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Consider sequence $a_{nm}$ such that $a_{nm} \overset{n}{\rightarrow} a_m$ and $a_{im} \leq a_{jm} \;\;\forall\; i<j$.

We also have $a_{m} \overset{m}{\rightarrow} a$ and $a_{p} \leq a_{q} \;\;\forall \;p<q$. Then, is it possible to prove that,

$\lim_n \lim_m a_{nm} = \lim_m \lim_n a_{nm} = a$

If not, then what are the conditions under which the limits commute?

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up vote 2 down vote accepted

Take $a_{nm}=2-(1/2)^n$ if $m$ is odd and $a_{nm}=2-(1/2)^{n+1}$ otherwise. Then for fixed $n$ the sequence $(a_{nm})_m$ oscilates, so you cannot consider the inner limit $\lim_m a_{nm}$.

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What are the conditions under which the limits can commute? – UnadulteratedImagination Jan 17 '13 at 7:10
Thanks. Another example would be $a_{nm} = (1-(1/2)^m)^n$. $\lim_m \lim_n a_{nm} = 0$ but $\lim_n \lim_m a_{nm} = 1$ – UnadulteratedImagination Jan 17 '13 at 13:20
@user167291 I am afraid that your proposed sequence cannot be used as counterexample: in fact, for fixed $m$ your sequence $(a_{nm})$ is decreasing as function of $n$. Regarding sufficient conditions, i didn't think yet. – Matemáticos Chibchas Jan 17 '13 at 22:37

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