\ldots),\cr {\bf b}_2&=(a_1+a_2,a_3,a_4,a_5,\ldots),\cr {\bf b}_3&=(a_1+a_2+a_3,a_4,a_5,\ldots),\cr &\vdots\cr {\bf b}_j&=(a_1+a_2+\ldots+a_j,a_{j+1},a_{j+2},a_{j+3},\ldots)\qquad(j\geq1).\cr}$$ 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 $j\geq1$ one has $\lim_{k\to\infty}{\bf b}_{j.k}=\alpha$.

That's all one can say about the sequences ${\bf b}_j$, $\>j\geq1$. In particular; this has nothing to do with the completeness of ${\mathbb R}$.

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  • $\begingroup$ 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 $j\to\infty$. 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. $\endgroup$ – Stefanie Jan 4 at 21:19

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