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Show that if $\lim\sup a_n = -\infty$, then $a_n\to -\infty$, and if $\lim\sup a_n = \infty$ there exists a subsequence of $a_n$ that $\to\infty$. What if $\lim\inf a_n = \pm\infty$? We're working with the extended real number system, $\bar{\mathbb{R}} = \mathbb{R} \cup\{\infty,-\infty\}$.

My work: $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m = -\infty$$ is known. This means that there exists some $n'\in\mathbb{N}$ such that $\sup_{m\ge n'}a_m = -\infty$. So, for all $m \ge n'$, we have $a_m \le -\infty$. This can happen iff $a_m = -\infty$ for all $m\ge n'$. Hence, $a_n\to -\infty$.

Is this proof alright? I'm concerned about steps where I've written $a_m=-\infty$ and such, even though we know that we are working with the extended real number system. Is this okay to do? Just some uneasiness since I haven't used $=$ with infinities before.

Now assume that $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}$$ I can intuitively see that the least upper bound of all tail sequences of $a_n$ is $\infty$, so maybe $a_n\to\infty$ itself? I'm not sure because the problem explicitly asks to produce a subsequence of $a_n$ (or show that one exists) that goes to $-\infty$. How do we do this?

Similarly for the case when $\lim\inf a_n = \pm \infty$, what can we say? I'd guess that $\lim\inf a_n = \infty$, then $a_n\to\infty$ and if $\lim\inf a_n = -\infty$ then there is a subsequence of $a_n$ that goes to $\infty$.

Here's my definition for diverging to $+\infty$: $$x_n\to +\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n > c)$$ $$x_n\to -\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n < c)$$

Second Attempt:

$$\lim\sup_{n\to\infty} a_n = -\infty \implies \lim_{n\to\infty} \sup\{a_k:k\ge n\} = -\infty$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (\sup\{a_k:k\ge n\} < c)$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_k < c, k\ge n)$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_n < c) \implies \lim_{n\to\infty} a_n = -\infty$$ Similarly, $$\liminf_{n\to\infty} a_n = \infty \implies a_n\to\infty$$

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  • $\begingroup$ I think the definition for diverging to $+\infty$ is that for any $M\in\mathbb R$, there exists an $N\in\mathbb N$ so that $n>N$ implies that $a_n>M$. (Similar for diverging to $-\infty$.) $\endgroup$
    – boink
    Commented Jan 19, 2021 at 6:54
  • $\begingroup$ What does $a_m\leq-\infty$ mean for a real number? $\endgroup$ Commented Jan 19, 2021 at 9:07
  • $\begingroup$ @MichaelBurr $a_m = -\infty$ perhaps? There is nothing smaller than $-\infty$ so $\endgroup$ Commented Jan 19, 2021 at 9:11
  • $\begingroup$ math.stackexchange.com/questions/2486599/… and math.stackexchange.com/questions/2079296/… were helpful. $\endgroup$ Commented Jan 19, 2021 at 14:10

2 Answers 2

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No, your proof is not correct. There exists in general no $n'\in\mathbb{N}$ such that $\sup_{m\ge n'}a_m = -\infty$, but there exists a subsequence $(n_k)$ so that $a_{n_k}\to-\infty(k\to\infty)$ and $\sup_{m\ge n}a_m \to -\infty (n\to\infty)$. The first comes from the fact that $\limsup a_n=c$ means your biggest cluster point is $c$ and cluster points are exactly the limits of converging subsequences. So $\limsup$ is the biggest extended real number a subsequence converges. The second can either be implied by the first or you can use $\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{m\geq n} a_m$, which comes from the fact that $(\sup_{m\geq n} a_m)_{n\in\mathbb N}$ is a decreasing sequence so you can exchange $\inf$ and $\lim$ in $$\limsup_{n\to\infty} a_n=\inf_{n\in\mathbb N}\sup_{m\geq n} a_m=\lim_{n\to\infty}\sup_{m\geq n}a_m. $$

What you have to show (as boink already mentioned in the comments) is that for every $M\in\mathbb R$ (especially for big negative $M$) exists a $N\in\mathbb N$ so that for all $n\geq N$ holds $a_n<M$. So choose any $M$ and now let's find such a $N$. Your first step to consider the sequence $(\sup_{m\ge n}a_m)_{n\in\mathbb N}$ was in the right direction. As already mentioned above we have $\sup_{m\ge n}a_m \to -\infty (n\to\infty)$. This means we find a $N'$ so that for every $n\geq N'$ holds $\sup_{m\ge n}a_m <M$. This implies that for every $n\geq N'$ holds $a_n<M$. This number is the $N=N'$ we were looking for and we have finished the proof.

