I am trying to prove that a Cauchy sequence must converge. To do this I used the fact that a sequence converges if and only if its lim inf equals its lim sup, as follows:
Consider a Cauchy sequence $\{a_n\}$, then for all $\epsilon>0$, there exists $N\in\mathbb Z^+$ such that for all $m,n>N$, we have $|a_m-a_n|<\epsilon$. We seek to show that $\liminf_{n\to\infty}a_n=\limsup_{n\to\infty}a_n$.\begin{align*} &-\epsilon<a_m-a_n<\epsilon\\ &\therefore -\epsilon+a_n<a_m<\epsilon+a_n\\ &\therefore -\epsilon+a_n\leq \liminf_{m\to\infty}a_m\leq a_m \leq \limsup_{m\to\infty}a_m\leq\epsilon+a_n\\ &\therefore |\liminf_{m\to\infty}a_m-a_n|\leq \epsilon \text{ and } |\limsup_{m\to\infty}a_m-a_n|\leq \epsilon\\ &\therefore |\liminf_{m\to\infty}a_m-a_n| + |a_n-\liminf_{m\to\infty}a_m|\leq 2\epsilon\\ & \therefore |\liminf_{m\to\infty}a_m-\limsup_{m\to\infty}a_m|\leq2\epsilon \end{align*} Hence we conclude $\liminf_{m\to\infty}a_n=\limsup_{m\to\infty}a_n$, and therefore by part (a) every Cauchy sequence of real numbers converges.
Is my proof correct? The other proofs I read of this all seem much longer, including another one on this website which also uses lim inf = lim sup.