# Question #7c37f

Oct 23, 2016

Use the definition of the limit inferior to show that there must be infinitely elements of the sequence that are "close" to it.

#### Explanation:

Given a sequence $\left({b}_{n}\right)$, we say that $b$ is a cluster point of $\left({b}_{n}\right)$ if every neighborhood of $b$ contains infinitely many elements of $\left({b}_{n}\right)$.

If we translate this into the real numbers, that means that if $L = \text{lim inf} \left({a}_{n}\right)$, then we should be able to create an arbitrarily small interval $\left(L - \epsilon , L + \epsilon\right)$ around $L$ and still have infinitely many ${a}_{n} \in \left(L - \epsilon , L + \epsilon\right)$.

To show this, we need to understand what the limit inferior of a sequence is. In the case of a sequence of real numbers, the limit inferior is an extension of the idea of a greatest lower bound. It is the greatest value such that all numbers in the sequence are eventually greater than that value.

$\text{lim inf} \left({a}_{n}\right) =$

$\max \left\{L \in \mathbb{R} | \exists N \in \mathbb{N} \text{ s.t. } n > N \implies {a}_{n} \ge L\right\}$

It's a lower bound which allows us to discard any number of finite elements of the sequence lower than it.

Now, how can we use this? Well, first of all, notice that $\left({a}_{n}\right)$ is bounded. This is important, as it guarantees that we actually have a limit inferior (imagine the sequence ${\left(- 1\right)}^{n} {n}^{2}$, no eventual bounds there!). So, as we are guaranteed of its existence, let's let $L$ be the limit inferior of $\left({a}_{n}\right)$.

Now, let's show that $L$ is a cluster point of $\left({a}_{n}\right)$ using an argument by contradiction.

Suppose that $L$ is not a cluster point of $\left({a}_{n}\right)$. Then there exists some $\epsilon > 0$ such that $\left(L - \epsilon , L + \epsilon\right)$ contains only finitely many elements of $\left({a}_{n}\right)$. Let ${a}_{{n}_{1}} , {a}_{{n}_{2}} , \ldots , {a}_{{n}_{k}}$ be the ordered list of elements of ${a}_{n}$ in $\left(L - \epsilon , L + \epsilon\right)$. Then, for all $n > {n}_{k}$, we have ${a}_{n} \ge L + \epsilon$. But $L + \epsilon > L$ and $L$ is, by definition, the greatest number with that property. Thus we have reached a contradiction, meaning $L$ must be a cluster point of $\left({a}_{n}\right)$. ∎