Is my proof that $\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{nm}=\lim_{n\rightarrow\infty}a_{nn}$ I am trying to see if
$$\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{nm}=\lim_{n\rightarrow\infty}a_{nn}$$
with some necessary conditions
My attempt proof:
Let  $\lim_{n\rightarrow\infty}a_{n}=a$ and  $\lim_{m\rightarrow\infty}a_{nm}=a_n$
So for all $\epsilon,\epsilon_1 >0$ there exist $N,N_1\in\mathbb{N}$ such that $|a_{n}-a|<\epsilon$ and $|a_{nm}-a_{nn}|<\epsilon_1$ for $n>N,N_1$
(The last line is by Cauchy Criterion)
What I want is $(a_{nn})\longrightarrow a$
for all $\epsilon_2>0$ find a natrual number $N_2$ such that for $n>N_2$,
$|a_{nn}-a|<\epsilon_2$
Now $|a_{nn}-a|=|a_{nn}-a_{nm}+a_{nm}-a|\leq|a_{nn}-a_{nm}|+|a_{nm}-a|$
the Cauchy Criterion guarantees the left-hand side and the convergence of $(a_{nm}) $guarantees the right-hand side, so $(a_{nn})$ must converge to $a$
Is my proof correct?
 A: The statement is wrong. Let $a_{nm}=\delta_{nm}$, that is the Kronecker delta, then
$$
\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{nm} = 0 \neq
1 = \lim_{n\rightarrow\infty} a_{nn}
$$
A: It is not right as it has been stated before me.
The problem relies on the usage of the Cauchy criterion. It states that for a given sequence $(u_n)_{n \in \mathbb N}$,
$$\forall \epsilon > 0, \ \exists N \in \mathbb N, \ \forall n, m \geq N, \ |u_n - u_m| < \epsilon $$
Here, you weren't using it for the sequence $(a_{n_0m})_{m \in \mathbb N}$ at a fixex $n_0$, so it should give you
$$\forall \epsilon > 0, \ \exists N \in \mathbb N, \ \forall n, m \geq N, \ |a_{n_0m} - a_{n_0n}| < \epsilon $$
You can't correlate the $n_0$ and the $n$ as you did.
In general, I think that you should make it clearer what you suppose (which limits existence do you suppose ?), and what variables you work with (what are the n and the m, when are they declared ?), this should help you clarifying your proofs and avoiding mistakes like this.