Now consider $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}.$$ The $\lim\sup$ means the bigest cluster point of your sequence is $\infty$, which directly implies the statement, since you find for an arbitrary big $n$ a $m>n$ so that $a_m$ is arbitrary big. If you don't wanna handle with cluster points here, you can use the fact from above $$\infty=\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{m\geq n}a_m$$ and then apply that the decreasing sequence $(\sup_{m\ge n}a_m)_{n\in\mathbb N}$ converges to $\infty$ which means it already has to be $\infty$.

Your last two guess regarding $\lim\inf a_n = \pm \infty$ are correct.

What you wrote under "Second Attempt" is comletely correct.


To give you an alternative proof to the first part, you could also use the property $$ \lim\inf a_n\leq \lim\sup a_n,\tag{1}\label{tag} $$ where equality holds if and only if the limit $\lim a_n$ exists and $$\lim\inf a_n= \lim\sup a_n= \lim a_n.$$ So if $\lim\sup a_n=-\infty$ already holds, it follows form \eqref{tag} that $\lim\inf a_n=-\infty$ and the statement follows from the 'iff'-part.

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  • $\begingroup$ Thanks for your answer. Could you first clarify if $\lim_{n\to\infty} \sup_{m\ge n} a_m = \inf_{n\ge1}\sup_{m\ge n} a_m$? Are they the same thing? $\endgroup$ Commented Jan 19, 2021 at 9:17
  • $\begingroup$ Yes, your first comment is correct. $\endgroup$
    – mag
    Commented Jan 19, 2021 at 9:20
  • $\begingroup$ $\lim\sup a_n=c$ means your biggest cluster point is $c$. But cluster points are exactly the limits of converging subsequences. So $\lim\sup$ is the biggest extended real number a subsequence converges. $\endgroup$
    – mag
    Commented Jan 19, 2021 at 9:24
  • $\begingroup$ In that case, I'm not able to see where your first statement is coming from. Is there a proof for the fact that $\lim\sup a_n = \pm\infty$ implies that there exists a subsequence $a_{n_k}$ such that $\sup_{m\ge n_k} \to \pm\infty$? $\endgroup$ Commented Jan 19, 2021 at 10:12
  • $\begingroup$ Sry, I made something a bit confusing. It is not only a subsequence $(n_k)$ sucht that $\sup_{m\ge n_k}a_m\to-\infty$, but it holds for the complete sequence $\sup_{m\ge n}a_m\to-\infty$. The 'subsequence'-part belongs to the actual sequence $a_{n_k}$. I corrected it in the answer. $\endgroup$
    – mag
    Commented Jan 19, 2021 at 10:40
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The statement $\sup_{m\geq n'}\alpha_m =-\infty,$ for some $n'$ is false in general. For example if $\alpha_m = -m$ then for every $n'$ we have $\sup_{m\geq n'}=\alpha_{n'}=-n'>-\infty$ and $\alpha_n\to -\infty$ as $n\to \infty$. To prove that $\alpha_n\to -\infty$ when $\limsup_{n}\alpha_n = -\infty$ fix some $c\in\mathbb{R}$. Since $\limsup_n \alpha_n = - \infty$ there is $n'$ such that $$\sup_{m\geq n'}\alpha_m < c.$$ This means that for every $m\geq n'$ we have $\alpha_m<c$ which is equivalent to $\alpha_n\to -\infty$. Now, if $\limsup_{n}\alpha_n =\infty$ it does not necessarily implies that $\alpha_n \to \infty$. For example consider the sequence $$ \alpha_n=\begin{cases} n,\ n \text{ is even }\\ -1,\ n \text{ is odd} \end{cases} $$ Then, the subsequence $\alpha_{2n-1}$ is constant and equal to $-1$, hence $\alpha_n$ cannot converge to $\infty$. Although, it is true that there exists a subsequence $\alpha_{k_n}$ such that $\alpha_{k_n}\to \infty$. Since $\limsup_n \alpha_n =\infty$ for every $n$ you have that $\sup_{m\geq n}\alpha_m =\infty$. In particular, $$\sup_{m\geq 1 }\alpha_m >1.$$ This means, that there exists some $k_1\geq 1$ such that $\alpha_{k_1}>1$. Now, using again the fact that $$\sup_{m\geq k_1 +1}\alpha_m>2$$ we may find $k_2 >k_1$ such that $\alpha_{k_2}>2$. Continuing this way, recursively, we may find indices $$k_1<k_2<...<k_n<k_{n+1}<...$$ in ascending order such that $\alpha_{k_n}>n$ for every $n$. This means that the subsequence $\alpha_{k_n}$ converges to $\infty$. Similarly, if $\liminf_n \alpha_n=\infty$ then $\alpha_n\to \infty$ and if $\liminf_n \alpha_n =-\infty$ then there exists a subsequence $\alpha_{k_n}\to -\infty$.

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