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

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 remove the first element, or adding the first two elements and make the sum be the new first element. 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?

Answer 1

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 ${\mathbb R}$.