# Proof Verification: Cauchy Sequences are convergent.

Assume $(a_n)$ is a Cauchy sequence, then it is bounded. By Bolzanno-Weierstrass theorem, there is a convergent sub-sequence $(a_{n_j})$; denote its limit as $a$. Thus we have the following: for every $\epsilon>0$, there exists a $N_1\in \mathbb{N}$ such that $$n,m \geq N_1 \implies |a_n-a_m|<\frac{\epsilon}{2},$$ and $N_2$ such that $$n_j \geq N_2 \implies |a_{n_j}-a|<\frac{\epsilon}{2}.$$

Construct $N$ as follows: $N=\max\{N_1,N_2\}$, thus: $$|a_n-a|=|a_n-a+a_{n_j}-a_{n_j}|\leq|a_n-a_{n_j}|+|a_{n_j}-a|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$ whenever $n\geq n_j \geq N$.

I think my first two inequalities are correct, but I am not so sure about the last part. Anyone care to check my work?

• Hm, I think, that the construction of $N$ might be unnecessary. This is quite confusing. – Kurome May 16 '15 at 19:44
• seems fine to me. what generates your lack of certainty? – David Holden May 16 '15 at 19:44
• @DavidHolden The problem is how $N_2$ is chosen, it should be a condition on $j\geq N_2$, not $n_j\geq N_2$. – Gregory Grant May 16 '15 at 19:50
• For sub-sequences $n_j \geq j$ (the reason why is quite vague to me, it does seem intuitive though). Thus, $j \geq N_2 \implies n_j \geq N_2$. – Kurome May 16 '15 at 19:54
• @GregoryGrant ah,yes, i see what you mean – David Holden May 16 '15 at 19:57

You're not quite using the subscript notation correct. You don't want to say $n_j\geq N_2$, you want to say $j\geq N_2$.
• ... which means we should switch to $N=\max\{N_1,n_{N_2}\}$ – Hagen von Eitzen May 16 '15 at 19:45
• Think through an example, like define $b_j$ by $b_j=a_{n_j}=a_{2j}$. Convince yourself how $b_1$ is $a_2$, $b_2$ is $a_4$, etc... Does that help at all? – Gregory Grant May 16 '15 at 19:49