3
$\begingroup$

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?

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – RRL Apr 27 '19 at 7:37
  • 1
    $\begingroup$ Please see answer. $\endgroup$ – RRL May 1 '19 at 18:28
  • 1
    $\begingroup$ 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 $m \geqslant N$. $\endgroup$ – RRL May 2 '19 at 0:32
  • 1
    $\begingroup$ 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. $\endgroup$ – RRL May 2 '19 at 0:34
  • 1
    $\begingroup$ 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. $\endgroup$ – RRL May 2 '19 at 0:36
2
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I was very much confused about this definition..Now it is clear..@RRL $\endgroup$ – cmi May 1 '19 at 18:33
  • 1
    $\begingroup$ @cmi: You’re very welcome. $\endgroup$ – RRL May 4 '19 at 6:29
-1
$\begingroup$

I Have a question and maybe it will have a simple answer.

If we have a sequence $x_{i,j}$ that converges to $x_i$ for a fixed $j$. In addition, the sequence $x_i$ converges to $x$, can we conclude that $x_{i,j}$ converges to $x$?

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ This is not an answer $\endgroup$ – Menezio Apr 15 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.