# 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?

The statement is wrong. Let $$a_{nm}=\delta_{nm}$$, that is the Kronecker delta, then $$\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{nm} = 0 \neq 1 = \lim_{n\rightarrow\infty} a_{nn}$$
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.