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 $b_\infty = \lim\sup x_n = \lim\inf x_n = a_\infty$.

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    $\begingroup$ I think one key characteristic of real Cauchy sequences is that they are bounded ....and, of course, you don't need to know they converge to prove this. $\endgroup$
    – DonAntonio
    Feb 27, 2017 at 13:24
  • $\begingroup$ @Jack I posted that question and was very wrong in my assumptions. I wanted to post my other attempt and I wasn't sure whether I should have answered my own question or posted a new one. If I should have answered my own question in the original question I'll remove this and post it there! Sorry, I am new to this. $\endgroup$
    – student_t
    Feb 27, 2017 at 13:30
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    $\begingroup$ @Jack I wouldn't say it's a duplicate. Granted the question asked is whether the proof is correct, and the proofs was for the same theorem. However the proofs themselves are different, that one was basically incorrect and this one is basically correct. $\endgroup$
    – skyking
    Feb 27, 2017 at 14:07

1 Answer 1


Basically your proof is correct, but maybe you should clarify some steps and consider different wording on other.

That $c_N = b_N-a_N$ might need some explanation.

That $c_n$ decreases might sound obvious, but also there one may note that the $\sup$ over a subset is no larger than $\sup$ over the full set (to nitpick one might either have to know that the sets are non-empty or that $\sup\emptyset=-\infty$).

You don't have to write $-\epsilon < 0$ in the penultimate paragraph.

The last paragraph should probably be reworded. The symbols $a_\infty$, $b_\infty$ and $c_\infty$ sounds like very sloppy reasoning. We know that $c_n = b_n-a_n$ and $\lim c_n = 0$ so $\lim (b_n-a_n) = 0$.

One may need to complete the last with the fact that by definition $\limsup x_n = \lim b_n$, and that it exists (because cauchy sequences are bounded(*) and therefore $b_n$ is and it's monotone) and the corresponding for $\liminf$ this means that $0 = \lim(b_n-a_n) = \lim b_n - \lim a_n$.

(*) Cauchy sequences are bounded because take $\epsilon=1$ and you have an $N$ such that especially $|x_n-x_N| < \epsilon = 1$ for all $n>N$. So for $n>N$ we have $|x_n| < |X_N+1|$ and for $n\le N$ we have $|x_n| < \max(|x_1|, |x_2|, \cdots, |x_N|)$.

  • $\begingroup$ Thank you this was very helpful! For $c_N = b_N - a_N$, I have proved recently in another question that $c_n = b_n - a_n$ and then since $x_n$ is Cauchy then $x_j - x_i < \varepsilon$ means $\varepsilon$ is an upper bound but $c_N$ is the least upper bound so $c_N \leq \varepsilon$. Does this clarify? Should I add that. $\endgroup$
    – student_t
    Feb 27, 2017 at 14:35
  • $\begingroup$ I should've also defined $c_n = \sup\{x_j - x_i : i,j \geq n\}$. $\endgroup$
    – student_t
    Feb 27, 2017 at 14:45

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