If $x_n$ Cauchy then $\lim\sup x_n = \lim\inf x_n$

Question

Suppose $x_n$ is Cauchy, then $\lim\sup x_n = \lim\inf x_n$.

Can't use the fact that all Cauchy sequences converge, as we haven't shown this yet. Here is my attempt.

Since $x_n$ is Cauchy, then $x_n$ is bounded.

First define $b_n = \sup\{x_n, x_{n+1}, ...\}$. Note that this sequence is decreasing and bounded below. Thus $\lim b_n = \lim\sup x_n \leq \inf\{x_n, x_{n+1}, ...\}$.

Take $a_n = \inf\{x_n, x_{n+1}, ...\}$. Note that this sequence is increasing and bounded above. Thus $\lim a_n = \lim\inf x_n \geq \sup\{x_n, x_{n+1}, ...\}$.

Since $\lim\sup x_n \leq \inf\{x_n, x_{n+1}, ...\} \leq \sup\{x_n, x_{n+1}, ...\} \leq \lim\inf x_n$. But $\lim\sup x_n \geq \lim\inf x_n.$

Thus $\lim\sup x_n = \lim\inf x_n$

I don't think this is the right approach, because I didn't even use Cauchy property.

Answer 1

Notice that the inequality $\limsup x_n \le \inf\{x_n,\dots\}$ is not correct. Indeed for $x_n = (-1)^n$, it would read $1 \le -1$.

You can try the following: suppose that $\liminf x_n = a < \limsup x_n = b$ and extract two subsequences that realize these limits, i.e. pick $\{x_{n_k}\}, \{x_{n_j}\}$ such that $x_{n_k} \to a$ and $x_{n_j} \to b$. Let $\epsilon$ be such that $a + \epsilon < b - \epsilon$ and let $K, J$ be large enough so that $x_{n_k} \le a + \frac{\epsilon}{2}$ for every $k \ge K$ and $x_{n_j} \ge b - \frac{\epsilon}{2}$ for every $j \ge J$.

Then for every $k \ge K$ and $j \ge J$ $$|x_{n_j} - x_{n_k}| \ge b - \frac{\epsilon}{2} - a - \frac{\epsilon}{2} = b - a - \epsilon \ge \epsilon.$$

Do you see how this contradicts the assumption that $\{x_n\}$ is a Cauchy sequence?