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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?

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2 Answers 2

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

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  • $\begingroup$ Thanks for the example. Still I wonder at what condition does the statement hold true? $\endgroup$
    – Dqrksun
    Dec 21, 2022 at 14:09
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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.

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  • $\begingroup$ Clear explanation, thank you! $\endgroup$
    – Dqrksun
    Dec 21, 2022 at 14:06

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