[Math] How to infinity be an accumulation point

analysisinfinity

I can't wrap my mind around this: An accumulation point is a point of a set, which in every yet so small neighborhood (of itself) contains infinitely many points of the set, right?

So if (in my case) $-\infty$ is an accumulation point, then we could find a point of the set, that is finitely distant from infintity?!

Here's the task:
Let $(a_n)_{n∈\mathbb N}$ be a set in $\mathbb R$. For every $r ∈ \mathbb R$ exits an accumulation point $b ∈ R$ ∪ {−∞} of $(a_n)_{n∈\mathbb N}$
(+∞ und −∞ are allowed as accumulation points) with b < r.
Show, that −∞ is an accumulation point of $(a_n)_{n∈\mathbb N}$.

Best Answer

Which points of a metric (or topological) space are accumulation points depends on which metric (or topology) you are using. In the case of $\mathbb{R}\cup\{-\infty,\infty\}$, the usual metric $d(x,y)=\vert x-y\vert$ is not going to work, since it is not a metric on this space.

Since you have tagged the question as analysis rather than topology, I'll assume that you don't want to work in an explicitly topological framework. So instead, we use a sort of work around:

A sequence $\langle x_n\vert n\in\mathbb{N}\rangle$ converges to $\infty$ if for every $M$ there is a natural number $N$ such that $$\forall n>N \, (x_{n}>M).$$

In this case, the idea of getting "close to infinity" is therefore expressed as "gets arbitrarily large".

Similarly:

$\infty$ is an accumulation point of the sequence if for every real $M$ there is some $n$ such that $M<x_{n}<\infty$.

Note that if the sequence does not take the value $\infty$, this is equivalent to the sequence being unbounded above. Note also that for any $M$, there must in fact be infinitely many such $n$s.

Convergence to $-\infty$, and the property of having $-\infty$ as an accumulation point are defined analogously.

So, you have a sequence $\langle a_{n}\vert n\in\mathbb{N}\rangle$ which has arbitrarily small accumulation points. To show that $-\infty$ is also an accumulation point, you just need to show that it is unbounded below. But this is straightforward: for every $M<0$, there is an accumulation point $b<(M-1)$, and so the sequence contains a term below $b+1$ (or $b+\varepsilon$ if you prefer, so long as $\varepsilon>0$). Thus $M$ is not a lower bound for the sequence. Since $M$ is arbitrary, the sequence has no lower bounds, and therefore has $-\infty$ as an accumulation point.

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