Theorem: Any bounded sequence in $\mathbb R$ has a smallest cluster point $\underline x$ and a greatest cluster point $\overline x$ and they satisfy $\liminf x_{n} = \underline x$ and $\limsup x_{n}=\overline x$.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
To simplify notation, we write $\lim x_{n}$ for $\lim_{n \to \infty} x_n$ and $\limsup x_{n}$ for $\limsup_{n \to \infty} x_{n}$.
Let $(x_{n})$ be a sequence in $\mathbb{R}$.
- We define the limit superior and the limit inferior (if exist) of $(x_{n})$ by $$\limsup x_{n} :=\lim\left(\sup _{k \geq n} x_{k}\right) ,\quad \liminf x_{n} :=\lim \left(\inf _{k \geq n} x_{k}\right)$$
It follows that $$\limsup x_{n}=\inf \{\sup _{k \geq n} x_{k} \mid n \in \mathbb{N}\}, \quad \liminf x_n= \sup \{\inf _{k \geq n} x_{k} \mid n \in \mathbb{N}\}$$
- We call $a$ a cluster point of $(x_{n})$ if every neighborhood of $a$ contains infinitely many terms of the sequence.
My attempt:
Let $\overline x := \limsup x_{n}$ and $\overline x < \xi$. Then there is some $n$ such that $\overline x < \sup _{k \geq n} x_{k} < \xi$. So $x_k \le \sup _{k \geq n} x_{k} <\ \xi$ for all $k \ge n$ and thus there are finitely many $x_k$ in $(\xi - \epsilon, \xi + \epsilon)$ where $\epsilon := \xi - \sup _{k \geq n} x_{k}$. As such, $\xi$ is not a cluster point of $(x_n)$ and consequently no cluster point of $(x_n)$ is larger than $\overline x$. It remains to show only that $\overline x$ is itself a cluster point of $(x_n)$.
Lemma: If every neighborhood of $a$ contains at least one term of $(x_{n})$ different from $a$ itself, then $a$ is a cluster point of $(x_{n})$.
Given $\epsilon > 0$, there is some $n$ such that $\overline x < \sup _{k \geq n} x_{k} < \overline x + \epsilon$. So there exists some $k \ge n$ such that $\overline x < x_k < \sup _{k \geq n} x_{k}$. As such, $\overline x \neq x_k \in (\overline x - \epsilon, \overline x + \epsilon)$. By our Lemma, $\overline x$ is a cluster point of $(x_{n})$.
Showing that $\underline x :=\liminf x_{n}$ is the smallest cluster point of $(x_{n})$ is similar.