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This question is from a multi-step problem on iterated limits. First, we are given a doubly indexed array $a_{m,n}$ where $m, n$ $\epsilon$ $\mathbb{N}$.

Define $\lim \limits_{m,n \to \infty} a_{m,n} = a$ to mean that for all $\epsilon > 0$ there exists an $N$ $\epsilon$ $\mathbb{N}$ such that if both $m, n$ $\geq$ $N$, then $\lvert a_{m,n} - a\rvert$ $<$ $\epsilon$.

The iterated limits are shown as: $$\lim \limits_{m \to \infty} \bigl( \lim \limits_{n \to \infty} a_{m,n} \bigr)$$ $$\lim \limits_{n \to \infty} \bigl( \lim \limits_{m \to \infty} a_{m,n} \bigr)$$

I've already proved in a previous part that if $\lim \limits_{m,n \to \infty} a_{m,n} = a$, and for each fixed $m$ $\epsilon$ $\mathbb{N}$ $\lim \limits_{n \to \infty} a_{m,n} = b_m$, that $\lim \limits_{m \to \infty} b_m = a$.

Prove that if $\lim \limits_{m,n \to \infty} a_{m,n}$ exists, and both iterated limits exist, that all three limits must be equal.

It seems that the result follows from what is proved in the previous part. Does the fact that $\lim \limits_{m \to \infty} \bigl( \lim \limits_{n \to \infty} a_{m,n} \bigr)$ exists imply that for each fixed $m$ $\epsilon$ $\mathbb{N}$ $\lim \limits_{n \to \infty} a_{m,n} = b_m$? Or can we not assume that $b_m$ exists for all $m$ $\epsilon$ $\mathbb{N}$?

I feel like I am missing a step.

RRL
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nimo959
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1 Answers1

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The existence of a double limit $\lim_{m,n} a_{mn}$ does not imply that $\lim_{n \to \infty} a_{mn}$ exists.

An example is $a_{mn} = (-1)^n (1/m + 1/n)$ where the double limit is $0$, but $\limsup_{n \to \infty} a_{mn} = 1/m$ and $\liminf_{n \to \infty} a_{mn} = -1/m$

Assuming that both iterated limits exist means that all of the single limts exist. (How, otherwise, are the iterated limits defined?) It then follows that if the double limit exists, then all three (double and iterated) limits are equal.

The weaker assumption is that $a_{mn} \to b_m$ as $n \to \infty$ for all fixed $m$ and $a_{mn} \to c_n$ as $m \to \infty$ for all fixed $n$. If the double limit exists then it follows that both iterated limits exist and equal the double limit.

RRL
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  • Thank you. This is what I needed clarity on. That since $\lim \limits_{m \to \infty} \bigl( \lim \limits_{n \to \infty} a_{m,n} \bigr)$ exists, then by definition, each fixed $m$ $\epsilon$ $\mathbb{N}$ produces a well-defined sequence $b_m$ with $\lim \limits_{n \to \infty} a_{m,n} = b_m$. – nimo959 Oct 29 '19 at 16:35