This is important for me and somehow I can't think straight now so I need some help.
Suppose $a_{n,k}$ is a real double sequence and it is bounded. Is it possible to prove the claim $$ \lim_{k \rightarrow \infty} \sup_{n \ge 1} a_{n, k} = \sup_{n \ge 1} \lim_{k \to \infty} a_{n,k}$$
Thanks!
On
Suppose we can list the elements of $a_{n,k}$ in a set called $\Delta$ = {$a_{1,0} \dots a_{n,k} $} . This sequence is bounded so is the set of it's elements ,so $\Delta$ has an upper bound $sup$ $a_{n,k}$.Lets distinguish 3 cases.
$ a\cdot$) $a_{n,k}$ is a constant sequence.
The set $\Delta$ is a singleton denoted by {$a_{n',k'}$} and $\lim_{k\to\infty}$ = $a_{n',k'}$ so the above equality holds.
$ b\cdot$) $a_{n,k}$ is an increasing sequence.
$a_{n,k}$ converges.
Arranging the set $\Delta$ in ascending order {$a_{n'',k''} \dots a_{n',k'}$} ,$sup$ $\Delta$ is $a_{n',k'}$ so $\lim_{k\to\infty}$$sup$ $a_{n,k}$= $a_{n',k'}$
From the property of the upper bound. $$supa_{n,k}-\epsilon \le a_{n,k} \le supa_{n,k}$$
$$-\epsilon \le a_{n,k}-supa_{n,k} \le 0 \le \epsilon$$
$$\mid a_{n,k}-supa_{n,k}\mid \le \epsilon$$
This shows that $sup$ $a_{n,k}$ is the limit of the sequence, applying the supremum shows the equality.
$ c\cdot$) $a_{n,k}$ is a decreasing sequence.
$a_{n,k}$ converges.
Arranging the set $\Delta$ in ascending order {$a_{n'',k''} \dots a_{n',k'}$} ,$sup$ $\Delta$ is $a_{n',k'}$ so $\lim_{k\to\infty}$$sup$ $a_{n,k}$= $a_{n',k'}$
From the property of the lower bound. $$infa_{n,k}\le a_{n,k} \le infa_{n,k}+\epsilon$$
$$-\epsilon \le 0\le a_{n,k}-infa_{n,k} \le \epsilon$$
$$\mid a_{n,k}-infa_{n,k}\mid \le \epsilon$$
This shows that $inf$ $a_{n,k}$ is the limit of the sequence, applying the supremum doesn't show the equality.
Let $a_{n,k}=1$ if $k\leq n$, $a_{n,k}=0$ if $k>n$. Then $$ \sup_{n\geq 1}\lim_{k\to\infty}a_{n,k}=\sup_{n\geq 1}0=0 $$ while $$ \lim_{k\to\infty}\sup_{n\geq 1}a_{n,k}=\lim_{k\to\infty}1=1 $$