Let $(a_{n,m})_{n\in\mathbb N,\,m\in \mathbb N}$ be a double sequence taking real values. We may consider different types of limit:
- $\lim_{n\to\infty,\,m\to\infty} a_{n,m} = l_1\in\mathbb R\,$ if for every $\epsilon>0$ there exists $N_\epsilon\in\mathbb N$ such that $\min(n,m)>N_\epsilon\Rightarrow|a_{n,m}-l_1|<\epsilon\,$;
- $\lim_{||(n,m)||\to\infty} a_{n,m} = l_2\in\mathbb R\,$ if for every $\epsilon>0$ there exists $N_\epsilon\in\mathbb N$ such that $\max(n,m)>N_\epsilon\Rightarrow|a_{n,m}-l_2|<\epsilon\,$;
- $\lim_{n\to\infty}(\lim_{m\to\infty} a_{n,m}) = l_3\in\mathbb R$ if for every $\epsilon>0$
- for every $n\in\mathbb N$ there exist $L_n\in\mathbb R$ and $M_{\epsilon,n}\in\mathbb N$ such that $m>M_{\epsilon,n}\Rightarrow|a_{n,m}-L_n|<\epsilon\,$
- there exist $N_\epsilon\in\mathbb N$ such that $n>N_\epsilon\Rightarrow|L_n-l_3|<\epsilon$ ;
- $\lim_{m\to\infty}(\lim_{n\to\infty} a_{n,m}) = l_4\in\mathbb R$ if for every $\epsilon>0$
- for every $m\geq M_\epsilon$ there exist $\Lambda_m\in\mathbb R$ and $N_{\epsilon,m}\in\mathbb N$ such that $n>N_{\epsilon,m}\Rightarrow|a_{n,m}-\Lambda_m|<\epsilon\,$
- there exist $M_\epsilon\in\mathbb N$ such that $m>M_\epsilon\Rightarrow|\Lambda_m-l_4|<\epsilon$ ;
In What is the definition of double sequence $a_{mn}$ being convergent to $l$? it is shown that convergence 1. does not ensure that the sequence $(a_{n,m})_{n,m\in\mathbb N}$ is bounded.
Convergence 2. should be enough to guarantee that $(a_{n,m})_{n,m\in\mathbb N}$ is bounded.
My question is : convergence 3. together with convergence 4. (with $l_3\neq l_4$ in general) suffice to guarantee $(a_{n,m})_{n,m\in\mathbb N}$ bounded?
Notice that in the definitions of double limits 3. and 4. I have assumed convergence of the first limit for every value of the second index, in order to avoid problems similar to those appearing here What is the definition of double sequence $a_{mn}$ being convergent to $l$?
