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Is $\lim_{n \to \infty} \lim_{l \to \infty} \ a_{n,l} = \lim_{l \to \infty} \lim_{n \to \infty} \ a_{n,l}\;$?

What happens if I replace limits with lim sups?

Thanks!

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yeah, i was trying to confirm it for some variations. I guess if it doesn't work for limits it automatically fails for limsup as well, right? and the same counterexample can be used? –  UH1 Dec 16 '12 at 23:11
    
Sorry, misread it... –  David Mitra Dec 16 '12 at 23:12
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4 Answers 4

up vote 22 down vote accepted

$$\begin{array}{cc} 1&2&3&4&\ldots&\to&\infty\\ 0&1&2&3&\ldots&\to&\infty\\ 0&0&1&2&\ldots&\to&\infty\\ 0&0&0&1&\ldots&\to&\infty\\ \vdots&\vdots&\vdots&\vdots&&&\vdots\\ \downarrow&\downarrow&\downarrow&\downarrow&&&\downarrow&\\ 0&0&0&0&\ldots&\to&\text{OOPS!} \end{array}$$

This PDF has an extensive discussion of the double limit, the two iterated limits, and under what conditions they’re equal.

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+1 for OOPS! Good example. Better question: under what conditions do limits commute? –  Mario Carneiro Dec 16 '12 at 23:17
    
Thanks, @Chris. –  Brian M. Scott Dec 16 '12 at 23:44
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Brian already answered the first question very well, so I will answer the second. If $\lim_{x\to\infty} x_{n}$ exists, then $$\limsup{x_{n}}=\liminf{x_{n}}=\lim_{x\to\infty}x_{n}.$$ Consider what would happen if this were not the case. Then there would exist at least two different subsequential limits. However, since every converging sequence has each subsequential limit converge to the same limit, clearly $\lim_{x\to\infty}x_{n}$ would not exist. So, if the limits do, in fact, exist, then the double $\limsup$s would be equivalent to the double limits.

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For positive integers $m$ and $n$, let $(a_{m,n})$ be the sequence with $m$ $0$'s followed by all $1$'s. Then $\lim_{m\to\infty}\lim_{n\to\infty}=1$ but $\lim_{n\to\infty}\lim_{m\to\infty}=0$.

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The classical example to see what happens is to take $$ a_{m,n}=\frac{m}{m+n}. $$ Then $$ \lim_m\lim_n a_{m,n}=0,\ \ \lim_n\lim_n a_{m,n}=1. $$ For the sup we have, for any double sequence $b_{m,n}$, $$ \sup_m\sup_n b_{m,n}=\sup_{m,n}b_{m,n}=\sup_n\sup_m b_{m,n}. $$

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