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Let the sequence $\{a_{n,m}\}$ be a doubly indexed real sequence. I was wondering if the assumption that $$\lim_{n\to\infty} \lim_{m\to\infty} a_{n,m} = \lim_{m\to\infty} \lim_{n\to\infty} a_{n,m} = L,$$ where $L$ is real, would imply that the following "concurrent limit" would also exist and equal $L$. $$\lim_{n,m\to\infty} a_{n,m} \stackrel{?}= L.$$

I am defining $\lim_{n,m\to\infty} a_{n,m} = L$ to mean that for each $\epsilon > 0$ there exists $K > 0$ such that $n,m>K \Rightarrow |a_{n,m}-L|<\epsilon.$

My intuition tells me that this conjecture is false, since the rapidity of convergence of $\lim_{m\to\infty} a_{n,m}$ to $b_n$ may depend on $n$, so $m$ and $n$ may not be able to approach $\infty$ in a sufficiently similar manner. However, I could not construct any counterexample.

I also wonder whether the weaker statement that

$$\lim_{n\to\infty} a_{n,n} \stackrel{?}= L$$

must hold.

I would appreciate any helpful insights.

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    $\begingroup$ Consider a_{n,m}=\frac{1}{n-m} $\endgroup$
    – R.Jackson
    Jul 24 '19 at 22:25
  • $\begingroup$ That is a simple and clever counterexample. I did not think of making the concurrent limit undefined. Thank you for your insight! $\endgroup$ Jul 24 '19 at 22:59
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Counterexample to both conjectures. Let $$a_{n,m}=\delta_{nm}$$ (Kronecker delta). Then for all $n$, $$\lim_{m\to\infty}a_{n,m}=0$$ and similarly for all $m$ $$\lim_{n\to\infty} a_{n,m}=0.$$ But, for any $1>\epsilon>0$ and $K>0$, we have $$\lvert a_{K+1,K+1}-0 \rvert=1>\epsilon.$$ And indeed $$\lim_{n\to\infty}a_{n,n}=1.$$

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  • $\begingroup$ Thank you very much! $\endgroup$ Jul 24 '19 at 22:57

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