Suppose I have a double index sequence $\{x_{n,m}\}$ in $\mathbb{R}$. Am looking for a bijection $f:\mathbb{N}\times\mathbb{N} \to \mathbb{N}$ such that $\{x_{n,m}\}$ converges if and only if $\{x_{f(n,m)}\}$ converges.
In other words, I would like to transform a double index sequence into a single index sequence without affecting convergence. This is in order to use results about single index sequences for my double index sequence.
Is this possible? The usual diagonal enumeration of $\mathbb{N}\times\mathbb{N}$ won't work here because $k>f(n,m)$ does not imply $n_k>n$ and $m_k>m$, where $(n_k,m_k)=f^{-1}(k)$.
Note: By convergence of a double sequence I mean $x_{m,n} \to x$ if and only if for all $\epsilon>0$ there exist $N\in \mathbb{N}$ such that $m,n\geq N$ implies $\mid x_{m,n}-x \mid < \epsilon$.
Thanks a lot for your help.
Best Answer
First we need to note that the notation in the question makes no sense: If $f:\Bbb N^2\to\Bbb N$ is a bijection and $(x_{n,m})$ is a double sequence then " $\{x_{n,m}\}$ converges if and only if $\{x_{f(n,m)}\}$ converges" is impossible. Because there's simply no such thing as $x_{f(m,n)}$, since $f(m,n)\notin\Bbb N^2$.
There are various ways to state the natural question that seems likely to be what the OP really meant to ask about, for example
The answer is no, there is no such bijection. Given $f$, choose $k_n\in\Bbb N$ with $$f(k_n)=(0,n).$$
Now let $$x_{m,n}=\begin{cases} 1,&(m=0),\\0,&(m>0).\end{cases}$$ Then $x_{m,n}\to0$. But $x_{f(k_n)}=1\not\to0,$ and since $k_n\to\infty$ this shows that $$x_{f(k)}\not\to0.$$