I want to prove that: If $\forall \epsilon >0$, $\exists k \in \mathbb{N}$, such that $| u_{n+p}-u_n| <\epsilon $, whenever $n\geq k$, $p\in \mathbb{N}$, then $\{u_n\}$ is convergent.
Proof:
[After showing that the sequence is bounded with bounds $b \leq u_n \leq B$].
By Bolzano-Weierstrass theorem, the sequence being bounded, it has a convergent subsequence, say $\displaystyle\{u_{r_n}\}$ with limit $\ l$. Hence, the derived set $S$ of the set $\{u_n\} $ is non empty. We show that there is only one point in $S$.
Suppose, there is another point $l'$ in $S$. Then, there is another subsequence $\{u_{s_n}\}$ converging to $l'$.
Again, $\forall \epsilon >0, \exists b\in \mathbb{N}$, s.t $|u_{r_n}-u_{s_n}|<\epsilon$ for any $s_n, r_n\geq b$. Hence, they converge to the same limit, i.e. $l = l'$ .
We conclude $S = \{l\}$, i.e. $\limsup u_n= \liminf u_n$
Is this okay? I feel like there is definitely some mistake in my procedure.
Thank you.
Compactness of $[b,B]$ allows us to conclude that there is a subsequence $u_{n_k}$ that converges to some $u$.
Since $u_k$ is Cauchy, it follows that $u_n \to u$.
To see this, choose $\epsilon>0$ and pick $N_1$ such that for $k \ge N_1$ we have $|u_{n_k} -u| < \epsilon$, and $N_2$ such that
If $n \ge N_2$ we can find some $k \ge \max(N_1,N_2)$ such that $|u_n-u| \le |u_n-u_{n_k}| + |u_{n_k}-u| < 2 \epsilon$. Hence $u_n \to u$.