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Given a sequence of real numbers, $\{ x_n \}_{n=1}^{\infty}$, let $\alpha =$ limsup$x_n$ and $\beta = $ liminf$x_n$.

Prove that there exists a subsequence $\{ x_{n_k}\}$ that converges to $\alpha$ as $k \rightarrow \infty$.

Not sure how to start this without since I'm not given that the subsequence is bounded..

Martin Sleziak
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user12279
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  • Think about what it means to say $\limsup x_n=\alpha$, i.e. the definition of the $\limsup$ of a sequence. – JohnD Nov 25 '13 at 22:18
  • @JohnD Since $\alpha $ is a supremum of $x_n$, can I say that all values of $x_n$ are strictly smaller than $\alpha$ and similarly, all values of $x_n$ are strictly larger than $\beta$ since $\beta$ is an infimum? – user12279 Nov 25 '13 at 22:25
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    No, that's not correct. $\alpha$ is the lim sup, not the sup. The limsup is the supremum of all the limit points of the sequence, not the supremum of all the points in the sequence. – JohnD Nov 25 '13 at 22:37
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    Did my post below answer your question? If so, consider accepting it as this is good form and will encourage others to answer your future questions. – JohnD Nov 26 '13 at 17:10
  • @JohnD Thanks! Sorry, I had stepped away from my computer for a bit. Your answer makes sense, I think I just need a better grasp on limsup and liminf. – user12279 Nov 27 '13 at 14:19
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    No problem. Glad to help. It should be helpful to think of $\limsup$ as the "largest limit/cluster point" and $\liminf$ as "smallest limit/cluster point". – JohnD Nov 27 '13 at 14:31
  • You should probably include your definition of limit superior. Several equivalent definitions appear in literature. (See, for example, the two definitions mentioned in this post.) – Martin Sleziak Aug 20 '14 at 09:11

4 Answers4

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Since $\alpha=\limsup x_n$, by the definition of $\limsup$, there is some $x_{n_1}$ with $|x_{n_1}-\alpha|<{1\over 2}$. (That's the crucial step, so be sure you understand why.)

Similarly, there is some $x_{n_2}$ with $|x_{n_2}-\alpha|<{1\over 2^2}$. Continuing, for each $k\in\mathbb{N}$, there is some $x_{n_k}$ with $|x_{n_k}-\alpha|<{1\over 2^k}$.

Then $\{x_{n_k}\}\subset \{x_n\}$ and $x_{n_k}\to \alpha$ as $k\to\infty$.

JohnD
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    How can you be sure that each time you reduce the size of the interval around $\limsup x_n$, you can choose an element of the sequence with an index that is higher than the index of the element that you had previously chosen? – dazedviper Jan 27 '14 at 15:42
  • Because $alpha$ is by definition the $\limsup$ of the sequence, and thus is a limit point of the sequence. – JohnD Jan 27 '14 at 16:02
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    I don't understand what you're saying. $\alpha := \limsup x_n = \lim_{N \to \infty} sup{x_n \mid n \geq N }$ by definition. The fact that for each $N \in \mathbb N$, $sup{x_n \mid n \geq N }$ is a limit point of the sequence guarantees that there is an element of the sequence to the left of this particular supremum, but when you reduce the interval and get another supremum and repeat the whole thing again, there is no guarantee that you can choose another element of the sequence with an index that is higher than the previous element you chose that was added to your subsequence. – dazedviper Jan 27 '14 at 16:23
  • $\limsup x_n$ is the greatest limit point of $x_n$, so exploit the fact that it is a limit point of ${x_n}$: for any $\varepsilon>0$, there are infinitely $x_n$ within $\varepsilon$ of $\alpha$. See this. – JohnD Jan 27 '14 at 17:05
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    Question related to your answer: the definition of lim sup is that it is the limit of the sequence defined as $v_N = \sup{s_i:i>N}$. You could from this that $|v_{n_1} - \alpha| < \epsilon$ and go on to choose some $\epsilon$. But you have rather said that this $v_{n_1}$ necessarily belongs to the sequence $s_n$. That is equivalent to saying that the $\sup$ of a set exists in it, which may or may not be true.

    Also, isn't your proof essentially selecting a subsequence $s_{n_k}$ from the set $A_N = {s_i: i > N}$ so that:

    $\lim \sup {A_N} \geq s_{n_k} \geq \lim \inf A_N$

     – sprajagopal Aug 19 '14 at 13:50
  • We define lim sup as inf sup. Please, provide a proof of your "crucial step", it is non-trivial! – georgy_dunaev Mar 18 '23 at 16:23
  • @dazedviper, that is an excellent question! Why we do not select the same elements in the sequence and create a ‘non-subsequence’? I have tried to address this in another MSE answer, if you want to check. – Kadmos Oct 20 '23 at 10:36
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I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it. Let us define the terms first.

$$\lim\sup s_n :=\lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$

Let $A_N := \{s_n:n>N\},~~ v_N := \sup A_N.$

Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that

$A_i \subseteq A_j$ when $i>j$.

