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I got to thinking about sequences of Cauchy sequences. Here is a simple example. Let us define $b_n = (n,1,1,1,...)$ for $n\in\mathbb N$. So we have \begin{align} b_1&=(1,1,1,1,...)\\ b_2&=(2,1,1,1,...)\\ b_3&=(3,1,1,1,...)\\ \dots \end{align} Question: what is $\lim_{n\to\infty} b_n$? It looks like that is not a valid sequence. Does this question even make sense? (I am not sure how such a limit would even be defined.)

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A sequence in $X$ is actually a function $\mathbb N\to X$ or equivalently an element of the set $X^\mathbb N$.

If $X$ is a topological space then $X^{\mathbb N}$ can be equipped with a topology that corresponds in some way with the original topology on $X$, which gives birth to the possibility of convergence of sequences.

There are several candidates for the topology on $X^{\mathbb N}$.

In each of them a sequence of sequences might have a limit.

The question whether $\lim_{n\to\infty}b_n$ exists (so that the expression makes sense) depends heavily on this context.

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  • $\begingroup$ Thank you! So what about $X=\mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$? $\endgroup$
    – Stefanie
    Commented Jan 7, 2019 at 9:41
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    $\begingroup$ $\mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $\rho((a_n)_n,(b_n)_n)=\inf\{\overline d(a_n,b_n)\mid n\in\mathbb N\}$ where $\overline d(a_n,b_n$ stands for $\min\{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge. $\endgroup$
    – drhab
    Commented Jan 7, 2019 at 9:53
  • $\begingroup$ In your comment to your A, I think you meant $\sup$, not $\inf,$ in the def'n of $\rho$. $\endgroup$
    – DanielWainfleet
    Commented Jan 7, 2019 at 19:30
  • $\begingroup$ @DanielWainfleet Yes, you are right. Thank you for attending me. $\endgroup$
    – drhab
    Commented Jan 7, 2019 at 20:32

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