If the iterated limits are equal does the double sequence converge? If $(a_{m,n})$ is a double sequence (in $\mathbb{R}$ or $\mathbb{C}$) and $\lim_m\lim_n a_{m,n}=\lim_n\lim_m a_{m,n}=a$ then can we deduce $\lim_{m,n}a_{m,n}=a$? More specifically can we deduce $\lim_{n}a_{n,n}=a$?
I think this is true but I am not sure. My attempt at a rigorous proof:
Let $\epsilon>0$. Argument 1:
$$\exists N_{\epsilon}\in \mathbb{N}: i\ge N_{\epsilon}\implies \left|\lim_ja_{i,j}-a\right|<\epsilon\text{ and }\left|\lim_ja_{j,i}-a\right|<\epsilon$$
This is the definition of the limit (I take $N$ as the maximum of the two $n_0$).
Argument 2:
$$i\ge N_{\epsilon}\implies \lim_j\left|a_{i,j}-a\right|<\epsilon\text{ and }\lim_j\left|a_{j,i}-a\right|<\epsilon$$
This is an application of the limit laws.
Now fix $i\ge N$. Argument 3:
$$\exists M_{\epsilon,i}\in \mathbb{N}:j\ge M_{\epsilon,i}\implies \left|a_{i,j}-a\right|<\epsilon\text{ and }\left|a_{j,i}-a\right|<\epsilon$$
This follows from the fact that if $\lim_n\left|b_n\right|<K$ then for large $K>0$, $\left|b_n\right|<K$.
Argument 4: As $\epsilon>0$ is arbitrary, 
$$\lim_j\left|a_{i,j}-a\right|=0\text{ and }\lim_j\left|a_{j,i}-a\right|=0$$
This is the wrong argment in my opinion as I have 
$$(\forall \epsilon>0)(\exists N_{\epsilon}\in \mathbb{N}): \ i\ge N_{\epsilon}\implies (\exists M_{\epsilon,i}\in \mathbb{N})\ :j\ge M_{\epsilon,i}\implies \left|a_{i,j}-a\right|<\epsilon\text{ and }\left|a_{j,i}-a\right|<\epsilon
$$
and from this I deduce
$$(\forall \epsilon>0)(\exists M\in \mathbb{N})\ :j\ge M\implies \left|a_{i,j}-a\right|<\epsilon\text{ and }\left|a_{j,i}-a\right|<\epsilon
$$
for large $i$ meaning
$$(\forall \epsilon>0)(\exists N,M\in \mathbb{N}): \ i\ge N_{\epsilon}\text{ and }j\ge M_{\epsilon,i}\implies \left|a_{i,j}-a\right|<\epsilon\text{ and }\left|a_{j,i}-a\right|<\epsilon
$$
I basicaly lose the dependence on $i$. This is the difference for example between continuity and uniform continuity right?
So,
$$i\ge N\implies \lim_ja_{i,j}=\lim_ja_{j,i}=a$$
(which I doubt). From here we let $i\to \infty$ and obtain the result.
 A: No. Let $a_{m,n}=0$ if $m\ne n$, and $a_{n,n}=1$ for all $n$.
A: Completely revised to match the expanded question:
Some extra notation will help me to discuss this. For $i\in\Bbb N$ let $r_i=\lim_ja_{i,j}$, and for $j\in\Bbb N$ let $c_j=\lim_ia_{i,j}$:
$$\begin{array}{cc}
a_{0,0}&a_{0,1}&a_{0,2}&\dots&\to&r_0\\
a_{1,0}&a_{1,1}&a_{1,2}&\dots&\to&r_1\\
a_{2,0}&a_{2,1}&a_{2,2}&\dots&\to&r_2\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
\downarrow&\downarrow&\downarrow&\dots&\ddots&\vdots\\
c_0&c_1&c_2&\dots&\dots&?
\end{array}$$
Your Argument 1 says that $|r_i-a|<\epsilon$ and $|c_i-a|<\epsilon$ when $i\ge N_\epsilon$. Argument 2 doesn’t say anything more. Now you fix an $i\ge N_\epsilon$. Your Argument 3 is incomplete. What you can argue is that there is an $M_{i,\epsilon}$ such that $|a_{i,j}-r_i|<\epsilon-|r_i-a|$ and $|a_{j,i}-c_i|<\epsilon-|c_i-a|$ whenever $j\ge M_{i,\epsilon}$, which then implies (via the triangle inequality) that $|a_{i,j}-r_i|<\epsilon$ and $|a_{j,i}-c_i|<\epsilon$ whenever $j\ge M_{i,\epsilon}$. However, the conclusion that you reached is correct.
As you thought, Argument 4 is the real problem, and for the reason that you give: you’ve lost the strong dependence on $i$ of $M_{i,\epsilon}$. I think that this dependence is even more evident in the expanded justification that I just gave for your Argument 3: you need to get $|a_{i,j}-r_i|$ not just less than $\epsilon$, but less than $\epsilon-|r_i-a|$, which could be a lot less than $\epsilon$ if $r_i$ is barely within $\epsilon$ of $a$. (And yes, this is similar to the difference between continuity and uniform continuity.)
