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 ?


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.

  • $\begingroup$ Makes sense. And now that you point at it i remember thinking vaguely the subsequence def was fishy while writing it. Thanks ! $\endgroup$ – nicolas Dec 31 '12 at 7:19
  • $\begingroup$ @did Ah, I just realized why you needed to introduce M'', and hence that my solution is incomplete as stated :) $\endgroup$ – Calvin Lin Dec 31 '12 at 9:06
  • $\begingroup$ @CalvinLin Indeed, the case of a decreasing convergent sequence $(x_n)$ may serve to avoid false assertions. $\endgroup$ – Did Dec 31 '12 at 9:11
  • $\begingroup$ @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! $\endgroup$ – Calvin Lin Dec 31 '12 at 9:13

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