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If we define a cluster point of a sequence $\{a_{n}\}$ to be an extended real number $c$ (possibly $\pm \infty$ ) such that a subsequence of $a_{n}$, say $a_{n_{k}}$ converges to it, then proving that a sequence in $\mathbb{R}$ converges to an extended real $a \in \mathbb{R}$ iff set of cluster points is a singleton reduces to showing that a sequence converges iff all its subsequences converge to the same value.

Right now, I'm having difficulty with the "all the $a_{n_{k}}$ converge to the same value implies $a_{n}$ converges part", specifically in the case where we're dealing with convergence (technically divergence) to $\pm \infty$.

I'll share what I have done so far, but I think it's utter crap:

  • (Just showing the case for $\infty$, but I had the same exact difficulty [mentioned below the quoted text] for $-\infty$): In order to show that if all subsequences $\{ a_{n_{k}} \}$ of $\{a_{n}\}$ converge to $\infty$ implies that $\{a_{n} \} \to \infty$, we will assume that $\{a_{n} \} \nrightarrow \infty$ and show that $\exists$ subsequence $\{a_{n_{k}} \}$ that also $\nrightarrow \infty$.

Now, $\{a_{n} \} \nrightarrow \infty$ means that $\exists M \in \mathbb{R}$, $\forall N \in \mathbb{N}$ such that $\exists n \geq N$ giving us $a_{n} \leq M$, so there are infinitely many such $N \leq n$ such that this is the case.

Running through them, we have for $N = 1$, $n_{1} \geq 1$ and $a_{n_{1}}\leq M$; for $N = 2$, $n_{2} \geq 2$, and $a_{n_{2}}\leq M$. Continuing on in this fashion, we obtain a subsequence $\{ a_{n_{k}}\} \leq M$, $\forall k \in \mathbb{N}$. Then, $\limsup \{a_{n_{k}}\}\neq \infty$, since $\limsup \{a_{n_{k}}\}= \lim_{n\to \infty}[\sup \{ a_{l_{k}}| l \geq n\}]=M$.

Then, for $\liminf\{a_{n_{k}} \} = \lim_{n \to \infty} \{\inf\{a_{l_{k}}|l\geq n\}]$, we have two cases:

  1. $\liminf \{a_{n_{k}}\} = M$ as well, then $\lim_{n \to \infty}\{a_{n_{k}}\} = M \neq \infty$.

  2. If $\liminf \{a_{n_{k}}\}\neq M$, then $\lim_{n \to \infty} \{a_{n_{k}}\}$ still does not converge to $\infty$, since we need $\liminf = \limsup$.

Now, here I had a real problem. I was able to prove that $\{a_{n}\} \to a$ implies $\liminf = \limsup$ for [nonextended] real $a$, but proving it for $a = \pm \infty$ is giving me a lot of problems. So, my question is, could somebody please show me (in detail - pretend like you're talking to a clueless person, which you are) how explicitly to prove that $\{a_{n}\} \to \infty$ implies $\liminf = \limsup$ and $\{a_{n}\} \to -\infty$ implies $\liminf = \limsup$ (as separate cases - please don't just do it for either $\infty$ or $-\infty$ and say that the other case follows; remember, you're talking to a clueless person)

If you can do this for me, I will be eternally, eternally grateful!

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Suppose $a_n\to+\infty$. You have that $\inf_{k\geq n}a_k$ is increasing. Since $a_n\to+\infty $, for all $k$, there is a $n_k$ such that $a_{n_k}>k$ and thus $\inf_{m\geq k}a_{n_m}\geq k$ therefore $\inf_{k\geq n}a_k$ is not bounded and thus $\liminf_{n\to\infty }a_n=+\infty $. To conclude that $\limsup_{n\to\infty }a_n=+\infty $, juste remark that

$$\inf_{k\geq n}a_k\leq \sup_{k\geq n}a_k$$ for all $n$.

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If $a_n \to \infty$, then clearly $\lim \sup a_n = \infty$. To show $\lim \inf a_n = \infty$, note that for any finite real $C$ you can find $N$ so that $a_n \geq C$ for all $n \geq N$. Thus for any $C$, you have $\lim \inf a_n \geq C$. This shows $\lim \inf a_n = \infty$.

If $a_n \to - \infty$, then clearly $\lim \inf a_n = - \infty$. To show $\lim \sup a_n = - \infty$, note that for any finite real $C$ you can find $N$ so that $a_n \leq C$ for all $n \geq N$. Thus for any $C$, you have $\lim \sup a_n \leq C$. This shows $\lim \sup a_n = - \infty$.

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  • $\begingroup$ @JessyCat Yes, thanks. $\endgroup$ – user2566092 Sep 24 '15 at 15:53

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