# Limit of sequence of sequences

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.)

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.

• 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 $b_n$? Commented Jan 7, 2019 at 9:41
• $\mathbb Q$ 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. Commented Jan 7, 2019 at 9:53
• In your comment to your A, I think you meant $\sup$, not $\inf,$ in the def'n of $\rho$. Commented Jan 7, 2019 at 19:30
• @DanielWainfleet Yes, you are right. Thank you for attending me. Commented Jan 7, 2019 at 20:32