# Existence of the limit of a subsequence of a double sequence

My question is about the existence of a limit of a subsequence of a double sequence.

Let $a(m,n)$ be a map with domain $\mathbb{N}^2$ and codomain $\mathbb{R}$

The limit $\displaystyle \lim_{n \rightarrow \infty} a(m,n) = L$ means for all $\epsilon > 0$ there is a $N(\epsilon,m)$ such that $n > N(\epsilon,m)$ implies $a(m,n)$ is in an $\epsilon$-neighbourhood of $L$.

However if $m,n$ satisfy the condition $m \rightarrow \infty$ iff $n \rightarrow \infty$ but are otherwise independent then does this mean $\displaystyle \lim_{n \rightarrow \infty} a(m,n)$ does not exist ?

Response to answer by "I_really_want_to_heal_myself"

Thanks for your answer. It was interesting but I may not have explained my question fully enough.

I will label the statements involved as follows:

$A$: If $a(m,n)$ is not uniformly convergent on the domain of $m$ then $\displaystyle \lim_{n \rightarrow \infty} a(m,n) = L$ means for all $\epsilon > 0$ there is a $N(\epsilon,m)$ such that $n > N(\epsilon,m)$ implies $a(m,n)$ is in an $\epsilon$-neighbourhood of $L$.

$B$: $m \rightarrow \infty$ iff $n \rightarrow \infty$

$C$: $\displaystyle \lim_{n \rightarrow \infty} a(m,n)$ does not exist

My original question was intended to be does $A+B \rightarrow C$ ? It can be broken down into the following questions:

Q1) Does $A$ require $N(\epsilon,m)$ to be independent of $n$?

Q2) If the answer to Q1 is "yes" then $A$ and $B$ are incompatible. Does this mean $A+B \rightarrow C$? Limits of sequences are usually said to not exist when the sequence has an ongoing, non-damped, bounded oscillation, or the sequence is unbounded, but $A+B$ would be a different kind of failure of the definition of convergence.

Q3) If the answer to Q1 is "no" then your example shows there is at least one double sequence for which $A+B \rightarrow C$. Does $A+B \rightarrow C$ for all double sequences?

## 1 Answer

I'm not quite sure if I understood your question correctly, but knowing that $\displaystyle \lim_{n\rightarrow \infty} a(m,n) = L$ for all $m\in\mathbb{N}$ and that $m_n$ is a non-decreasing function of $n$ does not guarantee, that $\displaystyle \lim_{n\rightarrow \infty} a(m,n)$ exists. Consider the following example: $$L:=0$$ $$a(m,n):= \begin{cases} 0, & \text{if n>m^2} \\ 1, & \text{if n\leq m^2} \end{cases}$$ $$m_n:= \lfloor \sqrt{n} \rfloor \text{ \qquad for all n\in \mathbb{N}}$$ Clearly, $a(m_n,n)$ does not converge towards $0$, despite convergence for all $m\in \mathbb{N}$. If we suppose, that $\displaystyle \lim_{n,m\rightarrow \infty} a(m,n) = 0$ we get a contradiction, since for $\epsilon=\frac{1}{2}$ we cannot pick neither $N$ nor $(N,M)$ such that for all $n>N$ ( $n>N$ and $m>M$ in the second case) distance between $a(n,m)$ and $0$ is no greater than $\epsilon$. Same argumentation goes for $N$ such that $n+m>N$. I hope that the following example is quite clear and provides answer to your question.