$\therefore \sup A_i \leq \sup A_j \Rightarrow v_N$ is a non-increasing (monotonous) sequence. Thus, $\lim \sup s_n$ is the infimum of the set of $\{v_N\}_{N=1}^\infty.$

$$\inf v_N = \lim\sup s_n \tag{d1}$$

$$\therefore~~ \exists \ v_N: v_N < \lim \sup s_n + \epsilon \ \forall \epsilon > 0\tag 1$$

Using the definition of $v_N$,

$$\exists s_n: s_n > v_N - \alpha ~~ \forall \alpha > 0\tag 2$$

Using (d1) and (1),

$$\lim \sup s_n \leq v_N < \lim \sup s_n + \epsilon ~~ \forall \epsilon>0 \tag a$$

Using (2) and previous definition of $v_N$,

$$v_N \geq s_n > v_N - \alpha~~ \forall \ \alpha > 0 \tag b$$

Using (a) and (b),

$ \limsup s_n - \alpha \leq v_N - \alpha < s_n \leq v_N < \lim \sup s_n + \epsilon$ where $\alpha,\epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $\lim_{N\rightarrow \infty}v_N$, i.e. $\lim \sup s_n$.

Community
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sprajagopal
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$A:=\lim_{n \rightarrow \infty} \sup a_n=\lim_{n \rightarrow \infty} A_n$, where $A_n=\sup ${$a_k|k \ge n$};

Let $L$ be the set of all limits of convergent subsequences.of $(a_n)$.

1) We show that $\forall \epsilon >0$ there exists an $a_n$ s.t. $|a_n-A| <\epsilon$.

Since $\lim_{n \rightarrow \infty} A_n =A$, there is a $n_0$ s.t. for $m \ge n_0$

$|A_m -A|<\epsilon$.

2)Since

$A_{n_0}= \sup${$a_k| k\ge n_0$}, there is

a $k \ge n_0$ s.t.

$|a_k-A_{n_0}| <\epsilon$.

Set $m=k$ in 1), and we get

$|a_k-A| \le |a_k-A_{n_0}| +|A_{n_0}-A| <2\epsilon$.

2) Given that in any $\epsilon$ neighbourhood of $A$ there is an $a_n$, we can construct a subsequence $a_{n_k}$ that converges to $A$, i.e. $A \in L$. (JohnD's answer)

Peter Szilas
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  • I understand that $A_{n_0}$ is the sup of the set. So we can find an element such that |a_k - A_n_0| < \epsilon. But how can we construct a subsequence? – Oran Oct 05 '23 at 04:38
  • @Oran. Does your question concern the very last point 2) only, i.e. given that for every epsilon there is a_n such that |a_n -A|< epsilon ? If yes: Consider epsilon_k:=1/k, k=1,2,3,...., then |a_(n_k) -A|<1/k. – Peter Szilas Oct 05 '23 at 17:11
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I think none of the answers considered what happens if things are $+-\infty$ and also why we are able to get elements of higher index in each step we take an interval closer to $\limsup$. Why couldn’t an element be exactly $\limsup$ and we always choose the same element ?

But these two issues are easily rectifiable.

By the definition of infimum there is $M_K$ such that:

$$\inf_{ m} \sup_{n\geq m} x_n\leq \sup_{n\geq M_K} x_n<\max\left\{\inf_{ m} \sup_{n\geq m} x_n+\frac{1}{K},-K\right\}$$

These $M_K$ can be taken such that $M_K>M_{K-1}$, because in this case $\sup_{n\geq M_K} x_n \leq \sup_{n\geq M_{K-1}} x_n$. This inequality ensures we will actually obtain a sequence later on.

By the definition of supremum there is $N_K\geq M_K$ such that:

$$\min\left\{\sup_{n\geq M_K} x_n-\frac{1}{K}, K\right\} <x_{N_K}\leq\sup_{n\geq M_K} x_n$$

$$\therefore \min\left\{\inf_{ m} \sup_{n\geq m}x_n-\frac{1}{K}, K\right\} <x_{N_K}\leq\sup_{n\geq M_K} x_n$$

With these two inequalities in mind:

$$\min\left\{\inf_{ m} \sup_{n\geq m}x_n-\frac{1}{K}, K\right\} <x_{N_K}<\max\left\{\inf_{ m} \sup_{n\geq m} x_n+\frac{1}{K},-K\right\}$$

Eventually $M_K$ will become larger then any fixed $N_{K_o}$. Therefore $N_K>N_{K_o}$ and we actually have a proper subsequence by selecting adequate $x_{N_K}$.

Case I: $\limsup x_n=-\infty $:

$$x_{N_K}< -K\Rightarrow x_{N_K}\rightarrow \limsup x_n $$

Case II: $\limsup x_n=\infty $:

$$x_{N_K}> K\Rightarrow x_{N_K}\rightarrow \limsup x_n $$

Case III: $\limsup x_n\in \mathbb{R} $. In this case for sufficiently large $\hat{K}$, $-\hat{K}<\limsup x_n<\hat{K} $, but in this case, for $K > \hat{K}$:

$$\limsup x_n-\frac{1}{{K}} <x_{N_{K}}< \limsup x_n+\frac{1}{{K}}\Rightarrow x_{N_K}\rightarrow \limsup x_n $$

Kadmos
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