Prove (informally) every Cauchy sequence is a convergent sequence

Question

Since Cauchy sequence meets $|a_{n}-a_{m}|<\epsilon$ for arbitrary numbers $n, m > n_{0}$, doesn't just picking up some $w>n_{0}$ qualifies as a (informal) proof? Fixing $b=a_{w}$, and modify the above inequality to $|a_{n}-b|<\epsilon$ makes sense to me.

Answer 1

The problem with your argument is that you didn't consider dependancy of $n_0$ on $\epsilon$. As $\epsilon$ changes, $n_0$ also changes($n_0$ depends on $\epsilon$). Thus for a different $\epsilon$, the term $b=a_w$ might be invalid to get $|a_n-b| \lt \epsilon$.

Answer 2

$n_0$ depends on $ \epsilon$ ! So your $b$ depends on $ \epsilon$ ! With your $b$ you can not garantee that for $ \epsilon/2$ there is $n_1$ such that

$|a_{n}-b|<\epsilon/2$ for all $n >n_1$.