Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to prove that if $x_n$ is a sequence such that every subsequence $a_n$ has a subsubsequence $b_n$ whose $\limsup b_n \le M $ then $\limsup x_n \le M $


if I take as a subequence $a_N = sup_{n>N} x_n$ then it is monotone decreasing, so that $sup_{n>N} a_n = a_N $ therefore $\limsup a_n = inf_{N} sup_{n>N} a_n = inf_{N} a_N = inf_{N} sup_{n>N} x_n = \limsup x_n $

$b_k$ is a subsequence of $a_n$, so $\limsup a_n \ge \limsup b_k $, and since $a_n$ is monotone decreasing so is $b_k$, hence $\limsup b_k = inf_{K} sup_{k>K} b_k = inf_{K} b_K = inf_{K} a_{n_k} \ge inf_N a_N = \limsup a_n $

so $\limsup b_k = \limsup a_n = \limsup x_n $ and the result follows


I am not too sure of my demonstration, since I have not done this for some years... Are those correct statements ?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Your argument breaks down because the identities $y_N=\sup\limits_{n\gt N}x_n$ do not define, in general, a subsequence $(y_n)_n$ of $(x_n)_n$. Recall that a sequence $(a_n)_n$ is a subsequence of the sequence $(x_n)_n$ if and only if there exists an increasing function $\varphi$ such that $a_n=x_{\varphi(n)}$ for every $n$. Here, it may happen that $y_N$ is different of $x_n$ for every $n$.

An approach to prove the result is to assume that $\limsup\limits_nx_n\gt M$ and to construct a subsequence $(a_n)_n$ of $(x_n)_n$ such that no subsequence of $(a_n)_n$ has a limsup $\leqslant M$. To do so, note that $\limsup\limits_nx_n\geqslant M'$ for some $M'\gt M$. This implies that the set $\{n\mid x_n\geqslant M''\}$ is infinite for every $M\lt M''\lt M'$, say $M''=\frac12(M+M')$. Thus, a subsequence $(a_n)_n$ of $(x_n)_n$ is such that $a_n\geqslant M''$ for every $n$. In particular, $\limsup\limits_na_n$ is such that $____$ hence $____$. You are done.

share|improve this answer
    
Makes sense. And now that you point at it i remember thinking vaguely the subsequence def was fishy while writing it. Thanks ! –  nicolas Dec 31 '12 at 7:19
    
@did Ah, I just realized why you needed to introduce M'', and hence that my solution is incomplete as stated :) –  Calvin Lin Dec 31 '12 at 9:06
    
@CalvinLin Indeed, the case of a decreasing convergent sequence $(x_n)$ may serve to avoid false assertions. –  Did Dec 31 '12 at 9:11
    
@did Yes, I didn't understand why your solution was so 'convoluted' when I read it, which was why I posted mine (thinking it was different from yours). It was only after your comment that I realized the mistake. You should have pointed it out! –  Calvin Lin Dec 31 '12 at 9:13

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.