From Carothers, Exercise 25, p. 11:
If $\lim\sup_{n\to\infty} a_n = -\infty$ show that $(a_n)$ diverges to $-\infty$. If $\lim\sup_{n\to\infty} a_n = +\infty$ show that $(a_n)$ has a subsequence that diverges to $+\infty$. What happens if $\lim\inf_{n\to\infty} a_n = \pm \infty$?
The first two parts seem straightforward, but I'm having trouble with the third part:
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If $\lim \inf_{n\to\infty} a_n = +\infty$ then for each $a_n$ we can choose $M$ so that $m\geq M$ implies $\inf\{a_m, a_{m+1}, \ldots\} \geq a_n$. Because each element of the set is no less than its infimum, $m\geq M$ implies $a_m \geq a_n$ and so $(a_n)\to +\infty$.
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If $\lim \inf_{n\to\infty} a_n = -\infty$ then $m\geq M$ implies that $\inf\{a_m, a_{m+1}, \ldots\} \leq \inf\{a_n, a_{n+1}, \ldots\}$ and because $a_n \geq \inf\{a_n, a_{n+1}, \ldots\}$, there always exists some subsequent term in the sequence no greater than $a_n$. Letting $a_{q_n} =\inf\{a_n, a_{n+1}, \ldots\}$, the subsequence $(a_{q_n}) \to -\infty$
Could anyone please explain how these two properties both hold?
Best Answer
If $\liminf a_n = \infty$, then $\lim a_n = \infty$, because there is nothing "bigger" than $\infty$, and so $\limsup a_n = \liminf a_n = \infty$ and you're done.
The argument you give for why the convergence holds is correct. I think your confusion is because the $\liminf$ and $\limsup$ are more general than the usual limits, designed to deal with when a sequence has multiple subsequences converging to different points, such as $(-1)^n + \frac{1}{n}$. There is always a subsequence converging to both the $\liminf$ and $\limsup$ separately when they disagree, so nothing precludes this sort of wild behavior. In fact by considering examples like $(-1)^n n$ you can even see that both are possible simultaneously.
There is a little topology that can be done to make more formal your observation, if you know about topological spaces.
We can say what it means for a sequence to converge to $\infty$ as follows. We define a neighborhood of $+\infty$ to be the positive connnected component of the complement of any connected, closed, bounded interval, i.e. the right ray leaving a connected, compact set. Then $a_n \to \infty$ if $a_n$ converges in the topology where the points are standard points, or $\pm \infty$, and the open sets are the standard open sets, together with sets of the form $\{ \text{ray leaving a compact set}\cup\{ \infty\} \}$.
Now in this topology, $\mathbb{R} \cup \pm \infty$ is homeomorphic to $[-1, 1]$. And it is clear that if you are a sequence in $[-1,1]$ with $\liminf a_n \to 1$, then it must be the case that $\lim a_n = 1$.
If you're interested, this construction is an example of the "End Compactification" of a topological space. We adjoined points corresponding to the different "infinities" in the space, and make a new topological space. This is often a more useful way of thinking about divergence, because you can distinguish divergence from what might be called "oscillatory" behavior.