# Proving cauchy sequence is convergent sequence.

I know it's very petty question in real-analysis but I want to make the proof finely logical.

There is 3 steps of the proof, we all know well, that cauchy sequence is convergent sequence. The first is proving cauchy sequence is bounded. The second is that by Bolzano-Weierstrass theorem, there exists a cluster point. The third is showing the original cauchy sequence is convergent to the cluster point.

I think there is a logical gap in my text book, but i cannot make it fine.

Let $L$ be the cluster point of $\{a_n\}$ which is cauchy sequence. Then there exists a subsequence $\{a_{n_k}\}$ convergent to $L$.

Given $\epsilon>0$, $\exists N_1 \in \mathbb{N}, \forall k : [k>N_1 \rightarrow |a_{n_k} - L| < \epsilon/2$.

By cauchy condition, $\exists N_2 \in \mathbb{N}, \forall n,m : [n,m > N_2 \rightarrow |a_m - a_n| < \epsilon/2$.

$\textbf{N := max\{N_1, N_2\}}$, then for all $n>N$ $$|a_n-L| = |a_n - a_{n_{N+1}} + a_{n_{N+1}} -L| \leq |a_n - a_{n_{N+1}}| + |a_{n_{N+1}} -L| < \epsilon.$$

I think the bold text is the logical gap in the proof. If we set $N$ like the above proof, there is no verification that $n_{N+1} > N_2$. So, if we make this proof well, additory condition is required such as

$\textbf{N :=max\{N_1, N_2\}}$, then for all $n>N$ $\textbf{and$n_k>N$}$.

However the definition of convergence has only $n>N$ assumption to make it sense the inequality that $|a_n -L|<\epsilon$.

How do we set the index numbering to make the proof have no logical gap?

• $n_j\ge j$ is part of the definition of "subsequence". – David C. Ullrich Aug 9 '15 at 23:30

$$n_k \ge k, \forall k \in \mathbb N$$
Where $(n_k)$ is a strictly increasing sequence of naturals.
(inductive step: $n_{k+1} > n_k \implies n_{k+1} \ge n_k + 1 \ge k + 1$)