Suppose $x_n$ is a Cauchy sequence. Prove that the $\lim\sup x_n = \lim\inf x_n$.
I made some serious mistakes when first approaching this, wanted some feedback on the proof to see if there were any errors, that is why I posted another question. Again, cannot use the fact that all Cauchy converge since we haven't shown this result yet.
Define $b_n = \sup\{x_k : k \geq n\}$ and $a_n = \inf\{x_k: k \geq n\}$. Finally, denote $c_n = \sup\{x_i - x_j: i,j \geq n\}$. I've recently proved in another question that $c_n = b_n - a_n$. Note that $c_n$ is a decreasing sequence and $c_n \geq 0$ for all $n$
Suppose $\varepsilon > 0$, since $x_n$ is Cauchy then there is an integer $N$ so that $|x_i -x_j| < \frac{\varepsilon}{2}$ for $i,j \geq N$. Thus,
$b_N - a_N = c_N \leq \frac{\varepsilon}{2} < \varepsilon$. Since, $c_n$ is decreasing and bounded below by $0$, we have:
$-\varepsilon < 0 \leq c_n < \varepsilon$ for all $n \geq N$ and thus $|c_n| < \varepsilon$ .
Therefore, $c_n \rightarrow 0$ as $n \rightarrow \infty$ and so $c_\infty = 0$. Since, $c_\infty = b_\infty - a_\infty$, then $b_\infty = \lim\sup x_n = \lim\inf x_n = a_\infty$.