Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $\lim_m a_{nm}$.

share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.