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I am supposed to prove that all Cauchy sequences converge, using the fact that a sequence converges iff its lim sup equals its lim inf. I think the proof I have works, but the one given in the solutions is different, so I'd like to make sure. Also, this is for a calculus class, not real analysis, so I don't know about metric spaces or any of that. I believe the problem pertains solely to sequences in $\mathbb{R}$.

We have the Cauchy sequence $\{a_n\}$. By the definition of the Cauchy sequence, we have $$\left|a_m-a_n\right|<\frac{\epsilon}{3}$$ for a given $\epsilon>0$ and $m,n>N$. There exist the numbers $\inf_{n>N}\{a_n\}$ and $\sup_{n>N}\{a_n\}$, that is to say, the infimum and supremum of the sequence with the first $N$ terms truncated. By the definition of the infimum, we must have $$\left| a_{low}-\inf_{n>N}\{a_n\}\right|<\frac{\epsilon}{3}$$ for some $a_{low} \in \{a_n|n>N\}$, because otherwise $\inf_{n>N}\{a_n\}+\frac{\epsilon}{3}$ would be a greater lower bound for $\{a_n|n>N\}$ than the infimum, a contradiction. By the same reasoning, we also have $$\left|\sup_{n>N}\{a_n\}-a_{high}\right|<\frac{\epsilon}{3}$$ for some $a_{high} \in \{a_n|n>N\}$. Now, by the definition of the Cauchy sequence, we must have $$\left|a_{high}-a_{low}\right|<\frac{\epsilon}{3}$$ Expanding the absolute value terms gives $$-\frac{\epsilon}{3}<a_{low}-\inf_{n>N}\{a_n\}<\frac{\epsilon}{3}$$ $$-\frac{\epsilon}{3}<\sup_{n>N}\{a_n\}-a_{high}<\frac{\epsilon}{3}$$ $$-\frac{\epsilon}{3}<a_{high}-a_{low}<\frac{\epsilon}{3}$$ Adding these together gives $$-\epsilon<\sup_{n>N}\{a_n\}-\inf_{n>N}\{a_n\}<\epsilon$$ So $$\left|\sup_{n>N}\{a_n\}-\inf_{n>N}\{a_n\}\right|<\epsilon$$ which, we remember, holds for $n>N$. By the definition of the infinite limit, then, because we can make the difference of the infimum and the supremum arbitrarily close to zero by increasing $N$, we have $$\lim_{N \rightarrow \infty}\left( \sup_{n>N}\{a_n\}-\inf_{n>N}\{a_n\} \right)=\lim_{N \rightarrow \infty}\sup_{n>N}\{a_n\}-\lim_{N \rightarrow \infty}\inf_{n>N}\{a_n\}=0$$ So the lim inf of any Cauchy sequence equals its lim sup, and therefore it converges.

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  • $\begingroup$ Your proof is essentially correct, though a pedant could object that you didn't explain what $N$ is. $\endgroup$ – Thomas Mar 11 '13 at 13:04
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Basically, a Cauchy sequence can not have different values for lim-inf and lim-sup, assuming both exists. I.e. if $lim_{inf}=a$ and $lim_{sup}=b$, where $b-a > 0$, then there will exists $n,m > N$ and an $\varepsilon < \frac{b-a}{3}$ such that $\left |a_{n} - a_{m} \right |> \varepsilon $, just take $a_{n}$ from the subsequence 'going' to $lim_{inf}=a$ and take $a_{m}$ from the subsequence 'going' to $lim_{sup}=b$.

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