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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.

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    $\begingroup$ How did you get $\lim \inf x_n \geq \sup \{ x_n, x_{n+1},...\}$? That doesn't look correct. $\endgroup$
    – Kenny Wong
    Commented Feb 24, 2017 at 16:52
  • $\begingroup$ Note that the limit superior and inferior are only equal if the limit exists. Plenty results exist that relate the existence of the limit and the sequence being Cauchy. $\endgroup$
    – Marc
    Commented Feb 24, 2017 at 16:52
  • $\begingroup$ Exactly. The argument would also apply to bounded sequence, wouldn't it? $\endgroup$
    – user251257
    Commented Feb 24, 2017 at 16:53

1 Answer 1

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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?

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  • $\begingroup$ Yes, I see this makes sense. I wasn't feeling very confident that those inequalities held. Also for the example of $x_n = (-1)^n$, that sequence does not converge/not Cauchy, but I see the point. $\endgroup$
    – student_t
    Commented Feb 24, 2017 at 17:00
  • $\begingroup$ The reverse inequality should hold though correct? $\endgroup$
    – student_t
    Commented Feb 24, 2017 at 17:06
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    $\begingroup$ @danny: correct! $\endgroup$
    – Giovanni
    Commented Feb 24, 2017 at 17:10

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