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What is the definition of double sequence $a_{mn}$ being convergent to $l$?

I have this definition.

Definition: The double sequence $(a_{m,n})^∞_{m,n=1}$ is said to Converge to the real number $A∈ \mathbb R$ if for all $ϵ>0$ there exists an $N∈ \mathbb N$ such that if $m,n≥N$ then $∣a_{m,n}−A∣<ϵ$ and we say $A$ is the Double Limit of this double sequence written lim$_{m,n→∞}a_{m,n}=A$. If no such $A∈ \mathbb R$ satisfies this, then we say that the the double sequence $(a_{m,n})^∞_{m,n=1}$ diverges. I took help from here.

If I go by this definition then convergent double sequence $(a_{m,n})^∞_{m,n=1}$ may not be bounded. Example :

$a_{1n} = n$, $ a_{mn} = 1/m + 1/n$ for all $m \geq 2 $ and $n\in \mathbb N$

It seems odd to me. I feel that I am going wrong anywhere. Can anyone please tell me where I am being wrong?

cmi
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    The definition you have written is in fact the accepted definition of the double sequence. See, for example, Definition 19.4 in Bartle, The Elements if Real Analysis, 2nd Edition. This definition dates back to a paper by Pringsheim published in 1900. The iterated limits may not exist or be unequal even if the double limit exists, and there is no reason that the limit must be finite along a finite number of rows or columns. Unlike an ordinary sequence there is no theorem to the effect that every subset of the elements must be bounded. – RRL Apr 27 '19 at 07:37
  • So the convergent double sequence may not be bounded -- is a correct statement?@RRL – cmi Apr 27 '19 at 12:26
  • ??????????????@RRL – cmi May 01 '19 at 18:04
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    Please see answer. – RRL May 01 '19 at 18:28
  • There is no good resource of this topic...I searched a lot in Internet..I pick up some random article and start reading..https://www.google.com/url?sa=t&source=web&rct=j&url=http://www2.iugaza.edu.ps/ar/periodical/articles/volume%252014-%2520Issue%25201%2520-studies%2520-16.pdf&ved=2ahUKEwjin8X9_PrhAhUMqY8KHQ_XBh0QFjAAegQIBRAB&usg=AOvVaw22vAOjo_Bnw1nPapvanCE6-- this article confused me....Here boundness of convergent sequence is proved like we proved for normal sequence...You can have a look..@RRL – cmi May 01 '19 at 18:41
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    The article has an error. According to Definition 2.5 in the article, a double sequence $(a_{mn})$ is bounded if there exists $M > 0$ such that $|a_{mn}| \leqslant M$ for ALL $n,m \in \mathbb{N}$. This is too strong a statement to be satisfied if the sequence converges as $\lim_{(m,n) \to (\infty,\infty)}a_{mn}=a$. The proof of Theorem 2.6 is flawed. Assuming the double sequence converges to $a$ they say there is an integer $N$ such that $|a_{mn} - a| < 1$ and $|a_{mn}| < 1 + |a|$ for all $n \geqslant N$ AND $m \geqslant N$. – RRL May 02 '19 at 00:32
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    They then say the global bound is $\max(|a_{11}|, \ldots |a_{N-1,N-1}|, 1 + |a|)$ which is false. For example, they forgot about $a_{1,N}, a_{1,N+1},a_{1,N+2}, \ldots$ which are unbounded in your example even though the double limit exists. – RRL May 02 '19 at 00:34
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    Notice that reputable books like Bartle, Apostol, etc. talk about the definitions of double and iterated limits and the conditions where they are equal but they never mention such a property as global boundedness. As I said before the existence of the double limit implies the double sequence is EVENTUALLY bounded. – RRL May 02 '19 at 00:36
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    Not the best example, but suppose a regular sequence $(a_n)$ converges in the extended real numbers to a finite limit, then it is eventually bounded but not globally bounded if, for example, $a_1 = +\infty$. – RRL May 02 '19 at 00:40
  • You have invested so much time for me...Thank You so much...Thank You..@RRL – cmi May 02 '19 at 01:43

1 Answers1

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Suppose the sequence $(m,n) \mapsto a_{mn}$ converges in the Pringsheim sense. This means there exists $L$ such that for any $\epsilon > 0$ there exists $N$ such that $|a_{mn} - L| < \epsilon$ for all $m,n \geqslant N$.

With convergence in this sense, it is not necessary that the set $\{a_{mn}: m,n \in \mathbb{N}\}$ be bounded. Your example illustrates this.

On the other hand a convergent double sequence is bounded eventually. That is, there exists $N \in \mathbb{N}$ such that the set $\{a_{mn}: m \geqslant N,n \geqslant N\}$is bounded.

RRL
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