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

Theorem Every sequence {$s_n$} has a monotonic subsequence whose limit is equal to $\limsup s_n$. I think to show that there exist a monotonic subsequence is kind of straight forward but I could show that there exist such subsequences whose limit is $\limsup s_n$.

share|cite|improve this question
It is known that $$ \limsup\limits_{n\to\infty} s_n=\sup\{\lim\limits_{k\to\infty} s_{n_k}:\{s_{n_k}:k\in\mathbb{N}\}-\text{ convergent subsequence of }\{s_n:n\in\mathbb{N}\}\}\} $$ So you can easily extract such a subsequence – Norbert Jul 16 '12 at 8:43
You probably mean real sequence (=sequence of real numbers), but perhaps it would be better to mention it explicitly in the question. – Martin Sleziak Jul 16 '12 at 9:03

Let $L=\limsup_n s_n$. (Note that $L$ can be $\infty$.) Probably the simplest approach is to prove first that $\langle s_n:n\in\Bbb N\rangle$ has a subsequence $\langle s_{n_k}:k\in\Bbb N\rangle$ converging to $L$, and then show that $\langle s_{n_k}:k\in\Bbb N\rangle$ has a monotonic subsequence.

By definition $$L=\lim_{n\to\infty}\sup_{k\ge n}s_k\;,$$ so for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $$\left|L-\sup_{k\ge n}s_k\right|<\frac{\epsilon}2$$ for all $n\ge n_\epsilon$. By the definition of supremum there is a $k_\epsilon\ge n_\epsilon$ such that $$0\le\left(\sup_{k\ge n}s_k\right)-s_{k_\epsilon}<\frac{\epsilon}2\;,$$ and therefore

$$|L-s_{k_\epsilon}|\le\left|L-\sup_{k\ge n}s_k\right|+\left|\left(\sup_{k\ge n}s_k\right)-s_{k_\epsilon}\right|<\epsilon\;.$$

For $i\in\Bbb Z^+$ let $n_i=k_{1/i}$, so that $|L-s_{n_i}|<\frac1i$; clearly the sequence $\langle s_{n_i}:i\in\Bbb Z^+\rangle$ converges to $L$. However, it may not be a subsequence of $\langle s_n:n\in\Bbb N\rangle$, because the indices $n_1,n_2,n_3,\dots$ may not be increasing. To finish the argument, you must do two things.

  1. Show that $\langle n_i:i\in\Bbb Z^+\rangle$ has an increasing subsequence $\langle n_{i_j}:j\in\Bbb Z^+\rangle$; $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ is then a subsequence of $\langle s_n:n\in\Bbb N\rangle$ converging to $L$.

  2. Show that $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ has a monotonic subsequence. That subsequence will necessarily have the same limit, $L$, as $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ itself.

share|cite|improve this answer
@Martin: Thanks. I can’t believe that I wrote increasing for monotonic every single time. – Brian M. Scott Jul 16 '12 at 8:52
This kind of relegate the crux of the matter to step 2. at the end of the post. – Did Jul 16 '12 at 12:28
@did: The OP seemed less concerned about that part. I can always provide more detail if asked. – Brian M. Scott Jul 17 '12 at 19:50

Let $S=\limsup\limits_{n\to\infty} s_n$.

  • First we construct a subsequence $s_{n_k}$ such that $\lim\limits_{k\to\infty} s_{n_k}=S$ and $$k\le l \Rightarrow |S-s_{n_k}| \ge |S-s_{n_l}|.$$

In the other words, we construct a subsequence which converges to $S$ and has the property, that the every term of the subsequence is closer so $S$ than the previous one (or in the same distance from $S$).

To prove this we only use the fact that for every $\varepsilon>0$ there are infinitely many $n$'s such that $$|S-s_n|<\varepsilon.$$ (Maybe it might be simpler to treat the case that there are infinitely many $n$'s such that $s_n=S$ separately.)

  • If we have sequence with the above property, then one of the sets $\{k\in\mathbb N; s_{n_k}\ge S\}$ and $\{k\in\mathbb N; s_{n_k}\le S\}$ is infinite. We take the one, which is infinite, and we obtain a monotone subsequence converging to $S$.
share|cite|improve this answer

I shall prove a similar property of the infimum

Let $\left(a_n\right)$ be a bounded sequence of real numbers and $a=\limsup{a_n}$, $\epsilon>0$

The set $\left\{n\in \mathbb{N}|a-\epsilon<a_n\right\}$ is infinite while the set $\left\{n\in \mathbb{N}|a+\epsilon<a_n\right\}$ is finite (why?) and thus the sets $\left\{n\in \mathbb{N}|a-\epsilon<a_n\right\}$ and $\left\{n\in \mathbb{N}|a_n\le a+\epsilon\right\}$ are both infinite subsets of $\mathbb{N}$

Therefore, the set \begin{equation}\left\{n\in \mathbb{N}|a-\epsilon<a_n\le a+\epsilon\right\}=\left\{n\in \mathbb{N}|a-\epsilon<a_n\right\}\bigcap\left\{n\in \mathbb{N}|a_n\le a+\epsilon\right\} \text{ is infinite (why?)} \tag{1}\end{equation} (the reader should try to prove that if $X$ is infinite and $A,B\subseteq X$ are infinite and $B^c$ is finite then $A\bigcap B$ is infinite. Hint: $A=A\bigcap (B\bigcup B^c)$)

For $\epsilon=1,\frac{1}{2},...,\frac{1}{n}$ we can construct by (1) a subsequence $\left({{a}_{k_n}}\right)$ such as that $a-\dfrac{1}{n}<a_{k_n}\le a+\dfrac{1}{n}$ Then, \begin{equation}\forall n\in \mathbb{N}\ \ \left|{{a}_{k_n}}-a\right|<\dfrac{1}{n}\end{equation}and thus $a_{k_n}\to a$. Next note that every sequence has a monotone subsequence, the proof can be found here:

share|cite|improve this answer
The intersection of two infinite subsets of N is not always infinite hence your (1) needs a justification. // And you simply produce a sequence with limit a, without any guarantee of monotonicity (although monotonicity is the main problem and requires another idea). – Did Jul 16 '12 at 12:32
Indeed this step requires justification, I purposely left that to the reader! The monotonicity follows simply from the fact that every sequence has a monotone subsequence (a Lemma used to prove the Bolzano Weierstrass Theorem) and if a sequence conversges say to $a$ all of its subsequences will converge to $a$ – Nameless Jul 16 '12 at 12:38
Well, it is intersection of an infinite set and a cofinite set. (Nameless probably wanted OP to come up with it.)\\ The argument using the result on monotone subsequences is nice. – Martin Sleziak Jul 16 '12 at 12:44
Nameless: Your comment is odd on two counts. First, the previous version of your post could only have one effect on the OP, which was to believe the statement (1) was true because of the wrong reason I gave. This is misleading (on purpose ? I hope not!). Second, to use the result that every sequence has a monotone subsequence (something which you do not mention in your post...) seems to simply preempt the whole question. – Did Jul 16 '12 at 13:06

Your Answer


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.